PDA

View Full Version : sin(x) * cos(x) - x = y

grav
2007-Oct-08, 09:26 PM
sin(x) * cos(x) - x = y

In the equation above, is there a way to find x if y is known?

Disinfo Agent
2007-Oct-08, 10:50 PM
I haven't look into it carefully, but you can probably do it by approximation. Numerical methods, I mean.

2007-Oct-08, 11:06 PM
It can be written as:

sin(2x) - 2x = 2y

If this helps, I can write the proof for it.

grav
2007-Oct-08, 11:33 PM
It can be written as:

sin(2x) - 2x = 2y

If this helps, I can write the proof for it.Thanks. :) I figured that out the hard way. I used

sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...

and

cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + ...

and multiplied them. I came up with

sin(x)*cos(x) = x(1/1!) - x^3(1/1!2!+1/3!) + x^5(1!4!+1/3!2!+1/5!) - ...

which ended up just reducing to

sin(x)*cos(x) = x(2^0/1!) - x^3(2^2/3!) + x^5(2^4/5!) - x^7(2^6/7!) + ...

but that is just the same as

[(2x)/1! - (2x)^3/3! + (2x)^5/5! - (2x)^7/7! + ...]/2

= sin(2x)/2

from the earlier equation for sine. So substituting this into the equation, we now have

sin(2x)/2 - x = y

sin(2x) - 2x = 2y

as you have shown. Since the 2's can be drawn out after the fact, to simplify things quite a bit, then, it looks like I am now just looking for a solution for x knowing y with the new equation

sin(x) - x = y

2007-Oct-09, 12:17 AM
Well then,

- x^3/3! + x^5/5! - x^7/7! + ... = y

and x will be a transcendent number in terms of y. One I'm not sure how to calculate.

Ken G
2007-Oct-09, 12:26 AM
You can't, it's transcendental. You can only approximate it numerically, like Disinfo Agent said. If |y| is small, you can just keep the first term, and get x is roughly the cube root of -6y. But if that's not small, you're out of luck. Also, if y is positive, there's no solution.

grav
2007-Oct-09, 01:02 AM
Darn. I figured as much, but I just wanted to make sure. Thanks everyone. It still seems like there should be some way to do it, though. Maybe one day.

01101001
2007-Oct-09, 01:47 AM
How about the inverse function for f(x) = sin(x)*cos(x) - x being approximately f-1(x) ≈ -x ?

f(1000) = -999.53498
f(-999) = 999.026451

f-1(-1000) ≈ 1000
f-1(999) ≈ -999

The applet at Graphing Calculator For Inverse Functions (http://www.analyzemath.com/graphing_calculators/graphing_inverse.html), with sin(x)*cos(x) - x, zoomed out for the big picture, looks a whole lot like
y = -x .

parejkoj
2007-Oct-09, 02:11 AM
It still seems like there should be some way to do it, though. Maybe one day.

That's the problem with Transcendental equations: there is no analytic solution, nor can there be. It's like trying to write a polynomial (in the Rational numbers) with Pi as its root. Ya can't.

Unless someone knows of some special functions that can be appropriately manipulated to produce the inverse... Linear combination of Bessel Functions, perhaps? ;) That just offloads the solution to others who have already solved those special functions, though.

grav
2007-Oct-09, 02:41 AM
Well, originating from another thread, I have come up with a program that will give the solution for x instantly for any such equation with just a few iterations, depending upon the degree of accuracy required, if anyone's interested, and if it's not already known. I have actually used something similar a few times before to come up with solutions such as these, but never really thought about how useful this knowledge might be to others that might need it for something if they don't already know about it. Twenty iterations will give the solution to about one part in a million, thirty will give it to one part in a billion, and so on.

Ken G
2007-Oct-09, 04:41 AM
sin(x)*cos(x) - x, zoomed out for the big picture, looks a whole lot like
y = -x .

That's the other limit-- if y is very large, the sin[x] isn't doing much, it's a lot like adding a random number from -1 to 1 to y to give you x.

2007-Oct-09, 04:41 AM
You can't, it's transcendental. You can only approximate it numerically, like Disinfo Agent said. If |y| is small, you can just keep the first term, and get x is roughly the cube root of -6y. But if that's not small, you're out of luck. Also, if y is positive, there's no solution.

Yes. But is it possible to find an infinite summation to calculate to the level of accuracy desired?

so that would be x= y() +y() + .....

grav
2007-Oct-09, 04:50 AM
Yes. But is it possible to find an infinite summation to calculate to the level of accuracy desired?

so that would be x= y() +y() + .....That is actually was I was working on when I then realized my iteration (from post #10) would be of the same consequence, bringing the value of x out to greater accuracy with a greater number of iterations (about three digits of accuracy for every ten iterations). In fact, it works even better than attempting to find an infinite series of y to find x because the same iteration can be used for almost any equation such as this in the same way instead of trying to find the coefficient values for y in the infinite series for each particular case, and also since the accuracy doubles with each iteration while an infinite series might add up very slowly in some cases.

Ken G
2007-Oct-09, 05:14 AM
But is it possible to find an infinite summation to calculate to the level of accuracy desired?
Yes, that could be possible, but a transcendental solution is not "algebraic", which basically means you'll always need the .... . So it's just an algorithm with converging accuracy, and iteration can do that too. Incidentally, if you want to iterate sin x - x = y, you can write f(x) = sinx - y and iterate toward the fixed point f(x) = x. That's not guaranteed to work, and may not be fast, so a better approach is to replace f(x) with another function g(x) that solves g(x) = x for the same x as for f(x) = x, but such that g'(x) = 0 when f(x) = x. That will converge quadratically fast, instead of linearly. What does this is g(x) = x - [f(x)-x]/[f'(x)-1]. So here the function to iterate to a fixed point is g(x) = x - (sin x - y - x)/(cos x - 1) = (x * cos x - sin x + y) / (cos x - 1). That should converge to the solution g(x) = x in a flash.

tony873004
2007-Oct-09, 06:16 AM
I set this up in Excel.
Column A: 0, 0.01, 0.02... all the way to 6.28
Column B: =sin(a1)*cos(a1)-a1 ...

I made a scatter plot, and asked it to give me a polynomial trendline, and show me the equation. 5th and 6th degree gave a pretty good fit except for the ends of the functions. I don't know how to ask for a sinusoidal trend line, but I think it can be done. Although Excel doesn't go beyond 6th degree polynomials, IDL does. I imagine a higher degree polynomial will fit it much better, perhaps to parts per million. That way, even though it is still just an approximation, and a real ugly long formula, you might be able to get your desired result in 1 iteration. It's just a thought... I don't have much experience doing this stuff.

grav
2007-Oct-09, 10:57 PM
From the last couple of posts, it sounds like my method might actually be helpful in these cases. Cool. It's rare that I can come up with something that may not be at least well known. It's strange that I've actually been using this method off and on for years now without really giving it much of a second thought, when it turns out it might be useful to others.

Those posts are reasonable, but instead of iterating from a lower limit to an upper limit at steady intervals to find the closest approximation, we can cut the limit in half each time to find it with extreme preciseness and speed. For instance, building up from zero to 6.28 in intervals of .01 would still take up to 628 iterations, and wouldn't end up being too precise. By the time my method had run 628 iterations, the solution would be found to an accuracy of about one part in 2^628, or to over 200 digits. An accuracy of one part in a billion, however, can be found in just 30 iterations. It is similar to a game where you might write down an integer between one and a million, and I will be able to guess it in twenty tries or less, with the stipulation that you tell me whether your number is higher or lower than the one I guess each time. It is the same thing for trying to approach the known result.

In another thread, I used this method to find the distance an object freefalls when dropped from some distance after falling for a known duration of time. The formula for the time, knowing the original and final distances, was provided as

t = sqrt(D^3/2GM)*(pi/2 - asin(sqrt(d/D)) + sqrt((d/D)(1-d/D)))

where D is the original distance that the object was dropped from and d is the distance the object has fallen to, but one cannot retract the distance the object falls to, knowing the time of descent. Since I only knew the time of descent and wanted the distance fallen to, I ran a simulation using very small intervals of time at first, which could take millions or billions of iterations, depending upon the accuracy I wanted. But once I had the formula for time, I realized there was a tremendously quicker, easier, and much, much more precise way to do it. Here is the post for that.

Well, although I still have to run a computer simulation to find the final distance, that formula for time has already helped me out tremendously. Before, I had to divide the total time up into a million intervals and run them one by one to find the final distance after falling for some time. This took about five to ten seconds. But to get any real accuracy, I would have to divide it up into something like a billion intervals, especially if the falling object approached the center of the gravitating body. This, however, would take a couple of hours, and I'm not patient enough to wait that long each time, especially since it bogs down the computer so I can't do anything else.

But now, with that formula for time, I can find the total time it would take the falling object to reach the center of the body. Then, as long as the duration of time I want to find for is smaller than that, I know the final distance will lie somewhere between zero and the original distance. So I have the computer run it this way. First it finds the time to fall to half the distance. Then it finds the time for that plus or minus a quarter of the distance, depending upon whether the time from the first iteration was larger or smaller than the time I want. Then it adds or subtracts half of that, or an eight of the total distance, and does the same thing again, and so on. After just twenty iterations, I have an accuracy of about one part in a million of the original total distance. With thirty iterations, it's one part in a billion. So now, thanks to you guys, I can have my answer in an instant, with just tens of iterations, instead of millions or billions, with as great an accuracy as I could possibly need. Thanks again.

grav
2007-Oct-09, 11:32 PM
To demonstrate, here is the progression for the equation x - sin(x) = y. I reversed the order in the equation to make y positive. I am finding for y=.1 so that I can make the upper and lower limits 0<x<1, whereas one can see how the program progresses more easily.

1 0.5
2 0.75
3 0.875
4 0.8125
5 0.84375
6 0.859375
7 0.8515625
8 0.85546875
9 0.853515625
10 0.8544921875
11 0.85400390625
12 0.853759765625
13 0.8536376953125
14 0.85369873046875
15 0.853729248046875
16 0.8537445068359375
17 0.85375213623046875
18 0.853748321533203125
19 0.8537502288818359375
20 0.85374927520751953125

30 0.853750157169997692108

40 0.853750156641581270378

50 0.853750156640866286750

100 0.853750156640865774281

200 0.853750156640865774281

Ray C.
2007-Oct-11, 03:27 PM
What you probably want for a problem like this is Newton's method (http://en.wikipedia.org/wiki/Newton%27s_method). The link will give you more details than this post could. Briefly, you take an educated guess at the solution, follow the tangent to the function to the point where it crosses zero, and that zero crossing is your refined guess.

It does require some first-semester calculus, to get the slope of that tangent line. Starting from The_Radiation_Specialist's simplified version of the function in question:

sin(2x) - 2x - 2y = 0

the derivative of the left side is:

2*cos(2x) - 2

and so the iteration goes like this:

x[n+1] = x[n] - (sin(2x[n]) - 2x[n] - 2y)/(2*cos(2x[n]) - 2)

grav
2007-Oct-11, 11:45 PM
What you probably want for a problem like this is Newton's method (http://en.wikipedia.org/wiki/Newton%27s_method). The link will give you more details than this post could. Briefly, you take an educated guess at the solution, follow the tangent to the function to the point where it crosses zero, and that zero crossing is your refined guess.

It does require some first-semester calculus, to get the slope of that tangent line. Starting from The_Radiation_Specialist's simplified version of the function in question:

sin(2x) - 2x - 2y = 0

the derivative of the left side is:

2*cos(2x) - 2

and so the iteration goes like this:

x[n+1] = x[n] - (sin(2x[n]) - 2x[n] - 2y)/(2*cos(2x[n]) - 2)Oh, thanks, wow. That's pretty neat, and fast. It looks like the same thing Ken was saying a few posts back, then, but I didn't quite understand it at the time. I tried that iteration and it blew past mine like there was no tomorrow. :( It did in five iterations what mine took ninety to do, bringing it out to about twenty-six digits, and only got faster after that. Oh, well. I still prefer mine, though, for those of us that might be somewhat calculus illiterate. Just plug in the original formula and let it go, no calculus required. :)

Incidently, here are the first ten iterations for x - sin(x) = y with limits 0<x<1 and y=.1 using Newton's method. Not that it even needed all ten iterations for this many digits, apparently.

1 0.872679337009432649568911355278353645494505227156
2 0.854132307053426948734049678208105876675164965010
3 0.853750317079009696469242758076743529872721413351
4 0.853750156640894070113020019320666453924008521882
5 0.853750156640865774281393275486602504833248452898
6 0.853750156640865774281393274606461525424158369062
7 0.853750156640865774281393274606461525424158369060
8 0.853750156640865774281393274606461525424158369060
9 0.853750156640865774281393274606461525424158369060
10 0.853750156640865774281393274606461525424158369060

01101001
2007-Oct-12, 12:08 AM
I still prefer mine, though, for those of us that might be somewhat calculus illiterate. Just plug in the original formula and let it go, no calculus required.

Then the Secant method (http://en.wikipedia.org/wiki/Secant_method) may be for you. It is sort of a Newton's method that fakes the differential with finite differences.

hhEb09'1
2007-Oct-12, 12:23 AM
I still prefer mine, though, for those of us that might be somewhat calculus illiterate. O for p*te's sake.

Newton's method is just rise/run = slope, where the rise is the y value, the run is the difference in the x values, and the slope is calculated from the derivative. It's an approximation because the derivative slope only fits the function near that particular point, but as you can see, it does a pretty good job of getting you in the ballpark quickly.

How to calculate derivatives? We can mess around with limits, and show you why they work but the formula for polynomials is simple: when you're taking the derivative of a polynomial with many terms, the derivative is just the derivative of each term, added together. And the derivative of a constant times a power of x is just the constant times the derivative (a function that is multiplied by a constant K is going to have a slope that is K as great). The derivative of xn is just nxn-1

That's all you need to take the derivative of any polynomial. The derivative of sin(x) is cos(x), the derivative of cos(x) is -sin(x). The derivative of sin(2x) is 2cos(2x) -- the x's are coming twice as close together. More about that later.

Let's look at an example, using limits. The derivative of x2 is supposed to be the slope of x2, right? Let's pick two points on the line, and calculate a rise over run, rise/run, what we call the slope, the difference in the y-values divided by the difference in the x-values.

(y2-y1)/(x2-x1) is the slope between the two points (x1,y1) and (x2,y2)

Let's let d = (x2-x1), the difference in the x values, then the y values are:
y1=x12
and
y2=x22=(x1+d)2
so the difference in the y values is
(x1+d)2 - x12
or
2dx1+d2

When you divide that difference in y values by the difference in x values:
(2dx1+d2 ) / d = 2x1+d

So, 2x1+d is the slope of the line between the two points on the curve.

If you let the two points get closer and closer together, notice what happens. The slope doesn't disappear, it just gets closer and closer to 2x1, because the d disappears. And that's why the derivative of x2 is 2x. If you want the slope of the x2 curve at the point x1, it's 2x1. The 2 comes from the second coefficient in the binomial expansion.

Try it for taking the derivative of x3, it's pretty cool.

grav
2007-Oct-12, 01:32 AM
Then the Secant method (http://en.wikipedia.org/wiki/Secant_method) may be for you. It is sort of a Newton's method that fakes the differential with finite differences.Cool. That worked out quite well. Thank you. :) I guess I'll be replacing my method with that one in the computer program, then. Here's the first ten iterations for the same equation using the Secant method.

1 0.630799351644374002751352173982416012897134204728
2 0.816125542796573833244427441348064721933751870011
3 0.865306109359596564968370941977852095633825486644
4 0.853262057044922735607397842760134871928456974349
5 0.853744011355712632007577789281751626827422522736
6 0.853750159939479025947676905426654057292968122226
7 0.853750156640843490806629126180265464325368062178
8 0.853750156640865774281312472696118593003877571354
9 0.853750156640865774281393274606461527403459437222
10 0.853750156640865774281393274606461525424158369058

grav
2007-Oct-12, 01:58 AM
O for p*te's sake.

Newton's method is just rise/run = slope, where the rise is the y value, the run is the difference in the x values, and the slope is calculated from the derivative. It's an approximation because the derivative slope only fits the function near that particular point, but as you can see, it does a pretty good job of getting you in the ballpark quickly.

How to calculate derivatives? We can mess around with limits, and show you why they work but the formula for polynomials is simple: when you're taking the derivative of a polynomial with many terms, the derivative is just the derivative of each term, added together. And the derivative of a constant times a power of x is just the constant times the derivative (a function that is multiplied by a constant K is going to have a slope that is K as great). The derivative of xn is just nxn-1

That's all you need to take the derivative of any polynomial. The derivative of sin(x) is cos(x), the derivative of cos(x) is -sin(x). The derivative of sin(2x) is 2cos(2x) -- the x's are coming twice as close together. More about that later.

Let's look at an example, using limits. The derivative of x2 is supposed to be the slope of x2, right? Let's pick two points on the line, and calculate a rise over run, rise/run, what we call the slope, the difference in the y-values divided by the difference in the x-values.

(y2-y1)/(x2-x1) is the slope between the two points (x1,y1) and (x2,y2)

Let's let d = (x2-x1), the difference in the x values, then the y values are:
y1=x12
and
y2=x22=(x1+d)2
so the difference in the y values is
(x1+d)2 - x12
or
2dx1+d2

When you divide that difference in y values by the difference in x values:
(2dx1+d2 ) / d = 2x1+d

So, 2x1+d is the slope of the line between the two points on the curve.

If you let the two points get closer and closer together, notice what happens. The slope doesn't disappear, it just gets closer and closer to 2x1, because the d disappears. And that's why the derivative of x2 is 2x. If you want the slope of the x2 curve at the point x1, it's 2x1. The 2 comes from the second coefficient in the binomial expansion.

Try it for taking the derivative of x3, it's pretty cool.A lesson in calculus, eh? Nice. I'll have to read through what you wrote some more, but I can start to see what you mean about the slope, I think. Anyway, that last derivative for x^3 would just be 3x^2, right? The second term if one expands (x+1)^3. From what you were saying, that would actually come out to [(x+d)^3 - x^3]/d = 3x^2 + 3xd + d^2, but for infinitesimal d, only the first term remains, since the other two become infinitely small in comparison.

grav
2007-Oct-12, 02:19 AM
So does the derivative of x - sin(x) become 1 - cos(x), where 1 is the derivative of x and cos(x) is the derivative of sin(x)?

Fortunate
2007-Oct-12, 02:29 AM
So does the derivative of x - sin(x) become 1 - cos(x), where 1 is the derivative of x and cos(x) is the derivative of sin(x)?

Yes. The derivative of a sum or difference is the sum or difference, respectively, of the derivatives. You have to be careful, though, because the derivative of a product is not the product of the derivatives, and the derivative of a quotient is not the quotient of the derivatives.

grav
2007-Oct-12, 02:41 AM
Yes. The derivative of a sum or difference is the sum or difference, respectively, of the derivatives. You have to be careful, though, because the derivative of a product is not the product of the derivatives, and the derivative of a quotient is not the quotient of the derivatives.Thanks, I noticed that in Ray C's post, where the derivative of 2x - sin(2x) becomes 2 - 2*cos(2x). I know it works because I used it in the program, the same as Ken's. But I'm still trying to figure out where that extra 2 for the cos(2x) came from.

Fortunate
2007-Oct-12, 03:18 AM
Thanks, I noticed that in Ray C's post, where the derivative of 2x - sin(2x) becomes 2 - 2*cos(2x). I know it works because I used it in the program, the same as Ken's. But I'm still trying to figure out where that extra 2 for the cos(2x) came from.

The "extra" factor of 2 comes from something called the "chain rule." You need to learn some specific derivatives:

the derivative of a polynomial
Anxn + An-1xn-1 +...+ A1x + A0 is
nAnxn-1 + (n-1)An-1xn-2 +...+ 2A2x + A1. You also need to learn the derivatives of the basic trig functions and of ex and lnx. Then you need to learn the product rule, the quotient rule, and the chain rule.

grav
2007-Oct-12, 03:25 AM
The "extra" factor of 2 comes from something called the "chain rule." You need to learn some specific derivatives:

the derivative of a polynomial
Anxn + An-1xn-1 +...+ A1x + A0 is
nAnxn-1 + (n-1)An-1xn-2 +...+ 2A2x + A1. You also need to learn the derivatives of the basic trig functions and of ex and lnx. Then you need to learn the product rule, the quotient rule, and the chain rule.Yes, I was thinking about that, and thought they might have something to do with it. Richard (publius) helped me work through some of those a while back. I can still only remember bits and pieces, though, unfortunately (no pun intended :) ), nothing I could put together coherently anymore. I'll have to look back over them.

hhEb09'1
2007-Oct-12, 03:41 AM
Thanks, I noticed that in Ray C's post, where the derivative of 2x - sin(2x) becomes 2 - 2*cos(2x). I know it works because I used it in the program, the same as Ken's. But I'm still trying to figure out where that extra 2 for the cos(2x) came from.Two more subtle principles (subtle in that it is very common to mis-apply them) are the product rule (for products of functions) and the chain rule (for functions of functions).

If the object that you are taking the derivative of is a product of two functions, y = F(x)G(x) then the derivative of y which is often denoted y' is just F'(x)G(x) + F(x)G'(x). Without proving it, you can check it out with the example y = x5, which is just the product of x2 and x3.

The chain rule is even easier. If y = F(G(x)), then y' = F'(G(x)) G'(x). So, when y = sin(2x), F is the sine function and G is the multiply by 2 function. So, F' is the cosine function and F'(G(x)) = cos(2x) and G'(x) = 2 and y' = 2 cos(2x)

There is a rule for quotients, but it follows from those two rules. If y = F(x)/G(x), then y = F(x) (G(x))-1 and you can use both the product rule and the chain rule (the chain rule on the function G and the negative-one power function as in the last post). I could spell it out for you if you like. :)

There are a few functions like the trig functions where it takes a little more work to figure out (if you apply those rules to the infinite series you used above, you'll see that the sine/cosine functions fall out pretty neatly), but basically you should be able to take the derivative of anything.

publius
2007-Oct-12, 03:45 AM
I'm getting Deja Vu here..........Ah yes, I went through derivatives with grav here a while back in another thread.............. Here it is:

-Richard

grav
2007-Oct-12, 12:32 PM
The chain rule is even easier. If y = F(G(x)), then y' = F'(G(x)) G'(x). So, when y = sin(2x), F is the sine function and G is the multiply by 2 function. So, F' is the cosine function and F'(G(x)) = cos(2x) and G'(x) = 2 and y' = 2 cos(2x)
Easier, you say? It looks simple, sort of, but there's something about it that is still confusing to me, seeming to go against my everyday logic somehow. If cos(x) is the derivative of sin(x), then in finding it for sin(2x), why can't I just replace 2x with x' and find that cos(x') is the derivative of sin(x'), since x' could have been x just as easily, and then substitute the 2x back in to get cos(2x) instead of 2cos(2x)? In other words, if we have cos(.1) as the derivative of sin(.1) and cos(.2) as the derivative of sin(.2), then why isn't the derivative of sin(2*.1) the same as for sin(.2), becoming cos(2*.1) and cos(.2), instead of 2cos(2*.1) for the first one, where x=.1 ? I know they must be the same in this case, so am I looking at something cross-eyed or something?

grav
2007-Oct-12, 12:41 PM
It must have something to do with the 2y present in there, right? It's the only thing I can see. If y' = 2cos(2x), and if y' = 2y, then 2y = 2cos(2x), so y = cos(2x). Something like that. Does that make sense?

grav
2007-Oct-12, 12:47 PM
It must have something to do with the 2y present in there, right? It's the only thing I can see. If y' = 2cos(2x), and if y' = 2y, then 2y = 2cos(2x), so y = cos(2x). Something like that. Does that make sense?Well, let's see. If we find the derivative for y to be 1 - cos(x), then the derivative for 2y would be 2 - 2cos(x), except that x is now half as great as the x' we are using, so it becomes 2 - 2cos(2x). Am I getting close?

Fortunate
2007-Oct-12, 01:22 PM
...why can't I just replace 2x with x' and find that cos(x') is the derivative of sin(x'), since x' could have been x just as easily, and then substitute the 2x back in to get cos(2x) instead of 2cos(2x)?

The phrase "derivative of sin(x)", although we say things like that all the time, is, in some sense, an incomplete description of what is gonig on. Really, we take the derivative of a function with respect to a particular variable. Think of a derivative as a rate of change of one quantity with respect to another quantity. That the derivative of sinx is cosx means that the rate of change of sinx with respect to the variable x is cosx. When you substitute x' for 2x as you describe, the derivative or rate of change of x' with respect to x' is cosx', but since x' is changing twice as rapidly as x, the rate of change of cosx' with respect to x is twice as great, and is therefore equal to 2cosx'.

From another point of view, think of the derivative of sinx as a quotient: the limit as the changes become smaller and smaller and approach 0 of (change in sinx)/(change in x). The derivative of sin2x with respect to x is the limit of (change in sin2x)/(change in x). When you took the derivative of sinx' with respect to x', you found that limit of (change in sinx')/(change in x') = cosx', which means that limit (change in sin(2x))/(change in 2x) = cos(2x). But (change in 2x) = 2(change in x), so
(change in sin2x)/(2(change in x)) = cos(2x). Multiplying both sides by 2 tells us that
limit (change in sin(2x))/(change in x) = 2cos(2x). The quantity on the left side of this equation matches the quantity printed in red, a quantity that we had said was the derivative of sin(2x) with respect to x.

hhEb09'1
2007-Oct-12, 01:29 PM
Easier, you say?OK, yeah, I meant easier in the sense that the formula looks simpler. As I said, these are the ones that everybody screws up, they take practice.
It looks simple, sort of, but there's something about it that is still confusing to me, seeming to go against my everyday logic somehow.You're not the only one who has ever tried to take the derivative of F(G(x)) to be just F'(G(x)). And even after you finally think you understand it, you'll find yourself making the mistake sometimes.
If cos(x) is the derivative of sin(x), then in finding it for sin(2x), why can't I just replace 2x with x' and find that cos(x') is the derivative of sin(x'), since x' could have been x just as easily, and then substitute the 2x back in to get cos(2x) instead of 2cos(2x)? In other words, if we have cos(.1) as the derivative of sin(.1) and cos(.2) as the derivative of sin(.2), then why isn't the derivative of sin(2*.1) the same as for sin(.2), becoming cos(2*.1) and cos(.2), instead of 2cos(2*.1) for the first one, where x=.1 ? I know they must be the same in this case, so am I looking at something cross-eyed or something?The point is, for F(x), you've modified the way that the x is changing (by using the function G) and you have to somehow account for that.

Remember, the rise over run (slope) definition of the derivative for F(x):
F'(x) = (F(x2) - F(x1)) / (x2-x1)
If you just substitute G in everywhere like you're doing, you get
F'(G(x)) = (F(G(x2)) - F(G(x1))) / (G(x2)-G(x1))
But we want the run in the denominator to be (x2-x1), in order to calculate the derivative of y = F(G(x)),
so we multiply by another term:
y' = (F(G(x2)) - F(G(x1))) / (G(x2)-G(x1)) times (G(x2)-G(x1))/(x2-x1)
By cancelling the (G(x2)-G(x1)), that simplies to what we want:
y' = (F(G(x2)) - F(G(x1))) / (x2-x1)
But the red part is F'(G(x)) and the blue part is obviously G'(x). Viola, the chain rule:
y' = F'(G(x)) G'(x)

You can play with this one using F as the cube function, and G as the square function. In other words, y is the sixth power. We know what the derivative of y is from the simple polynomial rule, but it works out the same using the chain rule.

PS: After you're comfortable with the chain rule, and you believe the product rule, try using them together to come up with the quotient rule. That's the rule for finding the derivative of F(x)/G(x). Use the product rule combined with the chain rule (using the F(x)(G(x))-1 form I mentioned earlier).

grav
2007-Oct-12, 02:08 PM
Geez. That'll definitely take some getting used to. :) I'll work on it some more later, after looking back through that other thread Richard linked back to as well.

hhEb09'1
2007-Oct-12, 02:30 PM
y' = F'(G(x)) G'(x)

You can play with this one using F as the cube function, and G as the square function. In other words, y is the sixth power. We know what the derivative of y is from the simple polynomial rule, but it works out the same using the chain rule.We have F(x) = x3 and G(x) = x2, so y = F(G(x)) = F(x2) = x6
We want to find the derivative of y = x6

G'(x) = 2x
F'(x) = 3x2
F'(G(x)) = 3(G(x))2 = 3(x2)2 = 3x4
The chain rule says
y' = F'(G(x)) G'(x)
so
y' = 3x4 times 2x, which is 6x5

Same as the polynomial formula

hhEb09'1
2007-Oct-12, 03:39 PM
The chain rule says
y' = F'(G(x)) G'(x)I just thought of a great example to motivate the chain rule.

Let F(x) = x, and G(x) be anything you want. Then y = F(G(x)) = G(x)

But F'(x) = 1 so F'(G(x)) = 1

Do you see how the introduction of G modifies the behavior of the F(G(x)) function? The F'(G(x)) has to be multiplied by the G'(x) to get y'

Fortunate
2007-Oct-12, 04:53 PM
Geez. That'll definitely take some getting used to. :) I'll work on it some more later, after looking back through that other thread Richard linked back to as well.

Here are some examples:

The derivative of (x3 + 5x2 - 3x + 2)8 is 8(x3 + 5x2 - 3x + 2)7(3x2 + 10x - 3).

The derivative of sin(x5) is 5x4cos(x5).

The derivative of x2 + 2(sinx)4 is 2x + 8(sinx)3cosx.

The derivative of (x + sin(x2))4 is 4(x + sin(x2))3(1 + 2xcos(x2)).
In the last example, note that we had to use the chain rule twice. The second use resulted in having to multiply cos(x2) by the derivative of x2.

grav
2007-Oct-12, 10:06 PM
The phrase "derivative of sin(x)", although we say things like that all the time, is, in some sense, an incomplete description of what is gonig on. Really, we take the derivative of a function with respect to a particular variable. Think of a derivative as a rate of change of one quantity with respect to another quantity. That the derivative of sinx is cosx means that the rate of change of sinx with respect to the variable x is cosx. When you substitute x' for 2x as you describe, the derivative or rate of change of x' with respect to x' is cosx', but since x' is changing twice as rapidly as x, the rate of change of cosx' with respect to x is twice as great, and is therefore equal to 2cosx'.

From another point of view, think of the derivative of sinx as a quotient: the limit as the changes become smaller and smaller and approach 0 of (change in sinx)/(change in x). The derivative of sin2x with respect to x is the limit of (change in sin2x)/(change in x). When you took the derivative of sinx' with respect to x', you found that limit of (change in sinx')/(change in x') = cosx', which means that limit (change in sin(2x))/(change in 2x) = cos(2x). But (change in 2x) = 2(change in x), so
(change in sin2x)/(2(change in x)) = cos(2x). Multiplying both sides by 2 tells us that
limit (change in sin(2x))/(change in x) = 2cos(2x). The quantity on the left side of this equation matches the quantity printed in red, a quantity that we had said was the derivative of sin(2x) with respect to x.Oh, okay. Thanks for spelling it out for me like that. :) Suddenly it makes perfect sense. A derivative is really a ratio, or slope, which is always understood to be in respect to a change in x, as hh said as well, but it took me a while to get it.

So let's see. Taking the derivative of sin(4x^3), say, would become sin(4x^3)/[4(x+d)^3-4x^3] = cos(4x^3), so sin(4x^3)/[4*3dx^2] = cos(4x^3), where d is the change in x, so then, sin(4x^3)/d = 12x^2*cos(4x^3). Is that right? It looks like the general equation for the derivative of sin(n*x^m), then, would be just [m*n*x^(m-1)]*cos(n*x^m).

grav
2007-Oct-12, 10:20 PM
Really, though, I should have written that as d sin(n*x^m) / d x = [m*n*x^(m-1)]*cos(n*x^m), right? What is d^2 x? Is that the same as (d x)^2?

grav
2007-Oct-12, 11:12 PM
After you're comfortable with the chain rule, and you believe the product rule, try using them together to come up with the quotient rule. That's the rule for finding the derivative of F(x)/G(x). Use the product rule combined with the chain rule (using the F(x)(G(x))-1 form I mentioned earlier).Okay, let's see. We are trying to find the derivative for y = F(x)/G(x) = F(x)G(x)-1 = F(x)H(G(x)), where H(x) = x-1. Using the product rule, this becomes y' = F'(x)H(G(x)) + F(x) d (H(G(x))). So now I'll take the derivative of H(G(x)) using the chain rule, and I come up with y' = F'(x)H(G(x)) + F(x)H'(G(x))G'(x). So we have the derivative of y = F(x)/G(x) as

y' = F'(x)H(G(x)) + F(x)H'(G(x))G'(x)
= F'(x)G(x)-1 + F(x)(-G(x))-2)G'(x)
= F'(x)/G(x) - F(x)G'(x)/G(x)2
= [F'(x)G(x) - F(x)G'(x)]/G(x)2

How does that look?

hhEb09'1
2007-Oct-13, 03:22 AM
It looks like the general equation for the derivative of sin(n*x^m), then, would be just [m*n*x^(m-1)]*cos(n*x^m).Looks good to me, there.

The derivative of ex is just ex, that's how the constant e is defined, more or less. The value of its slope is the same as its value, at each point.

Now, let's introduce this notation: F' = dy/dx. The dy and dx are representing those infinitesimal quantities, but it's still rise over run. Now, in the formulae y = ex and dy/dx = ex, exchange x for y and y for x

x = ey and dx/dy = ey,
Solve the first one for y, and the second one for dy/dx, and use the first one to get rid of y in the second. Viola! the surprise derivative of another function. Although dy/dx isn't really a fraction (it's y'), it's form suggests a lot.
What is d^2 x? Is that the same as (d x)^2?And that's where it suggests too much :)

(F')' is the second derivative, sometimes written d2y/dx2

= [F'(x)G(x) - F(x)G'(x)]/G(x)2

How does that look?Good.

My brother calls that HoDeeHigh minus HighDeeHo divided by HoHo

grav
2007-Oct-13, 03:02 PM
The derivative of ex is just ex, that's how the constant e is defined, more or less. The value of its slope is the same as its value, at each point.

Now, let's introduce this notation: F' = dy/dx. The dy and dx are representing those infinitesimal quantities, but it's still rise over run. Now, in the formulae y = ex and dy/dx = ex, exchange x for y and y for x

x = ey and dx/dy = ey,
Solve the first one for y, and the second one for dy/dx, and use the first one to get rid of y in the second. Viola! the surprise derivative of another function. Although dy/dx isn't really a fraction (it's y'), it's form suggests a lot.Well, I found for the derivative for y = ex as

y' = [e[sup]x+dx) - ex]/dx
= [exedx - ex]/dx
= ex[edx - 1]/dx
= ex[(1 + dx) - 1]/dx, since edx approaches 1 + dx for small dx
= ex dx /dx
= ex

and I found the derivative for y where x = ey, so y = In(x), as

y' = [In(x + dx) -In(x)]/dx
= [(In(x) + dx/x) -In(x)]/dx, where In(x + dx) approaches In(x) + dx/x for small dx
= (dx/x)/dx
= 1/x

Using the chain rule for F(G(x)), where F(x) =In(x) and G(x)=ex, I get y = F(G(x)) = x, so the derative of F(G(x)) is y' = F'(G(x))G'(x) = e(In(x))(1/x) = x/x = 1. Or equally in this case, y = G(F(x)), so y' = G'(F(x))F'(x) = (1/ex)ex = 1. Of course, it not at all surprising that the derivative of x is 1, so I very much doubt that is what you were aiming at, but it was fun anyway. :) I'll reread your post and try again when I get back later.

And that's where it suggests too much :)

(F')' is the second derivative, sometimes written d2y/dx2So that's what a second derivative is? I've always wondered. Would that just be taking the derative of the derative from the first result, then? In other words, is (F')' the same as (F'(x))'? Like the derivative of 2x3 is 6x2, and the derivative of that is 12x, so the second derivative of 2x3 would be 12x?

My brother calls that HoDeeHigh minus HighDeeHo divided by HoHoThat sounds like a good way to remember it.

hhEb09'1
2007-Oct-13, 03:21 PM
Of course, it not at all surprising that the derivative of x is 1, so I very much doubt that is what you were aiming at, but it was fun anyway. :) I was just highlighting that the derivative of ln x is 1/x. Notice, 1/x (as x-1) might fit into the scheme for mxn but for one little problem. :)

so the second derivative of 2x3 would be 12x?That's all it is. Since the first derivative is slope of the curve, the second derivative is slope of the slope--how fast the slope is changing. In other words, a positive second derivative usually indicates a concave up curve. Since 12x is positive when x is positive, that tells you 2x3 is concave up when x is positive, concave down when x is negative.

Next big important fact: another way of saying that the derivative of x2 is 2x? The integral of 2x is x2.

grav
2007-Oct-13, 04:21 PM
I was just highlighting that the derivative of ln x is 1/x. Oh, okay then. I guess I got that much. :)

Notice, 1/x (as x-1) might fit into the scheme for mxn but for one little problem. :)Yes, I see. If y = mxn, then y' = mnxn-1. So for y' = x-1, n would have to be zero for n-1 to come out to -1, but since the derivative is also multiplied by that, we should just get zero instead of 1/x. Interesting. To even that out, we could make m equal infinity, but then we have y = ex = mx0 = m = infinity. So what does that mean? I'm guessing it must mean that ex cannot be expressed in the form mxn, then.

That's all it is. Since the first derivative is slope of the curve, the second derivative is slope of the slope--how fast the slope is changing. In other words, a positive second derivative usually indicates a concave up curve. Since 12x is positive when x is positive, that tells you 2x3 is concave up when x is positive, concave down when x is negative.That sounds something like the difference between velocity and acceleration. So would velocity would be ds/dt and acceleration would be, well actually, let's see. a = d(ds/dt)/dt, so, umm, yeah, how would that work out? Do I use the chain rule here or something? Why did you express the second derivative of y as d2y/dx2 before? Why not just d2y/dx? Wouldn't dx2 come out to 2x, giving twice the value for the second derivative of y?

Next big important fact: another way of saying that the derivative of x2 is 2x? The integral of 2x is x2.Ah. That would be the same as taking the summation of 2x and solving for a large x. In other words, the summation of x is 1+2+3+4+...+x = x(x+1)/2, where dx = 1, which approaches x2/2 when x is large, so for 2x, it is just x2.

TomT
2007-Oct-13, 05:07 PM
Just noticed this interesting discussion. Have a couple comments on the math.

sin(2x) - 2x = 2y

as you have shown. Since the 2's can be drawn out after the fact, to simplify things quite a bit, then, it looks like I am now just looking for a solution for x knowing y with the new equation

sin(x) - x = y

You can't draw out the 2 from sin(2x) as you have shown. sin(2x) - 2x = 2y is the simplest correct form, I think.

Well then,

- x^3/3! + x^5/5! - x^7/7! + ... = y

and x will be a transcendent number in terms of y. One I'm not sure how to calculate.

I think you have to write the series solution for sin(2x) not sin(x), so the solution is

2y = -(2x)^3/3! + (2x)^5/5! -(2x)^7/7! + ....

I've always been a nit picker when it comes to math :>).

grav
2007-Oct-13, 05:27 PM
Just noticed this interesting discussion. Have a couple comments on the math.

You can't draw out the 2 from sin(2x) as you have shown. sin(2x) - 2x = 2y is the simplest correct form, I think.

I think you have to write the series solution for sin(2x) not sin(x), so the solution is

2y = -(2x)^3/3! + (2x)^5/5! -(2x)^7/7! + ....

I've always been a nit picker when it comes to math :>).Thanks, TomT. I didn't divide the whole equation by two or anything, though, since I know that's what it looks like, but really just changed the formula from sin(2x) - 2x = 2y to sin(x') - x' =y', where x'=2x and y'=2y, and then simplified that to make just sin(x) - x = y the original formula to find for instead, not that it is the same thing, but I could then just double the original value of y to find x', and then half that to find the original value for x after the fact. So I was really just trying to make it easier on everybody by keeping the formula as simple as possible.

TomT
2007-Oct-13, 06:02 PM
Thanks, TomT. I didn't divide the whole equation by two or anything, though, since I know that's what it looks like, but really just changed the formula from sin(2x) - 2x = 2y to sin(x') - x' =y', where x'=2x and y'=2y, and then simplified that to make just sin(x) - x = y the original formula to find for instead, not that it is the same thing, but I could then just double the original value of y to find x', and then half that to find the original value for x after the fact. So I was really just trying to make it easier on everybody by keeping the formula as simple as possible.

I see. Makes sense.

frankuitaalst
2007-Oct-13, 09:02 PM
Hi , I'm just curious in what field of application this formula occurs ?

hhEb09'1
2007-Oct-13, 09:44 PM
That sounds something like the difference between velocity and acceleration. So would velocity would be ds/dt and acceleration would be, well actually, let's see. a = d(ds/dt)/dt, so, umm, yeah, how would that work out? Do I use the chain rule here or something? Why did you express the second derivative of y as d2y/dx2 before? Why not just d2y/dx? Wouldn't dx2 come out to 2x, giving twice the value for the second derivative of y?Yes, velocity is the derivative of the position function, and acceleration is the derivative of the velocity function, making acceleration the second derivative of the position function. One of the original motivations for the whole concept.

The symbolism d(ds/dt)/dt is correct, and it would also be expressed as d2s/dt2. Possibly more accurately, it would also be d2/dt2(s), where the d2/dt2 is just considered the second derivative operator, like the '' in s'' so dt2 is more like (dt)2 but they leave off the parentheses.

That's why it is dangerous to take the notation too far :)
Ah. That would be the same as taking the summation of 2x and solving for a large x. In other words, the summation of x is 1+2+3+4+...+x = x(x+1)/2, where dx = 1, which approaches x2/2 when x is large, so for 2x, it is just x2.That is the discrete analogue, it's true

If you want to hone your physical intuition in these things, you can consider the area under the function 2x to be "growing" by a thin slice that is 2x by dx, then you just integrate (antiderivative) that. If you want to find the formula for the volume of the sphere, just take the surface area 4pi r2 times a small depth dr, and integrate that. But that's just suggestive, you'll see there are fine points to that later.

grav
2007-Oct-13, 11:43 PM
Hi , I'm just curious in what field of application this formula occurs ?You can see where I used it about halfway through this post (http://www.bautforum.com/questions-answers/65567-time-freefall-2.html#post1085856). I was supplied the formula to find the time of descent of a falling body from a starting distance to a final distance, but I wanted the final distance knowing the time.

grav
2007-Oct-14, 12:00 AM
Yes, velocity is the derivative of the position function, and acceleration is the derivative of the velocity function, making acceleration the second derivative of the position function. One of the original motivations for the whole concept.

The symbolism d(ds/dt)/dt is correct, and it would also be expressed as d2s/dt2. Possibly more accurately, it would also be d2/dt2(s), where the d2/dt2 is just considered the second derivative operator, like the '' in s'' so dt2 is more like (dt)2 but they leave off the parentheses.

That's why it is dangerous to take the notation too far :)That is the discrete analogue, it's true

If you want to hone your physical intuition in these things, you can consider the area under the function 2x to be "growing" by a thin slice that is 2x by dx, then you just integrate (antiderivative) that. If you want to find the formula for the volume of the sphere, just take the surface area 4pi r2 times a small depth dr, and integrate that. But that's just suggestive, you'll see there are fine points to that later.Well, let's see. We would just add the volumes for infinitesimal distances starting at the center of the sphere and moving outward to r, so the integration for the volume of a sphere would be just the summation of (4pi dr2)[1^2 + 2^2 + 3^2 + ... + n^2] * dr with n=r/dr, where 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6, which approaches n^3/3 for large n (or infinitesimal dr, since n=r/dr), so the integration is (4pi dr2)[((r/dr)3/3)] * dr = 4pir3/3.

hhEb09'1
2007-Oct-14, 12:05 AM
Well, let's see. The integration for the volume of a sphere would be just the summation of (4pi dr2)[(1^2 + 2^2 + 3^2 + ...)/dr2] with dr=1, where 1^2 + 2^2 + 3^2 + ... = r(r+1)(2r+1)/6, which approaches r^3/3 for large r (or infinitesimal dr), so the integration is (4pi dr2)[(r3/3)/ dr2] = 4pir3/3.No, no, no, use antiderivatives. :)

grav
2007-Oct-14, 12:25 AM
No, no, no, use antiderivatives. :)Oh, of course.

The antiderivative of 3x^2 is x^3, so the integration of x^2 is x^3/3. But I can't do it that way, it's too easy. :)

hhEb09'1
2007-Oct-14, 12:37 AM
But I can't do it that way, it's too easy. :)So, what you need now is some hard calculus problems :)

Seriously, calculus makes things a lot easier, which means you can spend all your energy on harder problems.

A couple comments. The antiderivative is a function, but it usually is written with a constant C appended, for maybe obvious reasons.

The integral of x4 evaluated at x=2 is the area under the x4 curve, from x equal zero to 2. To find the area between 1 and 2, just subtract the antiderivative for 2 minus the antiderivative for 1.

The integral of x3 from -1 to +1 results in zero. That's a pesky but sometimes valuable feature--negative area.

grav
2007-Oct-14, 01:11 AM
So, what you need now is some hard calculus problems :)I was going to practice with Fortunate's examples in post #39, and it became almost immediately obvious the chain rule was used for all of them, once I thought about it for a second. So are those rules all I need to know? It is certainly helpful to know that an integration is just an anti-derivative as well. I did not know that, and I was wondering if I would know when to use a derivative for something. Now I think I do, finding for something like the volume of a sphere knowing the surface area by integrating, and finding the surface area knowing the volume by finding the derivative. I'll keep that in mind. Thank you for everything, hh (and Fortunate and others). I really appreciate it tremendously. :D

hhEb09'1
2007-Oct-14, 02:29 AM
So are those rules all I need to know? Certainly the basics.

What's the derivative of sin(x)/cos(x) ?

What's the derivative of log10 x ?

What's the antiderivative of ln x ?

01101001
2007-Oct-14, 02:57 AM
It is certainly helpful to know that an integration is just an anti-derivative as well.

Yeah. Mathematicians find it so helpful that they labeled it The Fundamental Theorem of Calculus (http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus).

grav
2007-Oct-14, 12:54 PM
Certainly the basics.

What's the derivative of sin(x)/cos(x) ?

What's the derivative of log10 x ?

What's the antiderivative of ln x ?That first one gets the "HighDeeHo" treatment, so becomes

y = sin(x)/cos(x)
y' = [G(x)F'(x) - F(x)G'(x)]/G(x)2
y' = [cos(x)2 - sin(x)2]/cos(x)2
y' = 1 - [sin(x)/cos(x)]2
or y' = 1 - [1 - cos(x)2]/cos(x)2
y' = 2 - 1/cos(x)2

That next one is

y = log10(x)
y' = [log10(x + dx) - log10(x)]/dx
which for small dx approaches
y' = [(log10(x) + dx/2.302585x) - log10(x)]/dx
y' = (dx/2.302585x)/dx = 1/2.302585x

That number threw me off for a minute until I realized it was just In(10) = 2.302585, so

y' = 1/(In(10)x)

For that last one I had to cheat, I'm afraid. I don't know what derivative will give In(x) in order to find the antiderivative in the first place, so I used summations again. I find the summation (integration) of In(x) to be [In(x) - 1]x for large x (or infinitesimal dx), so the derivative of [In(x) - 1]x is In(x).

grav
2007-Oct-14, 01:07 PM
I find the summation (integration) of In(x) to be [In(x) - 1]x for large x (or infinitesimal dx), so the derivative of [In(x) - 1]x is In(x).I just realized something else as well. If the derivative of [In(x) - 1]x = In(x)x - x is In(x), and derivatives can be added, whereas the derivative of -x is -1, then the derivative of In(x)x is In(x) + 1.

grav
2007-Oct-14, 01:45 PM
I just realized something else as well. If the derivative of [In(x) - 1]x = In(x)x - x is In(x), and derivatives can be added, whereas the derivative of -x is -1, then the derivative of In(x)x is In(x) + 1.Oh, and this can be verified through the product rule, where the derivative of In(x)x becomes

y = In(x)x

F(x) = In(x), G(x) = x

y' = F'(x)G(x) + F(x)G'(x)
y' = (1/x)(x) + (In(x))(1)
y' = 1 + In(x)

So now that I know this, I can use it in the future (if I remember it) and just work backwards to get that the antiderivative of In(x), for example, is the same as the antiderivative of (In(x) + 1) - 1, which becomes In(x)x for In(x) + 1 and -x for -1, giving me In(x)x - x.

hhEb09'1
2007-Oct-14, 04:36 PM
That first one gets the "HighDeeHo" treatment, so becomes

y = sin(x)/cos(x)
y' = [G(x)F'(x) - F(x)G'(x)]/G(x)2
y' = [cos(x)2 - sin(x)2]/cos(x)2You missed a minus sign! and remember the other trig identities at the end, like sec(x) = 1/cos(x)

That number threw me off for a minute until I realized it was just In(10) = 2.302585, so

y' = 1/(In(10)x) That's the correct answer, except somehow ln turned into In. :)

A way to avoid going back to series approximations, is just to convert the original problem from base 10 to natural logs: log10x = logex / loge10
For that last one I had to cheat, I'm afraid. I don't know what derivative will give In(x) in order to find the antiderivative in the first place, so I used summations again. I find the summation (integration) of In(x) to be [In(x) - 1]x for large x (or infinitesimal dx), so the derivative of [In(x) - 1]x is In(x).And you can check that just by taking the derivative of [ln(x) - 1]x (note: I have changed the "I" to an "l" again :) ). In general, taking derivatives is a straightforward application of things we've already covered. Taking antiderivatives involves finding the "trick"--but it's easily checked. A lot of intro calc is studying such tricks, and practicing them.

Would you like to try one? :)

So now that I know this, I can use it in the future (if I remember it)Yep. A lot of it is just remembering what you already know :)

grav
2007-Oct-14, 10:06 PM
You missed a minus sign! and remember the other trig identities at the end, like sec(x) = 1/cos(x)Oh. I was thinking that since cos and sin had inverse properties but otherwise symmetrical, that each would be the derivative of the other. Apparently not, I guess. So the derivative of sin(x) is cos(x), but the derivative of cos(x) is -sin(x)? Hmm. I'll have to remember that. So starting again,

y = sin(x)/cos(x)
y' = [G(x)F'(x) - F(x)G'(x)]/G(x)2
F(x) = sin(x), G(x) = cos(x)
y' = [cos(x)2 + sin(x)2]/cos(x)2
y' = [cos(x)2 + (1 - cos(x)2)]/cos(x)2
y' = 1/cos(x)2 = sec(x)2

That's the correct answer, except somehow ln turned into In. :)Okay, thanks. I used to denote it ein(x) when I worked on this stuff on my own, so I could recognize it, since I always thought the calculator read ex and In as in "in" for the inverse of the logarithm, and even pronounced it that way to myself, as "in"(x). I think Grant corrected me on the ein part a while back, when I had gotten so used to it I began posting it that way, forgetting that it was my own notation. I still continued to use In, though, but now, looking at what you wrote, I suppose that it must be "ln" for "natural logarithm".

A way to avoid going back to series approximations, is just to convert the original problem from base 10 to natural logs: log10x = logex / loge10That's interesting. I also notice that loge(10) = 1 / log10(e), so that second derivative for log10(x) could be y' = 1/(ln(10)x) or equally y' = log10(e)/x .

Taking antiderivatives involves finding the "trick"--but it's easily checked. A lot of intro calc is studying such tricks, and practicing them.

Would you like to try one? :)Sure. Sounds like fun. :shifty: :)

hhEb09'1
2007-Oct-14, 10:22 PM
Oh. I was thinking that since cos and sin had inverse properties but otherwise symmetrical, that each would be the derivative of the other. Apparently not, I guess. So the derivative of sin(x) is cos(x), but the derivative of cos(x) is -sin(x)? Hmm.Not exactly inverse properties. Remember, cosine is the same as sine, with a 90 degree shift.
y' = 1/cos(x)2 = sec(x)2 Yep, the derivative of tangent is secant squared.
so that second derivative for log10(x) could be y' = 1/(ln(10)x) or equally y' = log10(e)/x Just remember that those "two" derivatives have the same constant.

Sure. Sounds like fun. :shifty: :)Find the antiderivative of x/sqrt(1-x2) :)

grav
2007-Oct-14, 11:35 PM
Find the antiderivative of x/sqrt(1-x2) :)
Well, that was quite the challenge. My first thought, since cos(x)^2 = sqrt(1 - sin(x)^2) and sin(x)^2 = sqrt(1 - cos(x)^2), was to substitute x with sin(z), giving y' = x / sqrt(1 - x2) = cos(z)/sin(z). But although I now know the derivative of that, I don't know the antiderivative. So I tried to reverse engineer each of the rules one by one, and finally settled on the chain rule. But first, I changed the original formula to y = z / sqrt(1-z2) and substituted in again with z = sin(x), so as not to confuse myself (as long as I don't confuse you ;) ). So that would be y' = F'(G(x))G'(x) = cos(x)/sin(x). Here I decided that if G'(x) = cos(x), then G(x) = sin(x). So F'(sin(x)) = 1/sin(x), therefore F'(x) = 1/x. The antiderivative of 1/x is ln(x), so F(x) = ln(x). So I get y = F(x)G(x) = ln(x)*sin(x). Substituting back in for x = asin(z), it becomes y = ln(asin(z))*sin(asin(z)) = ln(asin(z))z. Finally, changing z back to x, I get y = ln(asin(x))x. Is that correct?

EDIT-OOps. Wait a minute. I think I inversed the fraction somehow. Let me try again. :)

grav
2007-Oct-15, 12:01 AM
Okay, one more time.

y' = z/sqrt(1-z2)
z = sin(x)
y' = sin(x)/cos(x)
y' = F'(G(x))G'(x)
G'(x) = sin(x)
G(x) = -cos(x)
F'(G(x)) = 1/cos(x)
F'(x) = -1/x
F(x) = -ln(x)
y = F(x)G(x) = -ln(x) * -cos(x)
= ln(x) cos(x)
= ln(asin(z)) cos(asin(z))
z replaced with x
y = ln(asin(x)) cos(asin(x))
cos(asin(x)) = sqrt(1-x2)
y = ln(asin(x))*sqrt(1-x2)

How's that?

grav
2007-Oct-15, 12:31 AM
Alrighty then. Third time's the charm, I hope. I guess I need to slow down. Apparently I tried to put F(x) and G(x) back together using the product rule instead of the chain rule, whereas I was using the chain rule to find the values. So here we go again.

y' = z/sqrt(1-z2)
z = sin(x)
y' = sin(x)/cos(x)
y' = F'(G(x))G'(x)
G'(x) = sin(x)
G(x) = -cos(x)
F'(G(x)) = 1/cos(x)
F'(x) = -1/x
F(x) = -ln(x)
y = F(G(x)) = -ln(-cos(x))
= -ln(-cos(asin(z)))
z replaced with x
y = -ln(-cos(asin(x))
cos(asin(x)) = sqrt(1-x2)
y = -ln(-sqrt(1-x2))

But somehow, that doesn't look right either, especially since it would come out complex. I'll keep trying.

grav
2007-Oct-15, 12:48 AM
Well, I just took the derivative of sqrt(1-x^2) and found it to be -x/sqrt(1-x^2). :rolleyes: So I guess the antiderivative of x/sqrt(1-x^2) would be just -sqrt(1-x^2). Final answer. But where did I go wrong with my third attempt, and um, what's the trick?

hhEb09'1
2007-Oct-15, 01:26 AM
So here we go again.

y' = z/sqrt(1-z2)
z = sin(x)
y' = sin(x)/cos(x)
y' = F'(G(x))G'(x)You have to be careful here, because in general your variable is z still, so you'd have to multiply by x' (to continue the chain of the chain rule) which, as you've seen, complicates things.

This is where the notation comes in handy. You will always see the integrals expressed as integrals of a function f(x) times the differential dx

dy/dz = z/sqrt(1-z2)
dy = z/sqrt(1-z2) dz
So, you can integrate the left hand side with respect to y, because the differential is in y, and integrate the right with respect to z. The integral of dy is y, of course (with an additional constant)

You make a good guess in the next step
z = sin(x)
dz/dx = cos(x)
dz = cos(x) dx

Substituting:
dy = sin(x) / cos(x) times cos(x) dx
dy = sin(x) dx
since the variable and the differential match on the right
y = -cos(x) (plus a constant C of course)

y = -sqrt(1-z2) + C
or
y = -sqrt(1-z2)1/2 + C

which you can check by differentiation.

Well, I just took the derivative of sqrt(1-x^2) and found it to be -x/sqrt(1-x^2). :rolleyes: So I guess the antiderivative of x/sqrt(1-x^2) would be just -sqrt(1-x^2). Final answer. But where did I go wrong with my third attempt, and um, what's the trick?It gets worse. :)

grav
2007-Oct-15, 02:12 AM
You have to be careful here, because in general your variable is z still, so you'd have to multiply by x' (to continue the chain of the chain rule) which, as you've seen, complicates things.Yes, yes I have. As you said, the derivative of z, for example, which is 1, where z = sin(x), is not the same as the derivative of sin(x), which is cos(x), or sqrt(1-z^2). Considering this, it's surprising I got as close as I did.

This is where the notation comes in handy. You will always see the integrals expressed as integrals of a function f(x) times the differential dx

dy/dz = z/sqrt(1-z2)
dy = z/sqrt(1-z2) dz
So, you can integrate the left hand side with respect to y, because the differential is in y, and integrate the right with respect to z. The integral of dy is y, of course (with an additional constant)

You make a good guess in the next step
z = sin(x)
dz/dx = cos(x)
dz = cos(x) dx

Substituting:
dy = sin(x) / cos(x) times cos(x) dx
dy = sin(x) dx
since the variable and the differential match on the right
y = -cos(x) (plus a constant C of course)

y = -sqrt(1-z2) + C
or
y = -sqrt(1-z2)1/2 + C

which you can check by differentiation.
It gets worse. :)That's a neat trick. So you work through the differentials? Pretty nifty. I'll have to study that.

hhEb09'1
2007-Oct-15, 02:17 AM
OK, when you have sums/differences of squares, you can use trig identities. When you have odd powers of cosine/sine, you can make a substituion that makes gets rid of one factor, and turns the rest into a polynomial. If you have even powers, use the half-angle formulas. Mostly, you can look them up in a book of integrals :)

What's the antiderivative of (x+9)/(6x2-17x-14) ?
So you work through the differentials? Pretty nifty. That's sorta what gave Leibniz's followers a leg up, the notation suggested the line of attack. As Euler once said, sometimes it seems as if your pencil is smarter than you.

grav
2007-Oct-15, 11:03 PM
Well, I almost gave up on this one. All I had determined so far was that y' = (x+9)/(6x2 - 17x - 14) = (x+9)/[(2x - 7)(3x + 2)] = a/(bc), where a + b = c, so getting rid of one of the factors, y' = a/[b(a+b)] = a/[(c-a)c], and I was trying all sorts of ways to get a square of the variables and so forth, such as with ac(a+c)/[(c2 - b2)c2] and ab(b-a)/[b2(b2 - a2)], but still had no idea where to go from there. But then, on the way home today, I realized also that y' = (c-b)/bc, which becomes just 1/b - 1/c. Knowing that the derivative of ln(x) is 1/x, and b and c are both in terms of 1/x, I figured I could work with this.

For 1/b - 1/c, I can find the antiderivatives one at a time, first for 1/b and then for 1/c. Still thinking somewhat in terms of summations, I figured that the antiderivative is really an integration, and an integration is really a summation of infinitesimals. So if we were to say that the antiderivative in terms of a summation of 1/(n dx) from 1/dx (n=1) to 1/x (n=x/dx) is ln(x) for infinitesimal dx, then the antiderivative of 1/(x+z) is really just a summation of 1/(n dx +z) from 1/(dx + z) to 1/(x+z). That last one would just be the same as for 1/x, except it starts off a little higher in the progression, at 1/z instead of zero, and ends a little further up as well, at 1/(x+z) instead of 1/x. So the antiderivative of 1/(x+z) is simply ln(x+z), for zero to 1/(x+z), minus that amount which it started at, at 1/z, so minus that amount from zero to 1/z, which is ln(z), giving the antiderivative of y' = 1/(x+z) as y = ln(x+z) - ln(z). However, in finding the derivative for ln(x+z) - ln(z), the ln(z) part just cancels out as a constant anyway, so all we really need is the ln(x+z) part, I guess.

So I find the antiderivative of y' = (x+9)/(6x2 - 17x - 14), or y' = 1/b - 1/c = 1/(2x-7) - 1/(3x+2), to be the difference in the antiderivatives of those two, which are 1/b = 1/(2x - 7) = 1/[2(x - 7/2)] = (1/2)[1/(x - 7/2)], so y(b) = [ln(x - 7/2)]/2, and 1/c = 1/(3x + 2) = 1/[3(x + 2/3)] = (1/3)[1/(x + 2/3)], so y(c) = ln[(x + 2/3)]/3. Therefore, y = y(b) - y(c) = [ln(x - 7/2)]/2 - [ln(x + 2/3)]/3 + C, where C in this case should most probably be [ln(2/3)]/3 - [ln(-7/2)]/2. Whew. How was that?

grav
2007-Oct-15, 11:08 PM
The antiderivative of y' = (x+9)/(6x2 - 17x - 14) is y= [ln(x - 7/2)]/2 - [ln(x + 2/3)]/3 + C, but I just realized by working through it on my calculator here, that there is apparently a way to subtract ln's. Please hang on while I work that out.

grav
2007-Oct-15, 11:37 PM
I've finished working it out. It turns out that ln(x)/a - ln(y)/b = [ln(xb/ya)]/(ab), so the antiderivative of y' = (x+9)/(6x2 - 17x - 14) is y = [ln(x - 7/2)]/2 - [ln(x + 2/3)]/3 = [ln( (x - 7/2)3/(x + 2/3)2 )]/6 + C. Final answer.

grav
2007-Oct-15, 11:48 PM
Experimenting some more with the ln(x) function, it looks like ln(xn) = n ln(x), so [ln( (x - 7/2)3/(x + 2/3)2 )]/6 could also be expressed as [ln( (x - 7/2)/(x + 2/3)(2/3) )]/2 or [ln( (x - 7/2)(3/2)/(x + 2/3) )]/3, or just ln( (x - 7/2)(1/2)/(x + 2/3)(1/3) ), but I'm wondering how that would work out for the (x - 7/2)(3/2) and (x - 7/2)(1/2) for small x (x < 7/2), since that would make those parts complex.

EDIT- Oh, that's interesting. It doesn't matter if (x - 7/2)^(1/2) is complex, where x<7/2, because ln(x) for x<0 is complex anyway, and so the original (x - 7/2)^3 and (x - 7/2) would also both give a negative value for x in ln(x) with x<7/2, making ln(x) complex, so it would all work out the same. Much could probably be learned about the relation of ln(x), e, and i by studying this further, and probably has.

hhEb09'1
2007-Oct-16, 12:30 AM
Therefore, y = y(b) - y(c) = [ln(x - 7/2)]/2 - [ln(x + 2/3)]/3 + C, where C in this case should most probably be [ln(2/3)]/3 - [ln(-7/2)]/2. Whew. How was that?Pretty good! :)

Now for the short cuts. If you have a polynomial fraction, you can divide out until the numerator has smaller degree than the denominator. That leftover part is the tough nut, because the polynomial is easy (right?). But, as you've found out, (Ax+B)/((Cx+D)(Ex+F)) can be written G/(Cx+D) + H/(Ex+F) and you can find G and H just by adding them, so that G(Ex+F) + (Cx+D)H = Ax + B. So, GE + CH = A, GF+DH = B, and you can solve for G and H. This technique is called partial fractions, and is handy in a lot of places.

How to find the antiderivative of G/(Cx+D)? You're trying to integrate G dx/(Cx+D), so let z = (Cx+D), then dz = C dx.
Now, plugging in the substitutions, we get G dz/Cz, and the integral of that is G/C ln z
Which is G/C ln (Cx+D), plus the constant

In your problem the G and H were 1 and -1.

PS: Is my answer different from yours?

EDIT- Oh, that's interesting. It doesn't matter if (x - 7/2)^(1/2) is complex, where x<7/2, because ln(x) for x<0 is complex anyway, and so the original (x - 7/2)^3 and (x - 7/2) would also both give a negative value for x in ln(x) with x<7/2, making ln(x) complex, so it would all work out the same. Much could probably be learned about the relation of ln(x), e, and i by studying this further.The theory of complex variables would fill a book :)

grav
2007-Oct-16, 01:11 AM
(Ax+B)/((Cx+D)(Ex+F)) can be written G/(Cx+D) + H/(Ex+F) and you can find G and H just by adding them, so that G(Ex+F) + (Cx+D)H = Ax + B. So, GE + CH = A, GF+DH = B, and you can solve for G and H. This technique is called partial fractions, and is handy in a lot of places.Yes, that should come in very handy indeed. Thank you.

PS: Is my answer different from yours?Um, I think so. I get (G/C) ln[(Cx+D)/C] instead of just (G/C) ln(Cx+D). But I double-checked my answer this time by finding the derivative, so when I find the derivative for say, x=10, with y = (G/C) ln[(Cx+D)/C] = (G/C) ln(x + D/C) for y' = 1/(3x+2), so y = (1/3) ln(x + 2/3), with a very small dx like dx=.0001, I get y' = (1/3)[ln((x + dx) + 2/3) - ln(x + 2/3)] / dx = (1/3)[ln((10 + .0001 + 2/3) - ln(10 + 2/3)] / .0001 = (1/3)(.0937498) = 1/32.00006827, whereas y' = 1/(3x+2) for x=10 becomes 1/32, so it matches.

The theory of complex variables would fill a book :)Oh. No doubt. :)

hhEb09'1
2007-Oct-16, 01:24 AM
Um, I think so. I get (G/C) ln[(Cx+D)/C] instead of just (G/C) ln(Cx+D). Trick question.

(G/C) ln[(Cx + D)/C] equals (G/C) [ln(Cx+D) - ln C] because of the law of logarithms. (Turning division into subtraction, and similar tricks, were the reasons that people spent 40 years of their life devising logarithm tables for others to use.) That in turn is
(G/C) ln(Cx+D) plus -(G/C)ln C
That in red is a constant, and my answer includes an additional constant.

That's one of the two main reasons for including the constant--the solutions themselves may look different but be the same except for a constant. The other reason, you can use the constant to adjust the initial conditions for a function.

For instance, you know that an object has a acceleration of A, how far does it go in T seconds? The antiderivative of acceleration is velocity, and the antiderivative of A is At + C. That C represents its initial velocity at time t=0.

What's the antiderivative of velocity? I know you already know this one, I just want to see what insights you come up with. :)

PS: you did not need to use the infinitesimal dx in checking your answer--just use the chain rule, etc., and find dy/dx.

grav
2007-Oct-16, 01:57 AM
Trick question.

(G/C) ln[(Cx + D)/C] equals (G/C) [ln(Cx+D) - ln C] because of the law of logarithms. (Turning division into subtraction, and similar tricks, were the reasons that people spent 40 years of their life devising logarithm tables for others to use.) That in turn is
(G/C) ln(Cx+D) plus -(G/C)ln C
That in red is a constant, and my answer includes an additional constant.Well I'll be danged. That worked out just as well. :) So they are the same thing as far as the derivative is concerned, minus the constant, which doesn't figure in for that. I also found a constant of -(G/C) ln(D), so I guess the whole thing can also be derived as (G/C)[ln(Cx-D) - ln(C) - ln(D)] = (G/C) ln[(Cx+D)/(CD)], but since the C and D are both constants, we can use just (G/C) ln(Cx+D) + Constant. Cool.

For instance, you know that an object has a acceleration of A, how far does it go in T seconds? The antiderivative of acceleration is velocity, and the antiderivative of A is At + C. That C represents its initial velocity at time t=0.

What's the antiderivative of velocity? I know you already know this one, I just want to see what insights you come up with. :)Well, vt + C, but that's just off the top of my head. I'm not sure if I'll be coming up with any insights on that one, but I'll think about it some more, and see in what ways it can be derived.

PS: you did not need to use the infinitesimal dx in checking your answer--just use the chain rule, etc., and find dy/dx.Oh, yeah. Okay then, let's see. y = (1/3) ln(3x+2) = F(G(x)), where F(x) = (1/3) ln(x) and G(x) = 3x+2, so F'(x) = 1/(3x) and G'(x) = 3. So y' = F'(G(x))G'(x) = [1/(3(3x+2))]3 = 1/(3x+2). Yep. That'll do it. :)

hhEb09'1
2007-Oct-16, 02:12 AM
Well, vt + C, but that's just off the top of my head. I'm not sure if I'll be coming up with any insights on that one, but I'll think about it some more, and see in what ways it can be derived. Take the antiderivative of At + C

grav
2007-Oct-16, 03:02 AM
Take the antiderivative of At + CThat would be At2/2 + Ct + C', and I forgot to mention, the antiderivative of velocity is distance, of course.

hhEb09'1
2007-Oct-16, 03:20 AM
That would be At2/2 + Ct + C', and I forgot to mention, the antiderivative of velocity is distance, of course.Yes, so C is the initial velocity, and C' is the initial position. They're constants, in these cases, but they're essential to fitting the problem to real situations.

How much work is done raising an object of mass m from r[sub]1[/sup] to r[sub]2[/sup] through a gravity field that varies as g = GM/r2? Integrate the force GMn/r2 times the infinitesimal distance dr

Here's a differential equation, solve for the function y:

y'' + y = 0

And another:

y'' - y = 0

grav
2007-Oct-16, 03:24 AM
That would be At2/2 + Ct + C', and I forgot to mention, the antiderivative of velocity is distance, of course.And the antiderivative of distance would be At3/6 + Ct2/2 + C't + C''. So what would that be? :lol: Actually, though, I guess this would describe the distance travelled with a change in acceleration over time. I think that's called a jerk, is that right? So I guess it would be df = jt3/6 + aot2/2 + vot + do. How would that work out for an acceleration that changes over the distance travelled? What about a type of acceleration for a velocity that changes proportionally with distance travelled instead of time?

hhEb09'1
2007-Oct-16, 03:26 AM
What about a type of acceleration for a velocity that changes proportionally with distance travelled instead of time?Ha, anticipated! :)

grav
2007-Oct-16, 03:24 PM
How much work is done raising an object of mass m from r[sub]1[/sup] to r[sub]2[/sup] through a gravity field that varies as g = GM/r2? Integrate the force GMn/r2 times the infinitesimal distance drWell, the integration, or antiderivative of r-2 is -r-1, so I guess that would just be -GMn/r + C from dr to r. In this case, C=(pi2/6)GMn/dr. So for r=r1 to r2, it is [-GMn/r2 + C] - [-GMn/r1 + C] = GMn/r1 - GMn/r2 = GMn(1/r1 - 1/r2) = GMn(r2 - r1)/(r1 r2). I'm wondering about something, though. Should that be negative or positive? From earlier, if the antiderivative of t-2 is -t-1 for acceleration, then shouldn't A have become -At + C in our last example? Yet, the acceleration and the resulting velocity are both considered positive when acting in the same direction. If we figure it as (-A t1 +C) - (-A t2 + C), where t1 and t2 are times from some other specified T=0, then we get A(t2-t1) = At, where t here is the difference in time, which seems right also, depending upon how one looks at it, but we would lose C in the process. How should I think about this exactly?

Here's a differential equation, solve for the function y:

y'' + y = 0

And another:

y'' - y = 0Well, if I read you correctly, then going by what we had earlier, y = y'' dx2/2 + C dx + C', so y'' = 2(y - C dx - C')/dx2, and so for y'' + y = 0, I get

y'' + y = 0
2(y - C dx - C')/dx2 + y = 0
2y/dx2 + y = C dx + C'
y = (C dx + C')/(2/dx2 + 1)

For y'' - y = 0, then, that would become y = (C dx + C')/(2/dx2 - 1)

Or does that only apply when y is proportional to 1/x2? It seems that it should at least apply when y is proportional to 1/xn, where n=>2.

grav
2007-Oct-16, 04:26 PM
Also, I was just thinking. If we have A which is proportional to 1/t2, and the antiderivative of that is At + C, so that v = at + C, instead of claiming the C to be vo for v = vf, we can get v1 = a t1 + C and v2 = a t2 + C, so that v2 - v1 = (a t2 + C) - (a t1 + C) = a(t2 - t1), so that the C actually cancels out and becomes irrelevant.

hhEb09'1
2007-Oct-16, 04:37 PM
Also, I was just thinking. If we have A which is proportional to 1/t2, and the antiderivative of that is At + C, so that v = at + C, instead of claiming the C to be vo for v = vf, we can get v1 = a t1 + C and v2 = a t2 + C, so that v2 - v1 = (a t2 + C) - (a t1 + C) = a(t2 - t1), so that the C actually cancels out and becomes irrelevant.You have to be careful here. The antiderivative of A is At + C only if A is a constant. If it is itself a function of t, then the antiderivative has to take that into account (as in the chain rule examples for derivatives).

But yes when you integrate from one value to the next, the constants of integration do cancel. It's when you get the result v = At + C and you also have the knowledge that v(0)=3, for instance, that you can determine the value of C, and it is important to the interpretation of the problem.

I was working through your just previous post, but I thought I could answer this one first.

grav
2007-Oct-16, 04:54 PM
I'm wondering about something, though. Should that be negative or positive? From earlier, if the antiderivative of t-2 is -t-1 for acceleration, then shouldn't A have become -At + C in our last example? Yet, the acceleration and the resulting velocity are both considered positive when acting in the same direction.I guess it would have to do with, as with the origin of times of T=0, what we would consider to be the origin of distance. So for v = (r2 - r1)/(t2 - t1), the velocity is positive when r2>r1, but negative when moving toward the origin. So, considering the center of a body as the origin, that would make the work positive when moving toward the center and negative when moving away, I suppose, or the opposite, depending on which way it is specifically defined.

Zachary
2007-Oct-16, 04:57 PM
Waaah! This forum really needs latex! :s

I've just skim read this topic when i learn the chain rule way back ;) i used an explanation sort of like this:

So you've got your function f(x), and you want to find its derivative, but f(x) is a nasty function which you can't easily do straight up, a simple example would be sin(3x).

So you change f(x) to a function of another variable, say u. Eg, sin(3x) would be sin(u) where u=3x. sin(u) is a nice function and its derivative can easily be taken.

All well and good but then you have df(u)/dx. You can't take the derivative of a function in terms of u with respect to x! You need df(u)/du. But that's simple, because df/dx = df/du * du/dx. The du bits cancel (this is bit of a fudge but it works)

Basically you replace something nasty in your original function with u. You then take the derivative of the newer simple function and then multiply that by the derivative of the nasty bit you've replaced.

e.g. sin(3x) becomes sin(u). The derivative of sin(u) is cos(u) and the derivative of 3x is 3. So the derivative of sin(3x) = 3 * cos(u) = 3cos(3x).

Ugh now ive typed it out it doesn't look so simple; you're gonna love advanced calculus!

grav
2007-Oct-16, 05:26 PM
Ugh now ive typed it out it doesn't look so simple; you're gonna love advanced calculus!:)

Thanks, Zachary. Richard (publius) demonstrated it like that a while back, and while I appreciated it tremendously, I couldn't quite retain it. All those u's and v's and du's and dv's messed with me something awful for some reason. But y = F(G(x)) and y' = F'(G(x))G'(x) seems to have stuck, along with the rest of the rules, even the quotient rule, surprisingly. I think it has something to do with presenting it in functions of F(x) and G(x), which I can relate to (and perhaps because there are no infinitesimals to concern myself with). I thought about simplifying it to just y = F(G) and y' = F'(G)G', but even that wouldn't quite cut it for me, as I would still get confused until I think specifically about how F and G relate to F(x) and G(x) as functions again anyway. I guess sometimes it just helps to have things spelled out, and hh seems to have done that in just the right way for me.

hhEb09'1
2007-Oct-16, 08:01 PM
Well, if I read you correctly, then going by what we had earlier, y = y'' dx2/2 + C dx + C', so y'' = 2(y - C dx - C')/dx2, and so for y'' + y = 0, I getAs I say in the other post, you have to be careful here. You cannot treat y as a constant, or a particular function. In differential equations, we are trying to find the function y that makes the original equation true. Similar to algebraic equations, but here the solution is a function.

We might know or assume how a basic population changes, and how the rate of change is related to certain other quantities--basically a differential equation. Solving it gives us an explicit representation of the formula for the population itself based on our assumptions.

We probably should look at some more derivative/antiderivative examples before we get into differential equations. Just to develop feel.

y'' + y = 0
y = A cos(x) + B sin(x)

You can test it by taking the second derivative of y and adding it to y. A or B (constants) could be zero. We can get into techniques for finding these functions, later.
For y'' - y = 0, then,
Clearly y = A ex works, but so does y = B e-x, as well as their sum.

grav
2007-Oct-16, 08:43 PM
Ah. After reading through your first couple of paragraphs, I realized you might want the specific function that gives y = y'' and y = -y''. For the first, I figured ex is its own derivative, and so it would also be its own second derivative. With the second, the derivative of sin(x) is cos(x), but the derivative of cos(x) is -sin(x), so the second derivative of sin(x) is -sin(x). But I see you already posted those, so... :)

grav
2007-Oct-16, 08:47 PM
Oh, yes. I see after reading through your post some more that cos(x) would also give y = -y'', since the derivative of cos(x) is -sin(x) and the derivative of -sin(x) is -cos(x), and we can also add and subtract them as well and still get y = -y'', as you have shown.

publius
2007-Oct-16, 08:49 PM
I can tell hh/kilopi/whatever has a wee bit of experience teaching Principia Mathematica, and I'll just get out of the way. However, there is one thing I want to stress to you, Grav.

The laws of physics are generally expressed as differential equations. Finding solutions to problems in physics is about finding solutions to differential equations (and those equations get a lot more complex than the basics being discussed here).

For example, Newtonian gravity is a statement about what the acceleration of a mass is due to another mass. For a point source mass,

a = -GM/r^2. That is a differential statement. Solving for the motion of objects in a gravitational field is thus an example of solving a differential equation. Now, additional dimensions are a complication we won't worry with now, but for a straight line:

d^2r/dt^2 = -k/r^2. That's the radial free fall equation, a one dimensional problem. Back in that free fall thread, that was the governing equation, and that rather complex formula was a solution of this differential equation over a given range.

-Richard

grav
2007-Oct-16, 09:42 PM
Thanks, Richard. :) That would be an interesting one to try, finding the time of descent from r1 to r2 in a gravitational field, and would be quite an accomplishment for me, not having to work it out through tables and trial and error. But since I don't have a clue about where to start with that, I guess I'm still not ready yet.

grav
2007-Oct-16, 09:48 PM
Well, okay. I guess I would have to start with r1 - r2 = at^2/2 + vot, for constant acceleration where vo = 0, and integrate over the distance for a changing acceleration. Hmm. :think:

hhEb09'1
2007-Oct-16, 10:56 PM
All those u's and v's and du's and dv's messed with me something awful for some reason. Let's see how this goes then.

Do you now recognize this as an expression of the product rule:
d(uv)/dx = u dv/dx + v du/dx

Another way to write it is
(uv)' = uv' + vu'

The laws of physics are generally expressed as differential equations. Finding solutions to problems in physics is about finding solutions to differential equations (and those equations get a lot more complex than the basics being discussed here). Was it you or Ken G that made a recent post that stressed the other way of looking at F=ma, as a differential equation?

grav
2007-Oct-16, 11:15 PM
Let's see how this goes then.

Do you now recognize this as an expression of the product rule:
d(uv)/dx = u dv/dx + v du/dx

Another way to write it is
(uv)' = uv' + vu'
Oh, yes. I can see it much more clearly now that I can relate it directly to functions. It just fits the way my brain clicks, and the way I'm used to doing things. I still have trouble relating to the infinitesimals without thinking about them in terms of summations, though, but I'm getting there, I think.

Um, looking at what you wrote above, is u+v the derivative of u*v, and 2 the second derivative?
Oh, I guess so, since if u=v, then 2v= v+v is the derivative of v2, and 2 is the second derivative, then. Just a quick exercise, there.

publius
2007-Oct-16, 11:31 PM
Was it you or Ken G that made a recent post that stressed the other way of looking at F=ma, as a differential equation?

That must've been Ken, as I don't remember saying anything about that recently.

-Richard

publius
2007-Oct-16, 11:59 PM
Thanks, Richard. :) That would be an interesting one to try, finding the time of descent from r1 to r2 in a gravitational field, and would be quite an accomplishment for me, not having to work it out through tables and trial and error. But since I don't have a clue about where to start with that, I guess I'm still not ready yet.

Well, that link I posted in your free fall time, and Grant's post just cranked out the solution to that little differential equation (and orbits are about the solution to that equation in 2 or 3 dimensions).

We've learned enough right now to get to the last step, and I'll show something, which will be more of a physicist's way of looking at the meaning of some of these integrals. Our inverse square force equation is simply this:

d^2x/dt^2 = -k/x^2. Or to consider a more general case:

ma = m d^2x/dt^2 = F(x). That is, the force on our test mass is a function of *position*. The above is just specific type of that simple Newtonian statement. Again, I stress we're seeing how the laws of physics express things as differential equations.

Let's look at this general case (in 1D) and see if anything cute might fall out.

m d^2x/dt^2 = m dv/dt= F(x)

Now, a = dv/dt and v = dx/dt, but we're considering the x and v (and a if we like) to be functions of *time*. Well, they're all related, and we have only one degree of freedom, and we can just as easily think of v as being a function of position v(x). The test particle has some v at every x. And sometimes it's easier to solve for things as functions of other variables like this rather than time.

What is dv/dt thinking of v as a function of x? Well use the chain rule:

dv/dt = dv/dx * dx/dt. But dx/dt is just v(x), so we have:

dv/dt = v*dv/dx. Let's plug that in the above:

mv *dv/dx = F(x) --> mv dv = F(x) *dx

Now, let's simply integrate both sides:

1/2 mv^2 = Integral(F(x) dx) + C.

Hmmphh. Does that 1/2 mv^2 look familiar?! :) This integral seems to say something about the *kinetic energy* of our test particle has to do with integrating a force over a distance.

Now let me show a little trick that physicists like to do with integrations like this that gives our constant there more meaning (they do this except when it's easier or makes more sense to use the C way. :lol:).

hh/kilopi was showing you the *indefinite* integral "way" of doing this. Well a definite integral is an integration between points (and this has to do with interpreting the definite integral as the area under the graph of the function -- you have a range of integration. See the Fundamental Theorem of Calculus for this in all its glory).

Now, it turns out the a definite integal is just the difference between the anti-derivative of a function between the endpoints of integration. The "C" cancels in that subtraction. So let's do the integration as a definite one, explicitly using enpoints:

We integrate mv dv between two points. Our *initial* velocity
v0, and v(x), our desired velocity function. And we integrade F*dx from x0, the initial position of the object to our position function, x. Thus:

1/2 mv^2 - 1/2 mv0^2 = V(x) - V0, where "V" is a variable I'm using to mean the antiderivative of F(x). Re-arrange, and voila!

1/2 mv^2 - V(x) = 1/2 mv0^2 -V0.

What I've done there, with my limits of integration is to split that first constant, big C into two separate constants, but that have different meaning. One is the initial kinetic energy (function of initial velocity). Now, what is the V. Well let's replace V by -V, so that F = -dV/dx. V is just our potential energy function (provided F is conservative, which are complications we won't worry with here).

The minus sign is simply way to say the force points downhill, or toward *decreasing* potential energy, rather than increasing. Now we have

T + V = T0 + V0, where T is a term for kinetic energy (T and V are the traditional symbols for this, if you're wondering). Now, the term on the right is a *constant*, the sum of the initial kinetic and potential energies. That has a name, too: total energy, symbol E.

So T + V = E.

We have arrived at the energy integral formulation:

1/2 mv^2 = E - V(x)

{EDIT: Removed erroneous expression}

Now, you've already solved for V(x) for inverse square (be careful of the minus sign, now). You can directly get the time integral for radial free fall.

-Richard

publius
2007-Oct-17, 01:59 AM
Grav,

Sorry but I botched the energy integral expression there at the last. Here's the correct expression:

(dx/dt)^2 = 2/m *(E - V(x))

dt = dx/sqrt[2/m*(E - Vx)

Integrate that from t0 to t on the left, and x0 to x on the right, and you have the relation between x and t.

Now, as you saw in the free-fall time thread, that integration can be done for the inverse square case. However, you get t(x), and it cannot be inverted in closed form to get x(t).

Now, as an exercise you can go a simple constant force acceleration via the energy integral way, and see it is a rather roundabout way to get
x = 1/2(F/m) t^2 + v0 t + x0

The energy integral will generally be harder that simple cases where you can solve more directly. But in cases like the above, it is more simple.

-Richard

hhEb09'1
2007-Oct-17, 03:11 AM
Um, looking at what you wrote above, is u+v the derivative of u*v, and 2 the second derivative?No, sorry. (uv)' = uv' + vu' can be rewritten as d(uv)/dx = u dv/dx + v du/dx

We don't know what dv/dx is at this point (du/dx neither) because the u and v are just arbitrary functions. All we can say is that the derivative of their product is that sum, each of which terms is one function times the derivative of the other.

Oh, I guess so, since if u=v, then 2v= v+v is the derivative of v2, and 2 is the second derivative, then. Just a quick exercise, there.In general, that is not true. However, if we know that v(x) = x, then the theorem (product rule) says d(x*x)/dx = x dx/dx + x dx/dx

We do know that dx/dx is 1, so d(x*x)/dx = x + x = 2x

That is, the derivative of x2 is 2x

But in general the derivative of u2 is 2u du/dx, which is an example of the chain rule. If u=x then du/dx is 1, but in general du/dx is not going to be 1.

Look over publius's discussion of the definite integral--that's where you integrate from one value of the variable to another, and the answer is just the difference between the antiderivative evaluated for the two values.

A way to imagine that is the sum (integral) of an infinite number of thin slices whose height is the function (f(x)) and whose width is the infinitesimal dx. You add those all up, and you get the antiderivative. As 01101001 says (http://www.bautforum.com/questions-answers/65642-sin-x-cos-x-x-y-2.html#post1088642), that's the Fundamental Theorem of Calculus.

grav
2007-Oct-17, 03:37 AM
No, sorry. (uv)' = uv' + vu' can be rewritten as d(uv)/dx = u dv/dx + v du/dx

We don't know what dv/dx is at this point (du/dx neither) because the u and v are just arbitrary functions. All we can say is that the derivative of their product is that sum, each of which terms is one function times the derivative of the other. :doh: See what I mean? I've already overlooked the fact that u and v represent functions, not variables. I guess I'll just have to keep writing it out the way we have been.

hhEb09'1
2007-Oct-17, 03:48 AM
:doh: See what I mean? I've already overlooked the fact that u and v represent functions, not variables. I guess I'll just have to keep writing it out the way we have been.F, G, f, g, u, v, traditionally represent functions. But it can get weird, for instance F(t) = (x(t), y(t), z(t)) is a function of t, but its range is a set of points with x, y, and z coordinates, each of which is a function of t. Persevere, I'm going to use u and v some more.

(uv)' = u v' + v u' is the product rule right? So, we can rearrange this:

u v' = (uv)' - v u'

If we take the integral of both sides, we get an equality. The integral of (uv)' is just uv. The goal is to somehow rewrite the parts of our integral (the left hand side) so that it fits into the pattern of the right hand side, and is easier to integrate. This technique is called, integration by parts. Interested in seeing an example?

hhEb09'1
2007-Oct-17, 04:17 AM
u v' = (uv)' - v u'

In trying find the integral of f(x) = x ex, we run into a problem. We could guess the answer, but I'm going to apply the integration by parts.

Let u = x, and v' = ex
We have chosen u so that when we take the derivative on the right (top line of this post), it simplifies the expression. And we have chosen v' because when we take the antiderivative (v) on the right, it doesn't make the expression more complicated. Net effect, a simpler expression to take the antiderivative of.

So, v = 1, and u' = ex
The integral of (uv)' is uv, remember, so we get the antiderivative of
x ex
is
x ex - the antiderivative of (1 ex)
or
x ex - ex

You can check the answer by taking its derivative.

PS: u v' = (uv)' - v u'

∫u v' dx = ∫((uv)' - v u') dx

∫u v' dx = ∫(uv)' dx - ∫v u' dx

∫u v' dx = uv - ∫v u' dx

grav
2007-Oct-17, 11:17 PM
Wow. Thanks, hh! I think that'll just about get it right there. I can probably find the derivative or antiderivative of almost about anything with that trick in the bag, I believe. No more trial and error and reading into the numbers to find the result. Let me try one or two.

The integral of x2sin(x) is

u = x2, v' = sin(x)
uv' = (uv)' - u'v
I uv' = uv - I u'v
I x2sin(x) = x2(-cos(x)) - I 2x(-cos(x))
I x2sin(x) = -x2cos(x) + 2 I x cos(x)

But then I've got to find the integral of x cos(x), which is

u = x, v' = cos(x)
I uv' = uv - I u'v
I x cos(x) = x sin(x) - I sin(x)
I x cos(x) = x sin(x) - (-cos(x))
I x cos(x) = x sin(x) + cos(x)

So I x2sin(x) = -x2cos(x) + 2 I x cos(x)
I x2sin(x) = -x2cos(x) + 2x sin(x) + 2cos(x)

Now I'll try sin(x)2.

u = sin(x), v' = sin(x)
I uv' = uv - I u'v
I sin(x)2 = sin(x)(-cos(x)) - I cos(x)(-cos(x))
I sin(x)2 = -sin(x)cos(x) + I cos(x)2
2 I sin(x)2 = -sin(x)cos(x) + I cos(x)2 + I sin(x)2
2 I sin(x)2 = -sin(x)cos(x) + I (cos(x)2 + sin(x)2)
2 I sin(x)2 = -sin(x)cos(x) + I (1)
2 I sin(x)2 = -sin(x)cos(x) + x
I sin(x)2 = [x - sin(x)cos(x)]/2

:) :)

grav
2007-Oct-18, 03:05 AM
So let me try my hand at that time of descent thing, for the time for an object to fall from X to x in a gravitational field. I will integrate for the times of dt over infinitesimal distances of dx where the instantaneous velocity is v(x). So dt = dx/v(x).

The instantaneous velocity is

mv(x)2/2 = GMm(1/x - 1/X)
v(x) = sqrt[2GM(X-x)/Xx]
z=x/X
v(x) = sqrt[2GM(1-z)/Xz]

The integral for this, then, is

dt = dx sqrt[Xz/(2GM(1-z))]
dt = (dz*X) sqrt[Xz/(2GM(1-z))]
t = I dz sqrt[X3z/(2GM(1-z))]
t = sqrt[X3/(2GM)] I dz sqrt[z/(1-z)]

uv' = uv - u'v
u = dz, v' = sqrt[z/(1-z)]
I dz sqrt[z/(1-z)] = dz I sqrt[z/(1-z)] - dz' I sqrt[z/(1-z)]

Now here, I figured I dz is z, so z' = dz = 1 and z'' = dz' = 0. I'm not exactly sure about doing it that way, but continuing along...

I dz sqrt[z/(1-z)] = I sqrt[z/(1-z)]

and here I worked through the infinitesimals, probably more than I needed to, but anyway...

dy/dz = sqrt[z/(1-z)]

z = a2, so dz/da = 2a
dy/dz = sqrt[a2/(1-a2] = a/sqrt(1-a2)

a = sin(b), so da/db = cos(b)
dy/dz = sin(b)/cos(b)

(dy/dz)(dz/da) = dy/da = (sin(b)/cos(b))(2a)
= 2 sin(b)2/cos(b)

(dy/da)(da/db) = dy/db = (2 sin(b)2/cos(b))(cos(b))
= 2 sin(b)2

I 2 sin(b)2 = b - sin(b)cos(b)

I sqrt[z/(1-z)] = b - sin(b)cos(b)
= asin(a) - sin(asin(a))cos(asin(a))
= asin(a) - a sqrt(1 - a2)
= asin(sqrt(z)) - sqrt(z) sqrt(1 - z)
= asin(sqrt(x/X)) - sqrt[(x/X)(1-x/X)]

Now that would be integrated for dx from zero at the center of mass to x, so for the time falling from x to 0, with a starting velocity of v=0 at X. To find it all the way from X to 0, then, that would just be where z=1, or

I sqrt[z/(1-z)] = asin(sqrt(1)) - sqrt[(1)(1-1)] = asin(1) = pi/2

So to find the time to fall from X to x, then, we just take the difference in times, which finally becomes

t = sqrt[X3/(2GM)](pi/2) - sqrt[X3/(2GM)](asin(sqrt(x/X)) - sqrt[(x/X)(1-x/X)])
t = sqrt[X3/(2GM)][pi/2 - asin(sqrt(x/X)) + sqrt[(x/X)(1-x/X)] ]

hhEb09'1
2007-Oct-18, 10:42 AM
Wow. Thanks, hh! I think that'll just about get it right there. I can probably find the derivative or antiderivative of almost about anything with that trick in the bag, I believe.That covers a lot of them, but there are "impossible" ones. For instance the antiderivative of f(x) = e-x^2, involved in the gaussian distribution. Integrating the gaussian distribution from negative infinity to a gives you erf(a), the error function, the probability that a random choice of number will be less than a. But erf is approximated, tabulated, and digitized. :)

I x2sin(x) = -x2cos(x) + 2x sin(x) + 2cos(x)Plus a constant. :) -cos2x is the same as sin2x, except for a constant

As you've noticed, sometimes you have to use the techniques more than once.
I sin(x)2 = [x - sin(x)cos(x)]/2Cute. I knew I'd pick up some tricks from your attempts :)

Try this approach to the same problem: sin2x = (1 - cos 2x)/2 (the sine squared function is just an inverted cosine function with half the amplitude, half the period, and shifted up a half.)

Which reminds me, have you played with de Moivre's formula (http://en.wikipedia.org/wiki/De_Moivre's_formula)?

grav
2007-Oct-19, 01:21 AM
That covers a lot of them, but there are "impossible" ones. For instance the antiderivative of f(x) = e-x^2, involved in the gaussian distribution. Integrating the gaussian distribution from negative infinity to a gives you erf(a), the error function, the probability that a random choice of number will be less than a. But erf is approximated, tabulated, and digitized. :)So you're saying there's no antiderivative of e-x^2 using ordinary trig functions and e and such. Since those are also infinite series functions, it seems like there should be some way... Hmm...

Cute. I knew I'd pick up some tricks from your attempts :)Did you? :) Thanks. I had found that I sin2x gave I cos2x and I cos2 just gave I sin2x back in return, so I had to get a little innovative. Sin2x + cos2x = 1 is about the only trig function I know well, though, but it comes in handy quite often, I've noticed, so I work with it a lot.

Try this approach to the same problem: sin2x = (1 - cos 2x)/2 (the sine squared function is just an inverted cosine function with half the amplitude, half the period, and shifted up a half.)I haven't done this yet. I've been working on that e-x^2 thing. I'll do that now.

Which reminds me, have you played with de Moivre's formula (http://en.wikipedia.org/wiki/De_Moivre's_formula)?Never heard of it. I rarely work with complex numbers, but I've started reading through it.

hhEb09'1
2007-Oct-19, 02:28 AM
Never heard of it. I rarely work with complex numbers, but I've started reading through it.As you can see from that link, the formula is
(cos x + i sin x)n = cos nx + i sin nx
Euler's formula says
einx = cos nx + i sin nx
So for n=1 and x=π, we get eiπ = -1, or the famous
eiπ + 1 = 0
which combines probably the five most important constants in mathematics.

But, why I mentioned it, if n=2 then we get the result
cos 2x + i sin 2x = (cos x + i sin x)2
Expand the right hand side--since the real and imaginary parts have to be equal, it will result in a formula for cos 2x and sin 2x, the trigonometric double angle formulas.

How's that e-x^2 working out? :)

grav
2007-Oct-19, 02:41 AM
Try this approach to the same problem: sin2x = (1 - cos 2x)/2 (the sine squared function is just an inverted cosine function with half the amplitude, half the period, and shifted up a half.) Oh, okay. I was still trying to apply it with the reverse engineered product rule, the same as before, but I found I still had to use sin2x + cos2x = 1 in one way or another each time. Then I noticed the simplicity of what you really wanted me to do, so...

sin2x = (1 - cos 2x)/2
I sin2x = I (1 - cos 2x)/2
= I 1/2 - I (cos 2x)/2
= x/2 - I (sin 2x)/4
= (x - (sin 2x)/2 )/2
= (x - (sin x) (cos x) )/2

grav
2007-Oct-19, 03:10 AM
As you can see from that link, the formula is
(cos x + i sin x)n = cos nx + i sin nx
Euler's formula says
einx = cos nx + i sin nx
So for n=1 and x=π, we get eiπ = -1, or the famous
eiπ + 1 = 0
which combines probably the five most important constants in mathematics.That is intriguing. I would like to into that further at some point.

But, why I mentioned it, if n=2 then we get the result
cos 2x + i sin 2x = (cos x + i sin x)2
Expand the right hand side--since the real and imaginary parts have to be equal, it will result in a formula for cos 2x and sin 2x, the trigonometric double angle formulas.Well, let's see.

cos 2x + i sin 2x = (cos x + i sin x)2
cos 2x + i sin 2x = cos2x + 2i(sin x)(cos x) - sin2x

So for the imaginary part,

i sin 2x = 2i(sin x)(cos x)
sin 2x = 2(sin x)(cos x)

and for the real,

cos 2x = cos2x - sin2x
cos 2x = 1 - 2 sin2x

So it's kind of like separating the egg from the yolk? How convenient. :) Complex numbers might be very useful in this manner, then.

How's that e-x^2 working out? :)Hard to tell. :)
It's an all or nothing kind of thing. Maybe I'll try something like what you showed with the complex numbers for it.

hhEb09'1
2007-Oct-19, 02:41 PM
Hard to tell. :)
It's an all or nothing kind of thing. Maybe I'll try something like what you showed with the complex numbers for it.When you are satisfied, here is the wiki page on the Gaussian integral (http://en.wikipedia.org/wiki/Gaussian_integral), which isn't the antiderivative, but it shows how much work is involved in just computing the integral of the function from minus infinity to positive infinity. By dividing the function by that result, the area is one--and that makes it a probability distribution, the Gaussian distribution, or normal distribution.

By integrating that from 0 (adjusted by the constant (wiki: Error function (http://en.wikipedia.org/wiki/Error_function)), gives you the probability that a random number is between plus or minus some value. Used all the time in statistical tests. :)

grav
2007-Oct-20, 04:47 AM
How's that e-x^2 working out? :)I think I might be onto something. I've got precise solutions for x=1 and x=2 with summations so far using just e all the way. But the summations can differ slightly from the integrations. For instance, the summation of x becomes x(x+1)/2, but reduces to just x^2/2 for the integration. Can you give me the precise solution to as many digits as possible (twenty is plenty) for the integration of e-x^2 for x=1 and x=2 using that erf thing?

hhEb09'1
2007-Oct-20, 05:02 AM
Can you give me the precise solution to as many digits as possible (twenty is plenty) for the integration of 1/e[sup]-x^2[sup] for x=1 and x=2 using that erf thing?Did you mean to put that in the reciprocal, and include the minus sign as well?

grav
2007-Oct-20, 06:55 AM
Did you mean to put that in the reciprocal, and include the minus sign as well?No. Thanks. I've corrected it.

grav
2007-Oct-20, 07:28 AM
Well, so far I've been able to determine that

Sum [e-x^2] = (e-(n^2 + n) - 1)/(e-n - e) with x = 1 to n for n = 1 and n = 2. I'm still working on n>2. ;)

But I've also determined that

Sum [e-(x^2+x)] = e-(n^2 + n) Sum [e(2n + 2 - x)(x - 1)] with x = 1 to n for all n.

Kaptain K
2007-Oct-20, 11:11 AM
Welcome to the Bad Arithmetic and Mathematics Today forum! :whistle:

hhEb09'1
2007-Oct-20, 02:06 PM
Can you give me the precise solution to as many digits as possible (twenty is plenty) This wiki page (http://mathworld.wolfram.com/Erf.html) has some series representations of the error function.

The Microsoft Excel spreadsheet function ERF(x) returns the error function of x, ERF(a,b) just gives ERF(b) - ERF(a)

Unix and Linux have the erf function too, in the math libraries

grav
2007-Oct-24, 01:01 AM
Okay, I give. That last post I made didn't amount to much, once I realized that summations for x = 1 to n using integers only works for polynomials with exponents in the form xn, since when I find the derivatives, I can just take x/dx = n. In other words, it won't work for en except where dx=1 and n is very large, but then I can only find for z, where z = sqrt(pi)/2 (1 - erf(x)) anyway. So it's the same difference.

I figured the rules of calculus had been tried for this in every way as well, so I tried to build things from the ground up using summations, but then, as I started practically reformulating the same rules this way, I figured using the rules themselves in the first place was a much better way to go. But I applied them in every way I could think of, forwards, backwards, and upside down, and as I figured would probably be the case, it was to no avail.

I even tried de Moivre's and Euler's formulas, but again, to no avail. I still just kept getting erf(x) and erfi(x) which added together in different ways to produce the same result. Oh, well. I guess I'll just have to get used to erf(x) along with sine and cosine and all the other functions I hadn't realized how many there were until I read through this (http://algo.inria.fr/esf/index.html) and other links. All in all, though, I still learned a few things in the process and it was good practice.

hhEb09'1
2007-Oct-24, 09:09 AM
I even tried de Moivre's and Euler's formulas, but again, to no avail. I still just kept getting erf(x) and erfi(x) which added together in different ways to produce the same result.
You're not the first! but that does sound like you worked through a lot of background stuff.
Oh, well. I guess I'll just have to get used to erf(x) along with sine and cosine and all the other functions I hadn't realized how many there were until I read through this (http://algo.inria.fr/esf/index.html) and other links.Nice list. I am surprised that the gamma function didn't make it. There are many others. :)

All in all, though, I still learned a few things in the process and it was good practice.Yes, and I'm impressed. I still maintain, though, that you might make more progress if you'd give the older mathematicians some credit for having worked out some of the kinks :)

I am not saying, trust them completely :)

grav
2007-Oct-24, 11:05 PM
Nice list. I am surprised that the gamma function didn't make it.Me too, now that you mention it, especially considering they used it to explain a lot of the functions that are listed. That is that little inverted 'L' thingy, right?

There are many others. :)Hooray.:rolleyes: :)

I still maintain, though, that you might make more progress if you'd give the older mathematicians some credit for having worked out some of the kinks :)Oh, of course. It's amazing what has been accomplished and I have new respect for it every day, and I thank you for sharing it with me. I do tend to get a little side-tracked from time to time when I get something in my head or don't quite comprehend something, but by the time it's all said and done, even if I don't find anything any different, I have still gained a better understanding in the process.

hhEb09'1
2007-Oct-25, 03:20 AM
That is that little inverted 'L' thingy, right?Yes, the Greek capital gamma :)
by the time it's all said and done, even if I don't find anything any different, I have still gained a better understanding in the process.Absolutely, and I love that. The only thing that irks me (a tiny bit) is your readiness to announce that you've discovered something new that overturns three hundred years of "conventional" wisdom. I know it's possible, I've done it, but you can't do it by ignoring what has been done in the field--calculus, for instance.

You don't have to master the entire field to find errors, it's true. Follow your hunches (as I know you will), and use them to test the current results--find the weakness in your theory, what would it mean to your theory if what everyone else is saying is true and see where that goes. It's an amazingly productive way to procede.

grav
2007-Oct-25, 01:28 PM
Absolutely, and I love that. The only thing that irks me (a tiny bit) is your readiness to announce that you've discovered something new that overturns three hundred years of "conventional" wisdom. I know it's possible, I've done it, but you can't do it by ignoring what has been done in the field--calculus, for instance.Well yes, I know. Sorry. I sometimes get impatient and jump the gun a little. I really know in the back of my head that pretty much anything new and exciting I come up with is only new and exciting to me. It's probably been known for hundreds of years, experimented with in countless ways, and applied in areas I can't even have imagined. I want to stand on the shoulders of giants, that being my ultimate goal, so I can get a clear view of the big picture, but usually I can just barely stand up on my tip-toes long enough to peek over their shoulders and catch a glimpse. Whenever I come up with something, I know a lot of people on here are probably thinking something like "Yeah, okay. We already know this, so what's your point?" But others might not know about it so it might be new and interesting for them as well, and maybe even helpful.

Like when I thought I had come up with a way to do iterations earlier. My misunderstanding of Ken's post was due to my sheer ignorance of calculus at the time. But Tony also posted something that I did understand, and it was similar to how I was trying to find the final position of a falling body at first, by running it in very small intervals of time. If I wanted a more exact position, I had to run it again in smaller intervals, taking an immense amount of time. But after Richard linked to the formula for the time of descent and Grant worked it out in a way I could understand, I realized there was a much better and faster way. But I was still careful to write that I may have come up with something that is not at least well known. I know that Tony writes programs for N body systems using such time intervals, so I thought this might be very useful to him. But then I realized that it only works for two bodies in one dimension, or in two dimensions using two equations, like that one and the one for an ellipse, so it probably couldn't be used the way Tony would need it to anyway.

By the way, what did you discover? :)

grav
2007-Oct-26, 01:49 AM
Just so you'll know, when I said in the last post that I sometimes get impatient and jump the gun a little, I meant impatient with myself, not others. I just get a little ahead of myself sometimes is what I meant. But anyway, that's just me. :)

Getting back to the calculus, though, I've noticed that if one integrates from zero to r for a symmetrical shape, where r is the distance from the center to a point perpendicular (and centered) to each of the sides, the shape can be easily integrated from the perimeter to the area or area to volume. For instance, if one knows the circumference of a circle, then one can integrate from 2pir to 2pi(r^2/2) = pir^2 to find the area or integrate the surface area, 4pir^2 to 4pi(r^3/3) = (4pi/3)r^3 for the volume. For the perimeter of a square, which is 8r, where r is the distance from the center of the shape to the center and perpendicular to one of the sides, the integration is 8(r^2/2) = 4r^2 = L^2, where L is the length of a side. For a cube, it is 6*4r^2 = 24r^2 for the surface area, so 24(r^3/3) = 8r^3 = L^3 for the volume. For an equilateral triangle, if r=1, then the perimeter is 6*sqrt(3)r, so the area is 6*sqrt(3)(r^2/2) = 3*sqrt(3)r^2. That is something else I find new and interesting, but only to me, though, :) and anyone else that might not have thought about it in quite that way. But to most, it's probably just basic stuff. I'm sure you'll have much to add. I'm wondering how it would work out if the center of the sides aren't perpendicular to the center of the shape.

There's something else I want to do, though, too. Quite a while back I tried to find for the acceleration of gravity along the edge and a little bit away from the edge of an infinitesimally thin disk, but I kept coming up with infinities I couldn't resolve. I think I have a way to do that with the integration, by finding from -r to r (or from D-r to D+r) with arcs cutting through the disk at regular intervals of dr from the point at D, so finding for the inverse of the square of the same distance across each arc. Do you think that'll work out okay? I'll get to work on it but I'll definitely need you to check my result when I'm done.

publius
2007-Oct-26, 03:04 AM
Grav,

I'm impressed. You've got a good mathematical head on your shoulders, it's just needs to be disciplined. That what a formal education in this stuff does, but it can be done other ways, although it's harder.

You're not ready to tackle a (Newtonian) gravitational field integration yet. You've learned some of the principles, much there's more there. What you want to do is a volume integral for the potential, then take the gradient of that to get the field. But it turns out that integral, for a disc is very difficult. :)

You've discovered how to exploit symmetry to find areas and volumes, using a single variable of integration. There are general ways to find surface areas and volumes, and that involves multiple integrals -- integrating over several variables.

But, one of the first things you do before you get to the more general methods, is do things just like you discovered. And that is solids of revolution.

Imagine a curve, some y(x). Now rotate that curve around the x-axis and sweep out a solid. You can find both the surface area and the volume of that.

Another thing is "improper integrals". This is a definite integral where one or both of the limits is infinity. You integrate something from zero to infinity say, or from -infinty to +infinity. That might blow up, but in many interesting cases that have some physical meaning, it's very useful.

Anyway, take the y(x) = 1/x and make a solid of revolution out of it. It's a long funnel shape with a neck that tapers down to nothing, but is infinitely long. What is the surface area? Now, what is the volume?

Do that result make sense?

-Richard

grav
2007-Oct-26, 04:44 AM
Anyway, take the y(x) = 1/x and make a solid of revolution out of it. It's a long funnel shape with a neck that tapers down to nothing, but is infinitely long. What is the surface area? Now, what is the volume?Okay. I can find the area by taking the integration of little rectangles of height dx and width y(x), so that becomes A = I dx y(x) = I dx (1/x) = ln(x). The volume would just be little cylinders stacked up on top of each other with a height of dx and a radius of y(x), so that would be V = I dx pi y(x)^2 = I dx pi (1/x^2) = pi (sqrt(pi)/2) erf(x). [EDIT - oops.]

Do that result make sense?Not really. For one, neither of these includes x or x^2 in the result for the units of length. So what do I do, just tack on meters and meters squared or whatever unit of measure I am using? Secondly, the integration of 1/x is ln(x), but it doesn't quite fall in line with the n*x^(n-1) for the derivative rule, whereas x^0 gives 0*(1/x). I still don't get that. It's like I would have to take the derivative of x^0/0 = infinity to get it. But that might also explain some of my problems with infinities right there.

grav
2007-Oct-26, 05:02 AM
Not really. For one, neither of these includes x or x^2 in the result for the units of length. So what do I do, just tack on meters and meters squared or whatever unit of measure I am using?To answer that first one, I'm thinking the units should still be identifiable, so the only way I can see that working out for y(x) = z/x, where z is a constant, is if that were really something like y(x) meters = z meters^2 / x meters. Then for the integrations, we get A = I dx (z meters^2 / x meters) = I dx z meters (1/x) = ln(x) z meters, where z = 1 here, so A = ln(x) meters, and similarly for the volume. Does that sound right to you?

EDIT - Oh, I guess that would just be saying that the area that lies within the boundaries between y=0 and y(x) and x=0 and x=x at any point along the curve is a constant, such that y(x) * x = z, right? Pretty simple, really, now that I think about it.

grav
2007-Oct-26, 05:19 AM
Secondly, the integration of 1/x is ln(x), but it doesn't quite fall in line with the n*x^(n-1) for the derivative rule, whereas x^0 gives 0*(1/x). I still don't get that. It's like I would have to take the derivative of x^0/0 = infinity to get it. But that might also explain some of my problems with infinities right there.But what about that second deal there? The integration of 1/x should be x^0/0 = infinity. There's also a constant tacked onto that, but it would have to be something like C = ln(x) - infinity to give just ln(x) all in all, wouldn't it? So how does that work out, and how should I think about it?

grav
2007-Oct-26, 05:29 AM
Um, yeah. Oops. I just noticed I used the integration of 1/ex^2 instead of 1/x^2 for the volume. I must have that stuck in my head from working with it before. So that would be V = I dx pi (1/x^2) = - pi/x. Okay now, that's a little messed up there. :) There must be a constant tacked onto that as well or something to make it positive again, right?

publius
2007-Oct-26, 05:30 AM
Grav,

Well, for the surface area integral, note we have a solid of *revolution*. The cross section is circular. The circumference of a circle is 2pi*r. So the circumference of each circle at y = 1/x is 2pi*y = 2pi/x.

We integrate that from 1 to some x = a. The integral of that is
2pi*(ln(a) - ln(1)). That diverges as a --> infinity. Thus, the surface area of that solid is infinite. Actually, that's not the true surface area formula. :) To get the actual surface area, we need to know the arc length, which is another formula, but the arc length is *greater* than dx (the arc length of a curve is actually sqrt(1 + y'^2)dx). So we know the result is "greater than infinity", which means it's infinite. :)

Now, the volume is the integral of pi*r^2 = pi/x^2. Now, the integral of 1/x^2 is not the error function. That is the integral of e^(-x^2). What is the integral of 1/x^2? Well, we know power rule, d(x^n)/dx = n*x^(n-1). That works for all n save n = 0, where the ln comes in.

So the integral of 1/x^2 is simply -1/x. Thus the volume of the solid for x = 1 to x = a is

pi*(1/1 - 1/a). As a --> infinity, that becomes simply pi.

The surface area is infinite, but the volume is finite! So, how much paint would it take to paint the surface, an infinite or finite amount?

That surface is known as "Gabriel's Horn", and this is a famous "paradox".

-Richard

publius
2007-Oct-26, 05:34 AM
Grav,

That's good that you're worried about units. That's the way to think. Physicists are somewhat interested in this dimesional analysis. :lol:

The answer about units is simply in the definition of the function itself,
y = 1/x. If x has units of length, the '1' there must actually be a constant of dimensions of area to make y have units of length. Or, you can think of x and y are simply ratios to some unit length, and the result scales according to whatever your unit of length is.

-Richard

grav
2007-Oct-26, 06:41 AM
Grav,

Well, for the surface area integral, note we have a solid of *revolution*. The cross section is circular. The circumference of a circle is 2pi*r. So the circumference of each circle at y = 1/x is 2pi*y = 2pi/x.Oh, whoops. I did it again. Ouch. It must be getting late for me or something. I forgot you wanted the surface area and I apparently took the cross-sectional area, and only half of it at that. Geez. But yes, okay, the integration would be that of all of the circumferences stacked one on another. I'll get my head on straight eventually.

We integrate that from 1 to some x = a. The integral of that is
2pi*(ln(a) - ln(1)). That diverges as a --> infinity. Thus, the surface area of that solid is infinite. Actually, that's not the true surface area formula. :) To get the actual surface area, we need to know the arc length, which is another formula, but the arc length is *greater* than dx (the arc length of a curve is actually sqrt(1 + y'^2)dx). So we know the result is "greater than infinity", which means it's infinite. :) Now that's confusing. Since ln(1) = 0, then 2pi*ln(a) is the formula, right? But then, that starts at x=1. For x<0 we get a negative result, all the way up to -infinity, but the result for x=0 should just be zero surface area since there is no volume or surface then. So the constant should be positive infinity, but at ln(a) where a=infinity, the surface area is infinity plus infinity or something like that. Oh, good googly.

Now, the volume is the integral of pi*r^2 = pi/x^2. Now, the integral of 1/x^2 is not the error function. That is the integral of e^(-x^2). What is the integral of 1/x^2? Well, we know power rule, d(x^n)/dx = n*x^(n-1). That works for all n save n = 0, where the ln comes in.

So the integral of 1/x^2 is simply -1/x. Thus the volume of the solid for x = 1 to x = a is

pi*(1/1 - 1/a). As a --> infinity, that becomes simply pi.I was thinking that the constant there might be pi, but then I thought about when a is small, approaching zero, that would give -infinity. Does it change depending upon the lower limit that is applied? Is that what you mean about the arc length?

The surface area is infinite, but the volume is finite! So, how much paint would it take to paint the surface, an infinite or finite amount?

That surface is known as "Gabriel's Horn", and this is a famous "paradox".That is very interesting, but I'm still not really sure how to think about all of this yet. It's a little rough, so I'll have to try to work through it some more.

grav
2007-Oct-26, 07:24 AM
Aha. Minus those couple of silly mistakes with the integrations I made :rolleyes:, I've got it now, I think. I was thinking that the integrations themselves gave the solution for the area and volume, integrating from zero to x each time, so I was thrown by the -pi/x for the volume. But regardless of what the integrations give, that is only the first step, it seems. One must still subtract that for the integration of x=a from that of x=0. That also gives the added convenience of eliminating any additional constants to contend with. I guess I knew that, though, but forgot. So for x=0 to x=a, we get 2pi(ln(a) - ln(0)) for the surface area, giving 2pi(ln(a) - (-infinity)) = infinity + 2pi ln(a) for that, and -pi/a - (-pi/0) = infinity - pi/a for the volume. For x=1 to x=a, then, it is 2pi(ln(a) - ln(1)) = 2pi ln(a) for the surface area, and -pi/a - (-pi/1) = pi(1-1/a) for the volume. So now I understand the minus sign for fractions, since when a is large it actually makes the fraction smaller, so that the subtraction of a smaller fraction makes the result of the formula positive. And finally, for x=1 to x=a where a=infinity, we get 2pi ln(a) = infinity for the surface area and pi(1 - 1/a) = pi for the volume. In other words, what you said. That is some pretty wild stuff. But it looks like that would be all of it to me, then, the full formulas. So what did you mean by saying we needed to find the arc length?

grav
2007-Oct-26, 03:28 PM
Well, it looks like I somehow managed to louse up two formulas at the same time last night, and right after you said I had a good mathematical head on my shoulders too. How embarassing. :doh: I guess I showed you, huh? :) I'm like a gremlin. You can't feed me calculus after midnight or I'll start running around messing things up. :) (Although I know it was earlier when you posted that.) So does that mean I'm still not ready for that integration I was going to try? :sad:

Seriously though, as far as that surface area and volume is concerned, I've been trying to imagine it, and I can see the line for the curve becoming a straight line parallel to the y axis, so that the curve itself is infinitely long, and I can kind of see the surface area becoming infinitely large with it also, although it looks like that could go either way really, and I can see the volume getting smaller and smaller and converging, making it finite. That does blow my mind, though, that something with a finite volume can have an infinite surface area. So somewhere in the middle, then, there is a formula for the integration of x^n that defines the barrier between the finite and the infinite, between converging and diverging. I think I've asked that before. Do you know what the value of n would have to be for that? Obviously, somewhere between -1 and -2 anyway.

grav
2007-Oct-26, 04:13 PM
So somewhere in the middle, then, there is a formula for the integration of x^n that defines the barrier between the finite and the infinite, between converging and diverging. I think I've asked that before. Do you know what the value of n would have to be for that? Obviously, somewhere between -1 and -2 anyway.Okay, I guess I answered my own question again. It looks like x^n diverges right at n=-1, since if we take the integration of x^(-1+z), where z is a constant, we get an integration of (x^z)/z, so it blows up where z is infinitely small, or right at the integration of x^(-1). I can understand how this infinite result translates into ln(x) too now, since if we take an infinitely small z, say dz, then the result of (x^dz)/dz can also be stated as 1/dz + ln(x). So the infinite part is the constant. Taking the difference between two points of integration, then, say x and x', using the same infinitely small dz, we get (1/dz + ln(x')) - (1/dz + ln(x)) = ln(x') - ln(x), so the integration of x^(-1) can just be considered ln(x), once the constants cancel.

hhEb09'1
2007-Oct-26, 08:50 PM
Just so you'll know, when I said in the last post that I sometimes get impatient and jump the gun a little, I meant impatient with myself, not others. I just get a little ahead of myself sometimes is what I meant. But anyway, that's just me. :) We're cool there :)
That is something else I find new and interesting, but only to me, though, :) and anyone else that might not have thought about it in quite that way. But to most, it's probably just basic stuff. I'm sure you'll have much to add. I'm wondering how it would work out if the center of the sides aren't perpendicular to the center of the shape.As near as I can tell, understanding that principle is the key to being able to apply that part of calculus to real world problems.

I found that question interesting, so I decided to apply it to a right triangle whose angle with the x axis is fixed at theta θ. So, as the width x increases, the triangle grows by dx, but the increment in area is the length of the hypotenuse H times the width of the new sliver, the sine of θ times dx. So we should have the area of the triangle as
∫ H (sin θ dx)
But H sin θ is just y! (excitement, not factorial), y is the height of the triangle.
∫ y dx
I was a little surprised to see that, until I realized that that was just Cavalieri's Principle (http://www.jimloy.com/cindy/cavalier.htm)
So, to take the integration with respect to x, we have to convert y into a function of x, and y is just x times the tangent of θ
∫ x tan θ dx
Since θ is a constant, so is tan θ, and the antiderivative is just
1/2 x2 tan θ
But x tan θ is y, so we can convert back, and the area of the triangle is
1/2 xy
QED :)

That surface is known as "Gabriel's Horn", and this is a famous "paradox". And like most "paradoxes", it's not really a paradox. It would seem like you should be able to paint the surface with a finite amount of paint, if you could fill it with a finite amount of paint. However, most of the area of the surface (infinite) is out past that finite point where it has necked down to where a paint molecule could no longer fit inside. In other words, in order to "paint" the surface, you'd have to use infinitely thin coats of paint. So, it's just like painting an infinitely large ball with a finite amount of paint: paint a ball with a gallon of paint, and let the thickness of the paint decrease as the inverse of the radius squared, as the ball expands--you'd be able to coat any size ball with that gallon of paint. Same thing with Gabriel's Horn.

grav
2007-Oct-26, 10:50 PM
That's interesting, even though I'm not sure if I understood everything you did there, and so is that Cavalieri's principle you turned up. It looks like the same thing could be applied to a square or rectangle to produce a rhombus-like rhomboid with the same area. So I guess Gabriel's paradox could be applied to that as well, then. If we took, say a cube, and stretched the top surface all the way to the right of our view until it reaches infinity, then the left and right surfaces will now contain an infinite area each, but the top, bottom, back, and front will remain the same areas as before. Also, the volume will remain the same. So we would maintain the same finite volume while stretching the left and right surfaces to infinite area.

All in all, it would be sort of like placing two infinitely long sheets of metal on top of each other. The angle between the sheets is zero, so there is no volume between them, but infinite surface area. If we raise the end of one of the sheets to some height, however, the angle at the other end remains zero at infinity, but there is now some finite volume between them, and it would seem that the angle at the end at infinity should be something other than zero, then, but still infinitesimal. I think this might be similar to the arguments concerning Euler's fifth postulate for parallel lines, whereas, even with zero angle between them, they could still potentially diverge or converge at infinity.

publius
2007-Oct-26, 11:28 PM
hh,

Are you familiar with a device for measuring the area of arbitrary plane curve known as a "planimeter". When I was a teenager, I noticed one of those being used by government agencies for tax and other purposes. They would use aerial photographs, and if they needed to know the area of a plot of land, they would just use that cute device to trace around the circumference.

How did that work, I wondered. Well, after learning vector calculus, I figured it out, on my own, and was always quite proud of myself for doing it. I wrote a little program in QuickBASIC to take surveyor style plat information (length and compass bearing for the series of straight line segments they would use to bound a given plot) and calculate the area.

I called that the "planimeter integral", because that's exactly what the planimeter did as you moved it along. I was proud of myself, but that particular wheel had been invented back in the 1800s by a Swiss mathematician. That was before the modern vector calculus form of things, and I had benefit of that to make the lightbulb go off.

I'll go through it if anyone is interested. The trick is, knowing Stoke's Theorem, to find a function whose curl is always unity in the plane......... The result takes a rather cute form.

-Richard

hhEb09'1
2007-Oct-27, 05:21 AM
so is that Cavalieri's principle you turned up. It looks like the same thing could be applied to a square or rectangle to produce a rhombus-like rhomboid with the same area. Cavalieri (http://en.wikipedia.org/wiki/Bonaventura_Cavalieri) lived from 1598 to 1647, and his principle is a precursor to the calculus. It says that any objects of similar shape with the same size base and height have the same area/volume. So it not only applies to triangles and rhombuses but also pyramids, cylinders, and prisms.
hh,

Are you familiar with a device for measuring the area of arbitrary plane curve known as a "planimeter". I warr an ingg-in-ear. Spent many an hour hunched over the drafting table.

grav
2007-Oct-27, 12:17 PM
They would use aerial photographs, and if they needed to know the area of a plot of land, they would just use that cute device to trace around the circumference.That makes me think that if they traced the perimeter of any area, and then shade it in, they can then feed it into a computer to run lines across the shaded parts in regular intervals, then add up the length of all of the lines and multiply that by the interval to get the area.

(length and compass bearing for the series of straight line segments they would use to bound a given plot) and calculate the area. With straight line segments along the perimeter, they could divide they area up into triangles by running more straight lines across the area from point to point where the lines meet, then knowing the length of each straight line, they would know the lengths of the sides of the triangles, and can then find the areas for each of them, and finally add them together.

I'll go through it if anyone is interested. The trick is, knowing Stoke's Theorem, to find a function whose curl is always unity in the plane......... The result takes a rather cute form.I'm not sure what you mean here. Do you mean that you can find the area under a curve? That would be very useful. I used something once to find the circumference and area of a circle, by cutting it in half, and then running a triangle from each end of the straight line to the center of the arc on either side, then running triangles for each of the remaining four edges, then for each of the remaining eight edges, and so on. The result was an infinite series that found pi by adding the areas of the triangles, and an iteration that found the circumference. Is it something similar to that? Of course, none of these use calculus, so what you found is probably much simpler than any of that, I would think. But then again, if the formula for the curve is known, one can just use calculus anyway, so could you make an image of what you're describing?

publius
2007-Oct-27, 03:11 PM
Grav,

The planimeter was invented back in the 1800s, well before the notion of a computer was even a gleem in anyone's eye.

And it is calculus all right, Grav, nothing but calculus. It is a simple little device that does an integration mechanically, indeed by simply counting the distance a little wheel rolls. It uses a clever little trick to constrain that little wheel to travel a certain path relative to the pointer which is tracing the perimeter. You trace the pointer around the perimeter in one closed loop, then read the counter. That is proportional to the surface area enclosed by the path you just traced.

You can find the area of any closed curve in the plane. It is a consequence of Stoke's Theorem, which is from vector calculus. Stokes theorem relates a surface integral to a path integral around the boundary of the surface.

Path and surface integrals are a little more advanced notion than the simple integrals you're learning about now. But it's all built upon that foundation.

-Richard

publius
2007-Oct-27, 05:53 PM
Spooky. The above post timestap is the infamous 11:11. Play the spooky music.

-Richard

Kaptain K
2007-Oct-27, 08:15 PM
Spooky. The above post timestap is the infamous 11:11. Play the spooky music.

-Richard
Quick! Call George Noory!

grav
2007-Oct-27, 09:30 PM
Grav,

The planimeter was invented back in the 1800s, well before the notion of a computer was even a gleem in anyone's eye.

And it is calculus all right, Grav, nothing but calculus. It is a simple little device that does an integration mechanically, indeed by simply counting the distance a little wheel rolls. It uses a clever little trick to constrain that little wheel to travel a certain path relative to the pointer which is tracing the perimeter. You trace the pointer around the perimeter in one closed loop, then read the counter. That is proportional to the surface area enclosed by the path you just traced.

You can find the area of any closed curve in the plane. It is a consequence of Stoke's Theorem, which is from vector calculus. Stokes theorem relates a surface integral to a path integral around the boundary of the surface.

Path and surface integrals are a little more advanced notion than the simple integrals you're learning about now. But it's all built upon that foundation.

-RichardWell, I figured out a rudimentary way of doing it, at least, one of those deals which might look good on paper, but probably wouldn't do too well in practice. My girlfriend and I were just watching someone do that a while back and were wondering how that worked, so maybe I can impress her with this. :)

I was thinking something like a really powerful compass could keep the inner workings turned a certain way, with a ball underneath which would follow the perimeter. There would be two small axles inside that turned against the ball, one always aimed north and south and the other east and west. This would be just like those old computer mouses (mice?) worked. Both would be set to zero at the furthest point south on the perimeter. The one that is aimed north and south would count up from zero when going north and backwards when going the other direction. When going east and west, however, each time the inner east-west wheel turns one full revolution, say at one foot intervals, it would add whatever the north-south meter says at the time when going east and subtract it when going west. When travelling at an angle, with a ball underneath, each meter would still just turn according to the amount walked in each particular direction, just like a mouse, but the compass keeps it pointed a certain way at all times. By the time one has completely walked the perimeter, all they would have to do then is multiply whatever the east-west meter says times the interval of one foot to get the area. But actually, though, this idea isn't even complete, because I'm not sure about the mechanics of how one meter would add and subtract from the other. I'm sure the real version is much simpler than this anyway, but am I close?

[EDIT - Oh, yeah. The perimeter would also have to be walked clockwise for that to work the way I've got it set up. It would work counter-clockwise if the east-west meter adds going west and subtracts going east. But then again, if the east-west meter reads positive in either direction past zero, then it wouldn't matter which way one chose to walk, since one just takes the absolute value for the result anyway.]

publius
2007-Oct-27, 10:01 PM
Grav,

How it works is quite simple if you're familiar with vector calculus. There is a theorem there, called Stokes Theorem, which states that the surface integral of the curl of a vector function is indentically equal to the path integral of the function around the curve bounding that surface.

At this stage you're not familiar with surface integrals and path integrals, and that won't make much sense. It just won't click. But I'll show you how it plays out. In crude ASCII limited notation, Stokes Theorem is thus:

Integral (curl F) dot dS = Integral F * dr around C bounding S.

F dot dr is the type of integral you would do with a vector force field to determine the work done moving along a given path. A closed loop integral is one that goes around a closed path coming back to where you started. For a conservative force, that is always zero. So right there, we see a conservative force must be curl free. But that is not relevant to our purpose here.

Now, consider S in a plane, that is some plane surface patch bounded by some curve C in that plane. We choose coordinates so the plane is the x-y plane and the normal is z.

Now, here's the trick. If we have a function F, whose curl is identically 1 in the z direction, the left hand side becomes simply k dot dS, and since S is in the plane, the vector dS is always parallel to the z direction, and so the left hand side becomes simply the area of the surface.

And the path integral of that function F around C then must be equal to that area.

-Richard

grav
2007-Oct-27, 10:13 PM
Yep. That's what I said. ;)

publius
2007-Oct-27, 10:17 PM
Now, what is F? Well, a vector function F in three dimensions has three components, Fx, Fy, Fz, which are just three arbitrary functions of positions. That is three functions that may depend on three variables, x, y, z.

The curl operator involves (partial) derivatives of those three functions, and returns a vector which is related to the "vorticity" or "rotational" aspects of said F field. In Europe, curl is called "rot" for rotation, generally.

In operator form, one can see the curl as the cross product of the del operator against F. The z component of curl is d(Fy)/dx - d(Fx)/dy. The other two components involve Fz and derivatives of the others with respect to z.

So, if Fx = -y/2, Fy = x/2, and Fz = 0, we have a function whose curl is always one in the z direction. That's what we're after,

F = (-y/2, x/2, 0)

So our path integral on the right hand side is then,

Integral[ (-y/2, x/2) dot (dx, dy) ]

or 1/2 Integral[ x dy - y dx ]

That integral must be done counter-clockwise, coming from the right-hand rule, right-hand "sense" of things.

That is the Planimeter Integral. :)

-Richard

publius
2007-Oct-27, 11:05 PM
Now, let's look at the dot product form of that integrand more carefully:

2A = Integral[ (-y, x) dot (dx, dy) ]

What is this (-y, x) vector? Well the position vector in the plane is simply (x, y), call that R, so the length of this thing is equal to distance to the origin. It is some sort of complementary position vector, Q call it.

If R is along the x-axis, (d, 0), then Q = (0, d), along the y axis. If R is along the y axis, R = (0, d), the Q = (-d, 0). Q seems to be a vector in the plane perpendicular to R, and "leading it", in that right handed sense, rotating 90 degrees CCW.

Indeed, R dot Q = -xy + xy = 0.

So, the planimeter integration is the projection of the displacement vector, dR, along Q, perpendicular to R, around the perimeter of the area.

That is just slick.

-Richard

publius
2007-Oct-27, 11:24 PM
Hmmmphh, looking around (Googling), I see someone did his mathematics PhD thesis on planimeters. He wrote a fancy graphic program showing the mechanism working as you virtually drag the pointer around. I won't post the link, cause it will spoil the fun by showing how the mechanism works.

But to do the planimeter integration, we need to integrate this:

r (n_q dot dr ), where n_q is the unit vector along Q, perpedicular to R, and r is the magnitude, the length of R to the origin.

So imagine an arm with a pointer. We fix one end to an origin, then telescope the arm in and out as we move along the curve. The arm length is r, and we want to multiply that by the component of movement perpendicular to it.

That's one way to do it. The other way might be to play around with the angle of the wheel to compenstate for changes in length to the origin.

-Richard

publius
2007-Oct-28, 12:24 AM
There is also a purely geometric way to show how a planimeter works. I'm not familiar with it, but I've seen it. It's equivalent, but doesn't involve fancy smancy vector calculus stuff. But to me, the planimeter is Stokes Theorem in action.

Let's look the planimeter integral for a very simple case, the unit square, lower left corner at the origin. The area is simply 1 unit. A

The path integral is x dy - y dx along the perimeter, and our area is supposed to be 1/2 of that. Split into four pieces, along each side. The first path is a straight line from (0,0) to (1, 0). y is zero always, and dy is zero, so that contributes nothing.

The second leg is from (1, 0), to (1, 1). Here dx is zero, x is constant, 1, and Int(x dy) is just 1 times 1.

The third leg is from (1, 1) to (0, 1). Here y is constant, 1, dy = 0, x goes backwards from 1 to 1, so we get a minus sign, -(1 * -1) = 1. Note how the minus sign in the formula cancelled out that "backwards" movement. :)

So we now have two total units.

The final leg is from (0, 1) to (0, 0). x is zero and dx is 0, so we get nothing.

So our area is one-half of that, 2/2 = 1 unit, which it is supposed to be.

And you can do the same for a rectangle of with upper right corner
(a, b), and see the result is ab.

Now, generalize that to a N sided polygon, and N-gon. A result which I'll mention without proof (it can be done fairly easily), is that if let your origin be at one corner, the integration always cancels for the first and last legs from and to there, and so you only have to integrate N - 2 segments.

Place a triangle of any shape with one corner at the origin, and you'll find the only integration that counts is along the side opposite the origin.

For example, place a right triangle at the origin, hypotenuse sloping up -- that is the corners are (0,0), (a, 0), and (a, b) and the result will be 1/2 ab, the lower leg and hypot don't contribute.

As you saw above with the square, only the middle two contributed, there.

-Richard

grav
2007-Oct-28, 02:09 AM
So imagine an arm with a pointer. We fix one end to an origin, then telescope the arm in and out as we move along the curve. The arm length is r, and we want to multiply that by the component of movement perpendicular to it.Ah. That would be the mechanism, then, for my idea, where the distance along the east-west direction must be multiplied by that of the north-south as you go. I was trying to think of something like that, where a gear might expand and contract or a pulley might lengthen or something, where distance along one vector will add or subtract with the other meter faster or slower.

I guess my idea wasn't too far off anyway, was it? It would work, at least, if just the mathematics of it, not so much the device I described, which sounds like it would probably just fall apart as soon as it hit the ground, I know. :) It's still calculus, though, but it's basically just the Cavalieri principle at work.

publius
2007-Oct-28, 02:12 AM
It might be nice to show what I'm talking about :lol:

http://www.math.duke.edu/education/ccp/materials/mvcalc/green/index.html

This is a very nice derivation and discussion of the "Planimeter integral", and it has nice Java applets to show a planimeter in action. The type of Planimeter illustrated here is the polar planimeter. There are other types, but this one was the most common, and the type I first saw being used.

They describe it terms of Green's Theorem in the plane, which is just the plane piece of Stoke's Theorem.

Well, George Green,

http://en.wikipedia.org/wiki/George_Green

deserves it actually. He contributed much to the mathematics of EM theory. He was the self-taught son of a baker with one year of formal schooling who ran his father's mill. Just read the Wiki article.

He finally entered college at age 40 and went on to be a professor, but he got sick and died soon thereafter. He was a remarkable man. You'll note that Einstein spoke highly of him, as well as Schwinger. I have a book which is a compilation of Schwinger's lectures on Classical EM theory, and he makes much use of Green's work there.

The self-taught son of a baker running a mill, on his own, gave us much of the mathematical foundation of EM theory.

I wonder if a Green would be possible today? I doubt it. We've advanced so much. It's possible a Green could come along, but he would probably just reinvent the wheel, those he didn't know existed, and probably wouldn't break any new ground. But then again, it might be possible.

-Richard

publius
2007-Oct-28, 02:31 AM
Here's Schwinger's tribute to Green:

http://xxx.lanl.gov/PS_cache/hep-ph/pdf/9310/9310283v1.pdf

He's having fun there. He compares Green to Shakespeare and wonders why there are no conspiracy theories that the "miller of Nottingham" was not the real mathematician behind the name Green. Why, this Green, with one year of formal education couldn't have done all that. :lol:

-Richard

grav
2007-Oct-28, 06:19 AM
Well, I decided to go ahead and try that gravity along the plane of a flat disk thing. Below is an image for that. If the image appears blurry, bring the mouse onto the page after clicking on the thumbnail and another enlarge button will appear on the bottom of the page after a second or two. I am taking concentric circles of dz from the point at d from the center of the disk, but only integrating for those that cross the disk from z=-r to r. To find the point where one of the concentric circles from d crosses the circumference of the disk, I took

x^2 + y^2 = r^2 for the disk and

(d-x)^2 + y^2 = (d-z)^2 for the concentric circle, so

(d^2 - 2xd + x^2) + y^2 = d^2 - 2zd + z^2
-2xd + r^2 = -2zd + z^2
x = (r^2 + 2zd - z^2)/2d

y = sqrt[r^2 - (r^2 + 2zd - z^2)^2/(4d^2)]

The angle of the arc, then, is asin[y/(d-z)] above and below the x axis. So first we need to integrate for the mass and direction of acceleration for the mass along each arc, which is I 2 cos[asin[y/(d-z)]] (m/A) (d-z) dz . Now, I really had to put some thought into that one. I wasn't sure at first whether to transform y into z and find for only z with I 2 cos[asin[sqrt(r^2 - (r^2 + 2zd - z^2)^2/(4d^2))]] (m/A) (d-z) dz, or perhaps since cos(asin(x)) is sqrt(1-x^2), integrate for sqrt[1-(r^2 - (r^2 + 2zd - z^2)/(4d^2))] or something. But then I realized that I'm not actually integrating for z yet, but just for the angle at a constant z for each arc. So I am actually integrating for I 2 cos(L) (m/A) (d-z) dz with a constant z, but a changing angle L, so the result of that integration is simply

2 sin(L) (m/A) (d-z) dz
= 2 sin[asin(y/(d-z))] (m/A) (d-z) dz
= 2 [y/(d-z)] (m/A) (d-z) dz

Then I integrate again for the total acceleration across all of the arcs from -r to r, giving me

I 2 [y/(d-z)] (m/A) (d-z) dz * G/(d-z)^2

This time z is not a constant, so the entire integration in terms of z becomes

2G(m/A) * I sqrt[r^2 - (r^2 + 2zd - z^2)^2/(4d^2)] dz / (d-z)^2, for z = -r to r

This, then, is the integration for the acceleration of gravity toward a flat disk of radius r from a distance d from its center and along the same plane of the disk. But I have no idea where to go from here. Please help.

publius
2007-Oct-28, 06:50 AM
Grav,

Settle down. :) I can't exactly follow what you're doing because, well, it's non standard notation, and I'm getting old. :lol: But I'll let you in on a little secret. The gravitational field of the thin disc, in the near field, is a difficult problem. I think, but I'd have to check to make sure, it involves elliptic integrals (no closed form, analytic solution).

But grav, you're not exactly ready to tackle that. You need to learn how to do volume integrals, and the machinery and standard notation for that.

Second, you never integrate the field directly (unless it's really simple), you find the *potential*. The potential is a scalar, and the field is its gradient. It's much easier to do the volume integral for the potential, then trake the gradient to get the field.

IIRC, we can get the field easily enough on axis (that if the disc is centered at the origin in the x-y plane, the field along the z-axis above). Off axis, you get those messy elliptic integrals.

You've got a bit more to learn before you can tackle that.

-Richard

01101001
2007-Oct-28, 06:54 AM
If the image appears blurry, bring the mouse onto the page after clicking on the thumbnail and another enlarge button will appear on the bottom of the page after a second or two.

(That procedure for enlarging/zooming is browser-specific and would only be confusing to those whose browsers use different means.)

grav
2007-Oct-28, 01:35 PM
(That procedure for enlarging/zooming is browser-specific and would only be confusing to those whose browsers use different means.)Oh, yes. Thanks. I forgot. But if another browser doesn't have that button, would the image still appear blurry, or would it come out clear in the first place? Does the type of image matter for this?

grav
2007-Oct-28, 02:06 PM
Grav,

Settle down. :) I can't exactly follow what you're doing because, well, it's non standard notation, and I'm getting old. :lol: But I'll let you in on a little secret. The gravitational field of the thin disc, in the near field, is a difficult problem. I think, but I'd have to check to make sure, it involves elliptic integrals (no closed form, analytic solution).Yes, lots and lots of ellipticals. The image below showed what I got when I ran the integration on "The Wolfram Integrator" (http://integrals.wolfram.com/index.jsp), a wonderful thing to have handy.

But grav, you're not exactly ready to tackle that. You need to learn how to do volume integrals, and the machinery and standard notation for that. You're right. I need to find for the volume and then just thin the disk out. I think I've finally figured that out, although I might have had an inkling when I tried it before. It didn't make sense to me that the acceleration of gravity on the edge of a thin disk should approach infinity for a finite mass. But now that I've looked at that integration again, I can see why. Instead of the mass per area, I could take the density and multiply that by the thickness of the disk. But the thickness is just zero, so the density is infinite. So the integration itself might come out to something finite when multiplied by the zero thickness, but it would still then be multiplied again by an infinite density. So apparently, the acceleration of gravity has more to do with the density of a body than its mass per say. And if we figure that a body would then just have some finite density, instead of an infinite one, and then just multiply that by the zero thickness in the formula, instead of using mass per area, then the zero thickness might cancel out with the infinite result of the integration for the edge of the disk, providing a finite result, which then in turn just multiplies by the finite density, producing a finite acceleration overall. So I will try it again using some finite density for a volume like a cylinder or an oblate spheroid, whatever's easiest, and then just shrink down the thickness in the resulting formula accordingly and see what happens.

Second, you never integrate the field directly (unless it's really simple), you find the *potential*. The potential is a scalar, and the field is its gradient. It's much easier to do the volume integral for the potential, then trake the gradient to get the field.You've said that a few times, but I'm still not really sure what it means.

IIRC, we can get the field easily enough on axis (that if the disc is centered at the origin in the x-y plane, the field along the z-axis above). Off axis, you get those messy elliptic integrals.I found x and y in respect to the center of the disk, but I got ellipticals anyway. I've noticed that sometimes the integrals, like the one in the image below, can be reduced down. I wonder if that one can. Not that I'm going to even bother trying to wade through that mess or anything. :)

You've got a bit more to learn before you can tackle that.No doubt. :)

grav
2007-Oct-28, 04:10 PM
Second, you never integrate the field directly (unless it's really simple), you find the *potential*. The potential is a scalar, and the field is its gradient. It's much easier to do the volume integral for the potential, then trake the gradient to get the field.Oh, wait. I think I understand that after all. Isn't that what I did? I took those points across the disk that are equidistant from d by using concentric circles from d, thereby producing the arc. So all of the points across the arc have the same field potential toward d. I then integrated for the cosine of the angles of the points across the arc to get the potential directly toward the center of the disk from d. Then at the last, I integrated again with the resulting potential of each arc across the gradient. I agree, that's the easiest way to do it, especially considering I'm not even sure how to do it any other way. :)

I also see what you mean about following what I did when I look back. That z is not the z axis, but the x axis, and z is the distance from the center of the disk to the edge of the arc along the x axis. I probably should have picked another letter for that instead of z, like b or something. Also, that dz at the end of each integration is the differential for z, not d*z. I wrote zd for z*d. ;) I probably should have used D for the distance instead of d there also. I used z instead of x for the integration because x is the axis I'm using and I have seen z used for integrations as often as x, but I neglected to think that z can be an axis too, so sorry if I confused you there. I guess I need to look into starting to use more letters, but then, I don't want that to make things confusing either. :doh:

publius
2007-Oct-28, 05:57 PM
Grav,

It looked like to me you were trying to integrate the vector g directly up there.

Gravitational potential: What is the work done by gravity in moving a test particle in an arbitrary field? That is the notion of gravitational potential.

That's one of those path integrals I was talking about:

W = - Integral[ mg dot dr] from r1 to r2. The minus sign comes from defining this in the sense of the work done *against* gravity. The formal defintion of potential requires a reference, which is generally infinity (unless that doesn't work, which it sometimes doesn't when there's something that would involve an infinite amount of total mass, such as infinite wire or plane of mass).

And the is the potential, V(r), is the work done against the field moving a test particle in from in infinity to test point r in the field.

What is that for a point mass? Move a test particle in radially from infinity to r. Integrate -g dr from infinity to some r. What do you get? That is the gravitational potential of a point mass.

-Richard

grav
2007-Nov-02, 01:15 AM
Well, now I'm proud of myself now too. I've finally completed my first three dimensional problem using calculus, even though it took me a couple of days to figure out how to work through it properly. Of course, I've done plenty of integrations in a roundabout, trial and error sort of way, by running them out on the computer and drawing up graphs and such, and attempting to work through the numbers this way and that way to come up with a solution, but this may be the first time I can actually show how I achieved it, step by step, in a logical progression, using calculus. :dance: Maybe that'll redeem me from how I messed up that one for Gabriel's Horn. ;) Practice makes perfect.

I've come to realize that the best way (at least for me so far) to start off any three dimensional integration for a shape is with simply Int dx dy dz, in other words integrating for every infinitesimal point within the shape. Although I've always thought of a point as an infinitesimal sphere, that integration setup shows me that an infinitesimal point is generally to be thought of as cubic, which makes sense since all of the points need to fit snugly together in each of the three dimensions, and spheres don't do that too well. I figure I thought that because the gravity of a point is the same as the gravity of a sphere, but with an infinitesimal point, I guess it doesn't matter much what shape it would take as far as the gravity is concerned, except for the convenience of a cube when trying to integrate in three dimensions. Anyway, that's just a quick observation.

So what I wanted to do was just to integrate the gravity of a spherical body of radius 'r' for an object at some distance 'R' from the center of mass. So the acceleration of gravity from any infinitesimal point would be a = (GM/R'^2)*(x'/R') = GMx'/(R'^3), where x'/R' is the cosine of the angle of the point to that of the direction of the line through the center of mass, giving that amount of the acceleration which is directed toward the center. I am considering the center of the sphere with uniform density to be the origin, so the distances, x, y, and z are in reference to that, and x' becomes R-x, since I have also positioned the line through the center along the x axis. R' is the total distance between the point at a distance R from the center and to the point we are considering the gravity for. The integration for the sphere then becomes Int G D x' dx dy dz / (R'^3) = Int G D (R-x) dx dy dz / [(R-x)^2 + y^2 + z^2]^(3/2), whereas R' = [(R-x)^2 + y^2 + z^2]^(1/2) and D is the density of the sphere.

So first I will integrate along z. For any given x and y, these become constants, and so all constants aside, we are now integrating for [G D (R-x) dx dy] Int dz / [j + z^2]^(3/2), where 'j' is also a constant, then. I had to look up all of my integrals, by the way, but I figured that is probably the case for all but the simplest integrations for most people, unless one does them so much they can remember them, but I don't know enough yet anyway. That integration above comes out to Int dz / [j + z^2]^(3/2) = (z/j)/[j + z^2]^(1/2), where 'j' here is ((R-x)^2 + y^2), and the integration from -z to z becomes (z/j)/[j + z^2]^(1/2) - (-z/j)/[j + (-z)^2]^(1/2) = (2z/j)/[j + z^2]^(1/2), where +/-zmax = (r^2 - x^2 - y^2)^(1/2), so the full integration for z becomes [G D (R-x) dx dy] 2 (r^2 - x^2 - y^2)^(1/2) / [((R-x)^2 + y^2)((R-x)^2 + r^2 - x^2)^(1/2)].

Now I want to integrate that for y with a given x, making x constant. That becomes Int G D (R-x) dx dy 2 (r^2 - x^2 - y^2)^(1/2) / [((R-x)^2 + y^2)((R-x)^2 + r^2 - x^2)^(1/2)] = [ G D (R-x) dx 2 / ((R-x)^2 + r^2 - x^2)^(1/2)] Int dy (r^2 - x^2 - y^2)^(1/2) / ((R-x)^2 + y^2). The integration there becomes Int dy (j - y^2)^(1/2) / (k + y^2) = [(j+k)/k]^(1/2) atan [((j+k)y^2/(k (j - y^2)))^(1/2)] - atan [y/(j - y^2)^(1/2)], where j and k are constant. That looked complicated until I realized that +/-ymax = (r^2 - x^2)^(1/2), and j = r^2 - x^2 = y^2, so j - y^2 = 0. That makes the value inside both atan's infinite, so each atan itself is just pi/2, making the integration from -y to y equal to [((j+k)/k)^(1/2) - 1] (pi/2) - [((j+k)/k)^(1/2) - 1] (-pi/2) = 2 [((j+k)/k)^(1/2) - 1] (pi/2), where j = r^2 - x^2 and k = (R-x)^2, so 2 [((j+k)^(1/2) - k^(1/2))/k^(1/2)] (pi/2) = 2 [(((R-x)^2 + r^2 - x^2)^(1/2) - (R-x))/(R-x)] (pi/2). The full integration for y now becomes [G D (R-x) dx 2 / ((R-x)^2 + r^2 + x^2)^(1/2)] 2 [(((R-x)^2 + r^2 -x^2)^(1/2) - (R-x))/(R-x)] (pi/2) = 2 pi G D dx [((R-x)^2 + r^2 - x^2)^(1/2) - (R-x)] / [(R-x)^2 + r^2 - x^2]^(1/2) = 2 pi G D dx [1 - ((R-x) / ((R-x)^2 + r^2 - x^2)^(1/2)].

Finally, integrating for x, we get 2 pi G D [ Int dx - Int (R-x) dx / ((R-x)^2 + r^2 - x^2)^(1/2)] = 2 pi G D [ Int dx - Int (R-x) dx / (R^2 - 2Rx + r^2)^(1/2)] = 2 pi G D [ Int dx - Int (R-x) dx / [(2R)^(1/2) (R/2 + r^2/2R - x)^(1/2)]. Of course, the integration of dx is just x, so we are left with the integration of Int dx (j - x) / [(k - x)^(1/2) (2R)^(1/2)], where j = R and k = R/2 + r^2/2R, giving us Int dx (j - x) / [(k - x)^(1/2) (2R)^(1/2)] = (2/3) (k - x)^(1/2) (-3j + 2k + x) / (2R)^(1/2)= (2/3) (R/2 + r^2/2R - x)^(1/2) (-3R + R + r^2/R + x)/(2R)^(1/2) = (2/3) (R^2 + r^2 - 2Rx) (-2R + r^2/R + x) / 2R. For x=r, the integration is Int [dx - (R-x) dx /(R^2 - 2Rx + r^2)^(1/2)] = r - (1/3) (R^2 + r^2 - 2Rr)^(1/2) (-2R + r^2/R + r) / R = [3R^2 r - (R-r) (-2R^2 + r^2 +Rr)] / 3R^2 = [3R^2 r - (-2R^3 + R r^2 + R^2 r + 2R^2 r - r^3 - R r^2)] / 3R^2 = (2R^3 + r^3) / 3R^2. Doing the same thing for x=-r reduces to (2R^3 - r^3) / 3R^2, so the difference is [(2R^3 + r^3) - (2R^3 - r^3)] / 3R^2 = 2r^3 / 3R^2. So the entire integration now becomes just a = 2 pi G D 2r^3 / 3R^2 = (4pi/3) r^3 G D / R^2. Since the volume of the sphere is V = (4pi/3) r^3, and D is the density, whereas DV = M, then the integration reduces even further to just a = G M / R^2. Tada. :) :)

grav
2007-Nov-02, 01:29 AM
I could have made that a little more legible, I guess, but I just wanted to get it all down within a reasonable amount of time. I won't have to get into quite so much detail from now on anyway, I don't suppose, but just wanted to show it for that. Richard, I'm still not sure what you meant exactly, so how would you go about doing that same thing using the potential like you were talking about? Would it be easier than what I did?

grav
2007-Nov-05, 01:19 PM
This integration through calculus stuff is pretty neat. I have integrated for the gravity of a few shapes, but so far anything with an adjustable height, a height which relies on a variable that can be brought down to zero thickness to produce a thin disk such as with an oblate spheroid, cylinder, or diamond shaped on edge, for examples, has resulted in ellipticals. I'm getting the impression that this will be the case no matter how I go about it, or what original shape I try to use. So I'm thinking about just trying to take, say, the first ten orders of the ellipticals, if they converge quickly enough, and finding a very close approximation for a thin disk, at least, to give an idea of what the gravity should be at various points.

I could also find for a varying density, one that varies with distance from the center, by using something like (D/sqrt(x^2+y^2)) in the original integration instead of just D for a uniform density, which would be infinite at the center and fall off to zero at infinity with 1/R, or use a shape like Gabriel's horn with uniform density, which would be similar but probably not quite the same, or perhaps a little more realistically, D[1 - sqrt(x^2 + y^2)/r], which is D at the center and falls off to zero at the edge of the disk or other circular shape, which would be similar to that of the diamond shaped on edge thing (diamond shaped looking at it edge on).

I have also been trying to integrate for points on a rotating sphere, by attempting to find for the Doppler shift, directly toward or away along the line of sight from R to each point, and then integrating for the redshifted or blueshifted gravity from each point accordingly. Just for kicks, I am trying to see if I can find a form of Doppler, relativistic or otherwise, that might provide for the precession of the planets through the spin of the sun. So far, though, finding the vectors of force and the angles for the line of travel of the points to that of R is tougher than the integration itself, but I'm managing, little by little.

There is something I'm still wondering about, though. That integration for a stationary sphere I made a couple of posts back works out correctly for any points at a distance R which is outside of the sphere. When I took x(r) - x(-r), I ended up with ax = GM/R^2, but for points at R which are inside the sphere (R<r), the result should be different. It would seem that the true formula for the integration should work for any R, since it does not discriminate, as far as I can tell, whether R is greater or less than r to begin with, but it does not work out that way, only for R=>r. Why would that be? I did notice, however, that if we take that last integration for x(r) - x(-r), which is 2 pi G D [(2R^3 + r^3) - (2R^2 - r^3)] / 3R^2 = (4 pi / 3) r^3 G D / R^2 = V G D / R^2 = GM/R^2, and gives us the integration for outside of the sphere, and find instead for [x(r) + x(-r)]/2, then that becomes 2 pi G D [(2R^3 + r^3) + (2R^3 - r^3)] / 6R^2 = (4 pi / 3) G D R = (4 pi / 3) r^3 G D R / r^3 = V G D R / r^3 = GMR/r^3, which is correct for the inside of the sphere but not the outside. I have tried other ways of finding for the inside of the sphere, like taking the negative result for the square roots instead of the positive, but that didn't work out. So I'm wondering if this result is just a coincidence, or is there something more fundamental about it? Do we take the average for the limits across x to find for the acceleration along the x vector for the inside of any general shape, and the difference to find for the outside, or would this only work out for a sphere for some reason or another? I cannot think of any logical reason offhand why this should be the case, but if this works out the same way for any general shape, then there must be some mathematical reason for it to do so.

grav
2007-Nov-05, 01:42 PM
Okay, looking back at this thread (http://www.bautforum.com/against-mainstream/51537-dark-matter-solved.html), and going through just the "integration" for a thin disk there (disregarding any conclusions on dark matter for our purposes here, of course), it looks like at least a close approximation for a thin disk was found, along the lines of ln(x), using the posts provided and that of the gravity of thin wires Richard and I worked out in another thread. So I'm wondering if the ellipticals for a thin disk might possibly break down into something in terms of ln(x) overall. If so, maybe that can be reverse engineered or something, using the derivatives, to find a different form of integration for a thin disk.

publius
2007-Nov-06, 12:13 AM
Grav,

Sorry, but I sort of forgot about this thread. I was looking for an online reference of the general, off axis, thin disc problem, but couldn't find one.

EM material is more common, and Newtonian gravity and electrostatics are the same thing mathematically, just using a minus sign, and mass instead of charge, and G instead of 1/(4pi*epsilon).

I found online solutions of the on-axis electric version, but that's fairly easy, but nothing on the general case yet.

-Richard

publius
2007-Nov-06, 12:31 AM
Grav,

And oh yeah, about potentials. Here's a very quick summary:

The potential of a point mass is -GM/r. What is the derivative of that with respect to r? GM/r^2.

IOW, g = -dV/dr, where V is the potential. (Note this potential is not the energy directly, but the energy per unit test mass -- the energy potential is mV, where little 'm' is the mass of the test mass).

That's in one dimension. In general, the field is the *gradient* of the potential. In cartesian coordinates, the gradient is this operator, traditionally denoted by 'del', an upside down delta.

del = ( d/dx i + d/dy j + d/dz k), where i,j, and k are the unit vectors in the 3 directions as usual.

So, in general g = -del(V), where V is the potential. Now, the potential is a *scalar*, not a vector, and it's much easier to integrate the potential of a mass distribution than the vector field directly.

To get the potential you just integrate -G dM/r, where dM is the differential mass element, some rho*dV = rho*dx*dy*dz, where rho is the density.

You get some V(x, y, z). The g field is then the negative gradient of that. The 'x' component is dV/dx, etc.

As I mentioned above, the same mathematical thing applies to electrostatics, and that is more widely available than the gravitational version. But they are otherwise the same thing and learning one is learning the other.

-Richard

grav
2007-Nov-08, 12:08 AM
Well, I'm still stuck on that thin disk thing. I seem to be pretty determined to define it in terms of ln(x) instead of ellipticals. Anyway, it's still good practice. I've come across one particular manipulation for a cylinder so far which gives an integration that appears to fall directly in line with the formula in the other thread. Here it is.

Int G D (R-x) dx dy dz / ((R-x) + y^2 + z^2)3/2

Int dz / (j + z^2)3/2 = (zlimit / j) / (j + zlimit^2)1/2, where j = (R-x)^2 + y^2

zlimit = +/- h, where h is the height of the cylinder

[(h / j) / (j + h^2)1/2] - [(-h / j) / (j + (-h)^2)1/2] =
(2h / j) / (j + h^2)1/2

= Int G D (R-x) dx dy 2h / [((R-x)^2 + y^2) ((R-x)^2 + y^2 + h^2)1/2]

V = h * (pi r^2)

= Int 2 G D V (R-x) dx dy / [(pi r^2) ((R-x)^2 + y^2) ((R-x)^2 + y^2 + h^2)1/2]

M = D V

= Int 2 G M (R-x) dx dy / [(pi r^2) ((R-x)^2 + y^2) ((R-x)^2 + y^2 + h^2)1/2]

h = 0

= Int 2 G M (R-x) dx dy / [(pi r^2) ((R-x)^2 + y^2)3/2]

Int dy / (j + y^2)3/2 = (ylimit / j) / (j + ylimit^2)1/2, where j = (R-x)^2

ylimit = +/- (r^2 - x^2)1/2

[(ylimit / j) / (j + ylimit^2)1/2] - [(-ylimit / j) / (j + (-ylimit)^2)1/2] =
2 (ylimit / j) / (j + ylimit^2)1/2

= Int 4 G M (R-x) dx (r^2 - x^2)1/2 / [(pi r^2) (R-x)^2 ((R-x)^2 + (r^2 - x^2))1/2]

= Int 4 G M dx (r^2 - x^2)1/2 / [(pi r^2) (R-x) (R^2 - 2 R x + r^2)1/2]

Now normally that would also lead to ellipticals, but if we find only for when R = r, then we get

= Int 4 G M dx (r^2 - x^2)1/2 / [(pi r^2) (r-x) (2r^2 - 2rx)1/2]

= Int 4 G M dx (r+x)1/2 (r-x)1/2 / [(pi r^2) (r-x) (2r)1/2 (r-x)1/2]

= Int 4 G M dx (r+x)1/2 / [(pi r^2) (r-x) (2r)1/2]

Int dx (j+x)1/2 / (j-x) = 2 (2j)1/2 tanh-1[((j+xlimit)/(2j))1/2] - 2 (j+xlimit)1/2, where j = r

xlimit = +/- r

[2 (2r)1/2 tanh-1[((r+xlimit)/(2r))1/2] - 2 (r+xlimit)1/2] -
[2 (2r)1/2 tanh-1[((r + (-xlimit))/(2r))1/2 - 2 (r + (-xlimit))1/2] =

[2 (2r)1/2 tanh-1(1) - 2 (2r)1/2] -
[2 (2r)1/2 tanh-1(0) - 2 (0)]

tanh-1(b) = (1/2) [ln(1+b) - ln(1-b)] = (1/2) ln[(1+b)/(1-b)] for b=< 1

tanh-1(0) = 0
tanh-1(1) = ln(2/0) / 2

(2 (2r)1/2) [tanh-1(1) - 1] =
(2 (2r)1/2) [ln(2/0) / 2 - 1]

= [4 G M / [(pi r^2) (2r)1/2] [(2 (2r)1/2) [ln(2/0) / 2 - 1]]

= 4 G M [ln(2/0) - 2] / (pi r^2)

which is pretty much the same form (although the variables are used differently) and result that was obtained in the other thread. I was initially attempting to find for the gravity at the edge of an infinitesimally thin disk, and so I guess this is it. It is infinite. It probably isn't any different, then, than finding for the gravity at the edge of an infinitesimally small sphere, which becomes just a point, and is therefore also infinite. I'm still trying to see if I can find for other values of R (or ratios of R/r) in a similar fashion.

grav
2007-Nov-08, 10:48 PM
Grav,

And oh yeah, about potentials. Here's a very quick summary:

The potential of a point mass is -GM/r. What is the derivative of that with respect to r? GM/r^2.

IOW, g = -dV/dr, where V is the potential. (Note this potential is not the energy directly, but the energy per unit test mass -- the energy potential is mV, where little 'm' is the mass of the test mass).

That's in one dimension. In general, the field is the *gradient* of the potential. In cartesian coordinates, the gradient is this operator, traditionally denoted by 'del', an upside down delta.

del = ( d/dx i + d/dy j + d/dz k), where i,j, and k are the unit vectors in the 3 directions as usual.

So, in general g = -del(V), where V is the potential. Now, the potential is a *scalar*, not a vector, and it's much easier to integrate the potential of a mass distribution than the vector field directly.

To get the potential you just integrate -G dM/r, where dM is the differential mass element, some rho*dV = rho*dx*dy*dz, where rho is the density.

You get some V(x, y, z). The g field is then the negative gradient of that. The 'x' component is dV/dx, etc.

As I mentioned above, the same mathematical thing applies to electrostatics, and that is more widely available than the gravitational version. But they are otherwise the same thing and learning one is learning the other.

-RichardOh, okay, the potential energy (divided by mass). Well, I'm still not sure about all of your post, 'del' and so forth, as usual, but I tried to integrate for a sphere using the potential and didn't get very far. Here's what I got.

Int [-G D / ((R-x)^2 + y^2 + z^2)1/2] * [(R-x) / ((R-x)^2 + y^2 + z^2)1/2] dx dy dz

= Int -G D (R-x) dx dy dz / ((R-x)^2 + y^2 + z^2)

Int dz / (j + z^2) = (1 / j1/2) atan(zlimit / j), where j = (R-x)^2 + y^2 and zlimit = +/- (r^2 - x^2 - y^2)1/2

[(1 / j1/2) atan(zlimit / j)] - [(1 / j1/2) atan(-zlimit / j)] =
2 atan[(r^2 - x^2 - y^2)1/2 / ((R-x)^2 + y^2)] / ((R-x)^2 + y^2)1/2

= Int -2 G D (R-x) dx dy atan[(r^2 - x^2 - y^2)1/2 / ((R-x)^2 + y^2)] / ((R-x)^2 + y^2)1/2

Int dy atan[(j - y^2)1/2 / (k + y^2)] / (k + y^2)1/2

and that's it. I can't even find the integration for that last one. Am I doing something wrong? :neutral: Of course, I'm also still using vectors for this, while you mentioned scalars, so... :think: Okay, I'm lost. :sad: Is the original integration correct for the potential, as compared to the ax, ay, and az vectors for acceleration, or should it be something different?

EDIT - Maybe, being a scalar, the potential wouldn't depend upon the angles between the vectors and points? I'll try it again without using the angle in the original integration, then, which would then just become

Int -G D dx dy dz / ((R-x)^2 + y^2 + z^2)1/2

grav
2007-Nov-09, 02:06 AM
Okay, I'm slowly starting to put it together, I think. A scalar would be positive in any direction, as energy is due to the v^2 part of it, where v would have a positive or negative direction, but v^2 wouldn't. That 'del' thing for a scalar like energy would just be something similar to E = m del v= m(d/dx vx + d/dy vy + d/dz vz) = m(vx^2 / 2 + vy^2 / 2 + vz^2 / 2) = m v^2 / 2. The integration for just the scalar of -GM/R for a sphere becomes

Int -G D dx dy dz / ((R-x)^2 + y^2 + z^2)1/2

Int dz / (j + z^2)1/2 = ln(zlimit + (j + zlimit^2)1/2), where j = (R-x)^2 + y^2 and zlimit = +/- (r^2 - x^2 - y^2)1/2

= Int -G D dx dy [ln((r^2 - x^2 - y^2)1/2 + ((R-x)^2 + y^2 + (r^2 - x^2 - y^2)1/2) - ln(-(r^2 - x^2 - y^2)1/2 + ((R-x)^2 + y^2 + (r^2 - x^2 - y^2)1/2)]

= Int -G D dx dy [ln((r^2 - x^2 - y^2)1/2 + (R^2 -2Rx + r^2)1/2) - ln(-(r^2 - x^2 -y^2)1/2 + (R^2 -2Rx + r^2)1/2)]

Int dy ln[(j - y^2)1/2 + k] = ln(k - (j - ylimit^2)1/2)ylimit - ylimit + (k^2 - j)1/2 atan(ylimit / (k^2 - j)1/2) + k atan(ylimit / (j - ylimit^2)1/2) - (k^2 - j)1/2 atan(k ylimit / (k^2 - j)1/2 / (j - ylimit^2)1/2)

Int dy ln[-(j - y^2)1/2 + k] = ln(k - (j - ylimit^2)1/2)ylimit - ylimit + (k^2 - j)1/2 atan(ylimit / (k^2 - j)1/2) - k atan(ylimit / (j - ylimit^2)1/2) + (k^2 - j)1/2 atan(k ylimit / (k^2 - j)1/2 / (j - ylimit^2)1/2)

where j = r^2 - x^2, k = (R^2 - 2Rx + r^2)1/2, ylimit = +/- (r^2 - x^2)1/2

After subtracting for the limits of y for each integration and then subtracting each of these integrations from each other, we get

= Int -4 G D dx [k atan(ylimit (+)/(j - ylimit (+)^2)[/sup]1/2[/sup]) - (k^2 - j)1/2 atan(k ylimit (+) / (k^2 - j)1/2 / (j - ylimit (+)^2)1/2)]

but we notice that j = ylimit (+)^2 here, so (j - ylimit (+)^2)1/2 = 0 and atan(infinity) = pi / 2, and we have

= Int -4 G D dx (pi / 2) [k - (k^2 - j)1/2]

= Int -2 pi G D dx [(R^2 - 2R x + r^2)1/2 - ((R^2 - 2R x + r^2) - (r^2 - x^2))1/2]

= Int -2 pi G D dx [(R^2 - 2R x + r^2)1/2 - (R^2 - 2R x + x^2)1/2]

= Int -2 pi G D dx [(R^2 - 2R x + r^2)1/2 - (R - x)]

= Int -2 pi G D dx [(2R)1/2 (R/2 - x + r^2/2R)1/2 - R + x]

Int dx (j - x)1/2 = -(2/3) (j - xlimit)3/2, where j = R/2 + r^2/(2R) and xlimit = +/- r

Int dx -R = -R xlimit

Int dx x = xlimit^2 / 2

= -2 pi G D [(2R)1/2 (-2/3) (R/2 + r^2/2R - r)3/2 - (2R)1/2 (-2/3) (R/2 + r^2/2R - (-r))3/2 - R r + R (-r) + r^2/2 - (-r)^2/2]

= (-2 pi G D (-2/3) / 2R) [(R^2 + r^2 - 2R r)3/2 - (R^2 + r^2 + 2R r)3/2 - (-3/2) (2R) 2 R r]

= (2 pi G D / 3R) [ (R-r)^3 - (R+r)^3 + 6R^2 r]

= (2 pi G D / 3R) [(R^3 - 3R^2 r + 3R r^2 - r^3) - (R^3 + 3R^2 r + 3 R r^2 + r^3) + 6R^2 r]

= (2 pi G D / 3R) [-2 r^3]

= -4 pi G D r^3 / 3R

V = (4 pi / 3) r^3

= - G D V / R

M = D V

= - G M / R

So we get the same potential for a sphere as we would for an infinitesimal point at the center of the sphere with the same mass. This didn't seem to be too much easier to me than that other way I did it, though. :shifty: :) But I guess I now have options, at least. If I am having trouble finding an integration for the gravity one way, then I can try it with the other. Thanks, Richard.

Neverfly
2007-Nov-09, 02:09 AM
Yeah, thanks...

I've been following this thread and now I think I did myself a mischief...:doh:

publius
2007-Nov-09, 05:16 AM
Grav,

I am somewhat impressed with the above. You came up with the right answer at the end, but some things just didn't look completely kosher to me. You got the right answer, but things need polishing.

What you've discovered is integrating things in Cartesian coordinates that have spherical (or other non-rectangular symmetry) is general hard. That 1/r = 1/(sqrt(x^2 + y^2 + z^2) gets tedious.

Other coordinate systems, such as, you guessed it, spherical make things easier. However, it still gets hard in the general case.

One thing right off the bat that wasn't kosher is that E = m del v thing. That ain't right. Total energy E is the sum of kinetic and potential energies. Kinetic energy in vector form is 1/2 m (v dot v).

Gravitational potential is always *negative*, using the standard gauging. :) You get something postive by subtracting. When a particle falls, it moves to a more negative potential.

Potential is directly related to potential energy, but is a more abstract thing than that. For example, in EM, you'll come up a notion of a vector potential................ In GR, the metric itself can be seen as a tensor potential.

You've got talent and a love for this, which I like to see, but Rome was not built in a day. You've got to learn a lot more math and physics and gain some discipline. But you're a quick study, all right. You just need to not jump the gun so much sometimes.

-Richard

publius
2007-Nov-09, 05:31 AM
Grav,

Now, what did I mean when I said getting the potential is generally easier because it's a scalar?

What is a scalar? It's roughly a "single number". What is a vector. Roughly it's "more than one number". When you integrate the field, you are (unless there are symmetries you can take advantage of, which you can in simple cases) integrating a vector quantity.

What is the g field. It is a vector. Three numbers associated with each point in space. You are in general doing this integration:

dg = -(G/ R^2) r dm

where r is a unit vector along R, the distance between the differential mass element, dm, and the field point. That is a vector, three functions of position there.

With the potential, you're just after a single function of position. Once you have it, you get the vector g by taking derivatives of that scalar function.

-Richard

grav
2007-Nov-09, 02:00 PM
Grav,

I am somewhat impressed with the above. You came up with the right answer at the end, but some things just didn't look completely kosher to me. You got the right answer, but things need polishing.

What you've discovered is integrating things in Cartesian coordinates that have spherical (or other non-rectangular symmetry) is general hard. That 1/r = 1/(sqrt(x^2 + y^2 + z^2) gets tedious.

Other coordinate systems, such as, you guessed it, spherical make things easier. However, it still gets hard in the general case.Oh, maybe something like using polar coordinates for a sphere? Not that I'm familiar with that or anything. It's generally easier for me to keep track of the integration for straight lines. But I guess it would be something like using sin(x) and cos(x) around the circumference of a circle and integrating that across a cross-sectional area of the sphere to find for y and z, and then integrating that along x? It does seem that integrating for circumferences and then areas would be easier for a sphere, and it looks like it would only require two integrations also, knowing the circumference already, and knowing that the distance between each point on the circumference and a point at R would be the same all the way around. It seems that would only work for a stationary sphere, however, and maybe also an oblate spheroid or cylinder or even a rotating sphere when integrating for these "on axis".

One thing right off the bat that wasn't kosher is that E = m del v thing. That ain't right. Total energy E is the sum of kinetic and potential energies. Kinetic energy in vector form is 1/2 m (v dot v).Yes. I didn't try to relate the integration so much in terms of del like I had that, but just trying to think in terms of a scalar for the integration, something that doesn't rely upon the angle so much, but only the distance. But I was really just taking a shot in the dark, and it happened to work out. Looking back, I see I took the integration of v while it seems you had the derivative. So I guess that del thing should be something more like d/dE E = d/dE mv^2/2 = d/dE m [(vx^2/2 + vy^2/2 + vz^2/2)1/2]^2 = d/dE m (vx^2/2 + vy^2/2 + vz^2/2) = m (d/dx vx^2/2 + d/dy vy^2/2 + d/dz vz^2/2) = m (vx + vy + vz), and take the reverse of that for the integration, but then again, that doesn't look right either, so I'm still not sure. Does that look a little closer to what you're saying, at least? It looks like what I have there is just saying that the energy is just the integration of the momentum across each vector or something, but I would have to find v in terms of a ratio to the total speed or something somewhere to keep the units the same. I guess that's what you mean about the dot product. Aw heck, I don't know. I'll think about it some more. :sick:

You've got talent and a love for this, which I like to see, but Rome was not built in a day. You've got to learn a lot more math and physics and gain some discipline. But you're a quick study, all right. You just need to not jump the gun so much sometimes.Thanks. :) I don't know how not to jump the gun, though, although I think hh was also saying something similar, but more about my declarations. I don't always know when I have enough to run on and when I don't (I know I don't for that del thing yet, though ;) ), so I learn by doing. So far, though, I have only been running through the simplest integrations. It's good practice. There is actually so much more I want to do with it, but I'm taking my time, making sure I've got it down. With still what little bit I do know, I have no choice, but I try to innovate, to work around my ignorance, and push what I learn to the limit, to see how far it'll take me, tugging at it from different perspectives. But like they said in that movie "What abot Bob?", baby steps, right? I do that with even those tough looking integrations, just taking them one baby step at a time, writing and rewriting them for every little tranformation, sometimes using up several notebook pages, just to make sure I don't get lost somewhere in the middle. Look at how much Bob accomplished with that in the movie, too unsure to even try at first, and then eventually just diving right into it. But then again, he almost got blown up in the end. I wouldn't want that to happen to me. :)

grav
2007-Nov-09, 02:31 PM
Look at how much Bob accomplished with that in the movie, too unsure to even try at first, and then eventually just diving right into it. But then again, he almost got blown up in the end.And by his own mentor, no less. It seems he created a monster. But, but... you guys wouldn't do that to me, would you? :shifty: I'll try not to drive you too insane, I promise. :liar: :whistle:

hhEb09'1
2007-Nov-09, 07:07 PM
Oh, maybe something like using polar coordinates for a sphere? Not that I'm familiar with that or anything. It's generally easier for me to keep track of the integration for straight lines.The integration is the same, after you set it up. But instead of "stacking" little boxes ( dx dy dz), you stack wedges ( r2 sin θ dr dθ dφ ) of approx. uniform dimensions :) (Wiki page (http://en.wikipedia.org/wiki/Double_integral))

The area of a circle in polar coordinates would be the double integral where r goes from 0 to R, and θ goes from 0 to 2π, of the small area r dr dθ. The θ integral just gives you 2π r dr, and integrating that for r gives you 2π R2/2

I don't know how not to jump the gun, though, although I think hh was also saying something similar, but more about my declarations. I don't always know when I have enough to run on and when I don't (I know I don't for that del thing yet, though ;) ), so I learn by doing. Go, Bob, go! :)

grav
2007-Nov-10, 12:41 AM
Thanks, hh. I've just started reading through that link, but there's something in it I want to ask about real quick. About a third of the way through it, under 'normal domains of R^2', with the main heading "Formulae of reduction", they give an example there where they appear to come up with an integration for the area in the graph shown that becomes

Int (x + 1/2 - x^3 - x^4/2) dx

= x'^2/2 + x'/2 - x'^4/4 - x'^5/10

x' = xlimit = 0 to 1

= 1/2 + 1/2 - 1/4 - 1/10 = 13/20

But I'm wondering how they got that. When I do it the way I have been, with an integration like Int dx (1 - dy), or 1 - Int dx dy, depending upon how one chooses to look at it, then I get just

Int dx (1 - dy)

= x' - Int dx dy

Int y = y' = 0, x^2

= x' - Int dx x^2

= x' - x'^3/3

x' = 0, 1

= 1 - 1/3 = 2/3

I also ran it in small increments on the computer just to be sure, and it gives me the same thing. Am I looking at something the wrong way? They are integrating for the area shown in the graph, aren't they?

grav
2007-Nov-10, 03:56 AM
But I'm wondering how they got that. When I do it the way I have been, with an integration like Int dx (1 - dy), or 1 - Int dx dy, depending upon how one chooses to look at it, then I get just

Int dx (1 - dy)

= x' - Int dx dy

Int y = y' = 0, x^2

= x' - Int dx x^2

= x' - x'^3/3

x' = 0, 1

= 1 - 1/3 = 2/3

Oh, wait. The proper way to do that would have been

Int dx dy

{Int dy = y from x^2 to 1}

= Int (1 - x^2) dx

= Int dx - Int x^2 dx

{Int dx = x from 0 to 1}

{Int x^2 dx = x^3/3 from 0 to 1}

= (1 - 0) - (1^3/3 - 0^3/3)

= 2/3

grav
2007-Nov-10, 04:35 AM
Okay, I see what they must have done now. The original integration itself was confusing to me, but I see they must be taking a volume, where the length along any z direction is x+y. I wish they had been more specific about that. It's not in the graph and I'm have trouble even visualizing it. Anyway, here's the integration for that.

Int (x+y) dx dy

= Int x dx dy + Int y dx dy

{Int dy = y from x^2 to 1 = 1 - x^2}
{Int y dy = y^2/2 from x^2 to 1 = 1/2 - x^4/2}

= Int x (1 - x^2) dx + Int (1/2 - x^4/2) dx

= Int (x - x^3 + 1/2 - x^4/2) dx

{Int x dx = x^2/2 from 0 to 1 = 1^2/2 - 0^2/2 = 1/2}
{Int -x^3 dx = -x^4/4 from 0 to 1 = -1^4/4 - (-0^4/4) = -1/4}
{Int 1/2 dx = x/2 from 0 to 1 = 1/2 - 0/2 = 1/2}
{Int -x^4/2 dx = -x^5/10 from 0 to 1 = -1^5/10 - (-0^5/10) = -1/10}

= 1/2 - 1/4 + 1/2 - 1/10

= 13/20

Actually, that's basically just the same thing they had for that anyway, so I didn't need to write it out, but I wanted to show off my new forum-friendly style of writing out these integrations. :) It's faster to write out this way, yet still spelled out enough that it's easy to read and follow, at least I hope so. Compared to my other stuff, it is even much easier, without all that clutter, I'm sure, and I can keep track of what I'm doing even better now. So how does it look? Better?

publius
2007-Nov-10, 04:40 AM
Grav,

I was going to ramble a bit about the gradient. But Wiki did some rambling for me:

You've got a function f(r), where r is a position in some n-dimensional space. What is the differential of f?

df = df/dx *dx + df/dy *dy + df/dz *dz

df = (grad f) dot dr.

The gradient can be seen as a vector of the partial derivatives. It is the direction of quickest change of the function f at that point in space.

Oh what is a partial derivative, you ask. Good question. Long answer to be rigorous. Basically, you have function of two or more variables. The partials, denoted by a "funny looking" 'd'-looking symbol.

http://en.wikipedia.org/wiki/Partial_derivative

There is a lot more to all this good stuff............ A lot more. :)

-Richard

grav
2007-Nov-10, 05:17 AM
Okay, moving on, further down the link, those polar coordinates look like they would take some getting used to. I'm not used to thinking in circles. Err, well, then again... hmm.

Nope, I'm definitely a square.

publius
2007-Nov-10, 06:07 AM
Grav,

Now that you're familiar with double (or triple, or more) integrals, let's look at the big picture.

About that R^n notation. That bold 'R' denotes the continuum, the set of points on the real line. R^2 is the plane, two dimensions of the real line, R^3 is 3D space, and so on. Euclidean space, now mind you.

When we've got just one dimension, we've got this dx notion, this notion of a differential length along our dimension. In 2D we've got that, and two of them actually, but then we've got this notion of area, which I'll call dA. In 3D, we've got all of that, plus this notion of volume, some dV, we can call it.

Finding an area is just integrating dA. Now it turns out (under certain conditions which which the high priests of Principia Mathematica divine for us), this dA thing can be thought of as a dx*dy, and the integration of dA is equivalent to an iterated integral of dx*dy.

With that in mind, let's revisit the notion of the area under the graph of a function. Integral( f(x) dx). We can imagine that function is in a plane, and we want to integate dA between the curve and the x-axis.

That's just Int[dx dy]. We integrate y between y = 0 (the x-axis and y = f(x), the curve, and so get

Int dA = Int(dy dx) = Int (y dx), and there you go, back to the original f(x) dx integration.

But note that thinking of it as dA gives us more flexibility. Rather than doing the dy integration first, we can do dx first, if that suits us. There our inner bounds would be between x = constant and x = f(y).

-Richard

publius
2007-Nov-10, 06:14 AM
Grav,

Since you're such a quick study, let's see what you think of the following.

We've been considering areas under graphs and kid stuff like that, which you now see is a restricted from of the general dA for a plane.

Now, suppose we have a curve in the plane, and we want to determine the *arc length* of that curve between two points along it. That is, if we had a little bicycle wheel odometer and pushed along the curve, what would it read?

For instance, what would be the arc length along a parabola, given by y = x^2 between the endpoints at x = 0 and x = 1? That length is traditionally denoted by 's'.

We want to integrate something to get that, some 'ds' to get the total s along that curve. What would that be?

Just as we found the differential area under the curve was y dx, what would a differential length ds, along that curve be in terms of dx? Pythagoras leads the way. Think of small segment of that curve that can be approximated by a straight line. What is the length between the endpoints?

-Richard

hhEb09'1
2007-Nov-10, 05:01 PM
what would a differential length ds, along that curve be in terms of dx? Pythagoras leads the way. Think of small segment of that curve that can be approximated by a straight line. What is the length between the endpoints?Good one, I suggest you work this out, grav :)

grav
2007-Nov-10, 10:43 PM
Grav,

Since you're such a quick study, let's see what you think of the following.

We've been considering areas under graphs and kid stuff like that, which you now see is a restricted from of the general dA for a plane.

Now, suppose we have a curve in the plane, and we want to determine the *arc length* of that curve between two points along it. That is, if we had a little bicycle wheel odometer and pushed along the curve, what would it read?

For instance, what would be the arc length along a parabola, given by y = x^2 between the endpoints at x = 0 and x = 1? That length is traditionally denoted by 's'.

We want to integrate something to get that, some 'ds' to get the total s along that curve. What would that be?

Just as we found the differential area under the curve was y dx, what would a differential length ds, along that curve be in terms of dx? Pythagoras leads the way. Think of small segment of that curve that can be approximated by a straight line. What is the length between the endpoints?

-RichardCool. That's definitely not something I ever thought I would be doing on my own, at least not without writing it out in a computer program. But I'll still try integrating for it in the same way I would normally write it into the program, so it might not be like you intended, I'm not sure. So let's see, that would be something like

y = x^2, x' = x + dx, y' = (x + dx)^2

Int sqrt[(x' - x)^2 + (y' - y)^2] for the distance between points along the curve

= Int sqrt[dx^2 + (2x dx + dx^2)^2]

But the dx^2 in the second part of that falls out compared to 2x dx, being infinitesimal and all, so we have just

= Int sqrt(dx^2 + 4x^2 dx^2)

= Int sqrt(1 + 4x^2) dx

= Int 2 sqrt(x^2 + 1/4) dx

{ Int sqrt(x^2 + j) dx = (1/2)[x sqrt(x^2 + j) + j ln[x + sqrt(x^2 + j)] for x = 0 to 1, so (1/2)[sqrt(j+1) + j ln[1 + sqrt(j + 1)]] - (1/2)[0 + j ln[sqrt(j)]], where j = 1/4}

= 2 (1/2) [sqrt(5/4) + (1/4) ln[1 + sqrt(5/4)] - 0 - (1/4) ln(1/2)]

= sqrt(5)/2 + ln[1 + sqrt(5)/2]/4 - ln(1/2)/4

= 1.478942858

How's that? :)

Good one, I suggest you work this out, grav Yes, that was quite interesting. I had actually already worked most of this out before you posted this morning, but didn't have time to run back through it at the time. But I'm back now and ready for more. :)

publius
2007-Nov-10, 11:46 PM
Grav,

Excellent. I continue to be impressed. But let's clean up the notation a bit and write things in standard form.

First, you used x' and y' to mean the second endpoint. The prime notation suggests derivatives in this context, so let's don't use that -- use something like x1 and x2, y1 and y2.

Second, we don't need to worry about limits, doing the x + dx stuff, and taking limits. We're beyond that. Actually the dx there should be
*delta x*, dx is the limit as delta x goes to zero.

So all we need to note is, using Pythagoras (or we can call that the standard Euclidean *metric*, but we won't worry with that stuff now :) ) is

ds = sqrt( dx^2 + dy^2 ). Pull dx out, and

ds = sqrt[ 1 + (dy/dx)^2 ] dx, or sqrt[ 1 + (y')^2 ] dx

where I used y' to mean dy/dx. And that is the general formula for finding the arc length along some curve given by y = f(x)

And so, for a parabola y = x^2, dy/dx = y' = 2x and so our integration is indeed:

ds = sqrt(1 + 4x^2) dx from x = 0 to x = 1.

I'm bad to cheat, and look up a table of integrals to get that, such as here:

http://en.wikipedia.org/wiki/List_of_integrals_of_irrational_functions

Any standard math handbook will have a huge table of integrals, such as the exhaustive Abramowitz & Stegun:

http://en.wikipedia.org/wiki/Abramowitz_and_Stegun

But one doesn't want to cheat until one has worked a number of these out the long way to get the feel of it.

Good show, grav.

-Richard

publius
2007-Nov-11, 12:05 AM
Grav,

Ready for more, eh? Let's look at our arc length formula again in a slightly different light:

ds^2 = dx^2 + dy^2.

If we have our curve given by some y = f(x), we do as above. But what if we didn't. Rather than thinking of points on the curve as (x, f(x)), what if both x and y were *parameterized* by some other variable, call it 't'. Well,
dx = (dx/dt) dt, and dy = (dy/dt) dt, so

ds = sqrt[ (dx/dt)^2 + (dy/dt)^2 ] dt

Suppose a particle is moving along that curve, and t is time. x(t) and
y(t) are just its postition as a function of time, and so dx/dt and dy/dt are the components of its velocity. The square root is the speed, the magnitude of the velocity vector. So the above is just saying:

ds = v dt. Well, DUH!, we say. :)

With that in mind, suppose we are standing on the ground and fire a projectile at some intial velocity, (v_x, v_y) under the influence of a constant g field.

What is the total distance the particle will travel before it hits the ground. Here, ignoring air resistance, assuming uniform g, and letting x be the horizontal and y the vertical, and we fire at the origin, our equation of motion is simply:

x(t) = v_x * t

y(t) = v_y * t - g/2 * t^2

The particle will hit the ground at some t, call it t_f ( f is for final), what is x(t_f) and y(t_f), and what is the total arc length s(t_f) the particle travels before it hits the ground?

-Richard

grav
2007-Nov-11, 12:17 AM
Good show, grav.Thanks, Richard. :)

I'm bad to cheat, and look up a table of integrals to get that, such as here:

http://en.wikipedia.org/wiki/List_of_integrals_of_irrational_functions

Any standard math handbook will have a huge table of integrals, such as the exhaustive Abramowitz & Stegun:

http://en.wikipedia.org/wiki/Abramowitz_and_Stegun

But one doesn't want to cheat until one has worked a number of these out the long way to get the feel of it.Um, I've been looking up the integrals for most of these, except for the easy ones and those I've used so much I can remember them now. I've been steadily compiling a stack of tables, and your links just went into the pile. So, what is the long way, besides the trial and error way I used to attempt to find them, that is?

grav
2007-Nov-11, 03:39 AM
Grav,

Ready for more, eh? Let's look at our arc length formula again in a slightly different light:

ds^2 = dx^2 + dy^2.

If we have our curve given by some y = f(x), we do as above. But what if we didn't. Rather than thinking of points on the curve as (x, f(x)), what if both x and y were *parameterized* by some other variable, call it 't'. Well,
dx = (dx/dt) dt, and dy = (dy/dt) dt, so

ds = sqrt[ (dx/dt)^2 + (dy/dt)^2 ] dt

Suppose a particle is moving along that curve, and t is time. x(t) and
y(t) are just its postition as a function of time, and so dx/dt and dy/dt are the components of its velocity. The square root is the speed, the magnitude of the velocity vector. So the above is just saying:

ds = v dt. Well, DUH!, we say. :)

With that in mind, suppose we are standing on the ground and fire a projectile at some intial velocity, (v_x, v_y) under the influence of a constant g field.

What is the total distance the particle will travel before it hits the ground. Here, ignoring air resistance, assuming uniform g, and letting x be the horizontal and y the vertical, and we fire at the origin, our equation of motion is simply:

x(t) = v_x * t

y(t) = v_y * t - g/2 * t^2

The particle will hit the ground at some t, call it t_f ( f is for final), what is x(t_f) and y(t_f), and what is the total arc length s(t_f) the particle travels before it hits the ground?

-RichardWow. That was quite a whopper! I probably did it the hard way, but I just hope I got it right. I'll cross my fingers on this one. Okay, you didn't say at what angle the projectile was fired, so I'll just find for the projectile fired at any angle at speed v with vx = (cos L) v and vy = (sin L) v. You asked to find x(t_f) and y(t_f), but it seems y(t_f) would just be the given for when the projectile hits the ground at y(t_f) = 0. So y(t_f) = 0 = vy t_f - g t_f^2 / 2, so vy = g t_f / 2, therefore t_f = 2 vy / g, and x(t_f) = vx t_f = 2 vx vy /g = 2 (cos L) (sin L) v^2 / g = (sin 2L) v^2 / g. As for the length of the arc, I get

Int sqrt[dx^2 + dy^2]

= Int sqrt[(vx (t + dt) - vx t)^2 + ((vy (t + dt) - g (t + dt)^2 / 2) - (vy t - g t^ / 2))^2]

= Int sqrt[(vx dt)^2 + (vy dt - g (2 t dt + dt^2) / 2)^2]

= Int sqrt[vx^2 dt^2 + (vy dt - g t dt)^2]

= Int sqrt[vx^2 + vy^2 - 2 vy g t + g^2 t^2] dt

= Int sqrt[v^2 - 2 vy g t + g^2 t^2] dt

{ Int sqrt[jt^2 + kt + m] = (1 / (8j^(3/2))) [2 sqrt(j) (k + 2j t) sqrt(m + t(k+jt)) - (k^2 - 4jm) ln((k+2j t) / sqrt(j) + 2 sqrt(m + t(k + j t))], t = 0 to 2 vy / g, j = g^2, k = -2 vy g, m = v^2}

= (1 / (8 g^3)) [2 g (4 vy g - 2 vy g) sqrt(v^2 + (2 vy / g) (2 vy g - 2 vy g)) - (4 vy^2 g^2 - 4 v^2 g^2) ln((4 vy g - 2 vy g) / g + 2 sqrt(v^2 + (2 vy / g) (2 vy g - 2 vy g))] - (1 / (8 g^3)) [2 g (-2 vy g) v - (4 vx^2 g^2) ln(-2 vy + 2 v)]

= (1 / (8 g^3)) [4 vy g^2 v - (4 vx^2 g^2) ln(2 vy + 2 v)] - (1 / (8 g^3)) [-4 vy g^2 v - (4 vx^2 g^2) ln(-2 vy + v)]

= vy v / g - (vx^2 / 2g) [ln(2v - 2vy) - ln(2v + vy)]

= (sin L) v^2 / g + ((cos L)^2 v^2 / 2g) ln[(2v - 2vy) / (2v + 2vy)]

= (v^2 / g) [(sin L) + (cos L)^2 ln((v^2 - vy^2) / (v + vy)^2) / 2]

= (v^2 / g) [(sin L) + (cos L)^2 ln((cos L)^2 / (1 + (sin L))^2 ) / 2]

Please tell me I got it. I'm holding my breath. :sick:

publius
2007-Nov-11, 04:21 AM
Wow. That was quite a whopper!
..............
Please tell me I got it. I'm holding my breath. :sick:

Grav,

Hell, I don't know -- I just thought it up and haven't worked it out myself. :lol: Best start breathing again, it may be a while before I get one of those round tuit things. :whistle: Seriously, I had you figure out the arc length along a parabola, and since the trajectory of our particle is a parabola, it should be the same, just a slightly different form with the time (with a t plus a t^2 term) And it looks like you were on track there -- looks good to me, but I'll check when I get that round 'tuit.

And yes, t_f is exactly right, v_y(t) = 0. And I'm sure you noted that maximum horizontal "range", x(t_f), occurs when that angle is 45 degrees.

And about the angle. You didn't really need to worry about that, because I just set it up with a v_x and v_y directly rather than a total v and an angle -- that just eliminated worrying about v*cosO and v*sinO. But if you like to think about it that way, v and the angle, more power to you. You said you were a "square" kind of guy anyway, so I figured you'd think Cartesian components of v anyway.... :)

-Richard

publius
2007-Nov-11, 04:32 AM
Grav,

The integral of sqrt(a*x^2 + bx + c) is a goldanged doosie. I had forgotten about that and didn't realize that was such a humdinger. But you did it without blinking, looks like. Yes sir, you ain't no slacker.

-Richard

grav
2007-Nov-11, 04:47 AM
Grav,

The integral of sqrt(a*x^2 + bx + c) is a goldanged doosie. I had forgotten about that and didn't realize that was such a humdinger. But you did it without blinking, looks like. Yes sir, you ain't no slacker.

-Richard:)

publius
2007-Nov-11, 06:27 AM
Grav,

Yessiree bobtail, you're on a roll. Quick study, indeed. Let's go up one dimension.......

Note what we did with arc length. We've got a curve above the x-axis, and get a "curvy" ds from moving along a dx on the x-axis. Let's go up one dimension. Imagine a curved surface above the xy plane given by some
z = f(x, y). For example, z = x + y is a simple plane.

Can we find the surface area of such a curvy surface above the plane. Can we find something for dS (note that capital S, which is tradition for this notion of surface area) in terms of dA = dx*dy in the plane?

That is can we get S over some region in the plane by doing some double integration of dS = (something) dx*dy?

Note our arc length element, ds is just the hypotenuse of a right triangle. We can write ds*cosO = dx, where O is the slope of the tangent at that point.

Imagine a differential piece of surface area. Just as there is a tangent line to a curve, there is a tangent plane to a surface. Note that tangent plane will be cocked at some angle O to the XY plane, there. What is the projection of the *area*, there? Might it be some dS*cos O = dA = dy*dx?

Now, don't look until you've thought about it a while, but I'm gonna post the answer below, using "spoiler text" style. Just drag the mouse over the following to read it:

dS = sqrt[1+ (dz/dy)^2 + (dz/dx)^2] dy dx

where dz/dy and dz/dx are the partial derivatives of z = f(x, y). Now, ain't that just cool?

-Richard

publius
2007-Nov-11, 06:48 AM
Grav,

Note what we've been doing above with arc lengths, curved surface areas, and all that is geometry. Geo metry. Measuring "stuff". But we've used calculus to get a bit fancy, and are using *differentials*. So, we're sort of doing a differential geometry.............. :)

Well, we're taking little baby steps, sort of just getting started and seeing some basic ideas.

Full differential geometry takes all these concepts and generalizes them fully to deal with whatever coordinates we like in as many dimensions as we like. And our space we're working in doesn't have to be Euclidean.

Yes, the roots of General Relativity are found right in what we are doing, here.

-Richard

publius
2007-Nov-11, 05:03 PM
Grav,

Integrating even sqrt( a^2 + x^2) the "long way" is pretty darned involved. I was going to do it, then quickly realized it wasn't trivial at all. :)

Go here:

http://en.wikibooks.org/wiki/Calculus/Further_integration_techniques#Trigonometric_Subst itution

and look at the tangent substitution section to see how that is done. One uses a tangent trig substitution, then integrates by parts *twice*. But one isn't done there, and a little trick happens. One of the integrations by parts returns the original integral, and that allows you to solve for it.

You get something like this:

I = f(x) - k*I + g(x), where I is the desired integral.

You then solve for I.

-Richard

grav
2007-Nov-11, 11:48 PM
Uh oh. I must be getting into uncharted area there. I knew it would be along the lines of what you had there, but I had to look at your spoiler. But even then, I still couldn't figure out how to get it. The best I can figure, once I slowed down and thought through it some more over a cup of coffee, is that sqrt(dx^2 + dz^2) would be a small piece of an arc along x and z, so if we extend that piece of arc across y, for a small piece in that direction, the length across that small piece of the arc for y and z would be sqrt(dy^2 + dz^2). We can then find for infinitesimal areas of little sections of the total area with the dimensions sqrt(dx^2 + dz^2) * sqrt(dy^2 + dz^2) = sqrt[(dx^2 +dz^2) (dy^2 + dz^2)] = sqrt[dx^2 dy^2 + dx^2 dz^2 + dy^2 dz^2 + dz^4]. Now here I'm thinking dz^4 is small compared to components of dz^2, so dropping that, it reduces to sqrt[dx^2 dy^2 + dx^2 dz^2 + dy^2 dz^2] = sqrt[dx^2 dy^2 (1 + (dz / dy)^2 +(dz / dx)^2)] = sqrt[1 + (dz / dx)^2 + (dz / dy)^2] dx dy.

The thing is, though, I'm not absolutely sure if I should have dropped that dz^4. I would if I knew the rest of the variables were constants, and of course, taken one by one for each integration, that would be the case, so maybe so. It just seems strange that we should drop it for higher exponentials of the same infinitesimal variable, but not for for the same total exponent with different infinitesimal variables, that is, if that's what we do. If so, though, that would be a handy thing to know. That would mean that integrating for something like (dx+dy)^2 would break down into dx^2 + 2 dx dy + dy^2, and dropping out the higher exponents of the same variables, then, we would just be left with the integration of 2 dx dy. But then again, it may also be something about defining dz in terms of x and y to begin with that might cause only the higher exponent of dz to drop out, I'm not really sure.

grav
2007-Nov-12, 12:11 AM
Integrating even sqrt( a^2 + x^2) the "long way" is pretty darned involved. I was going to do it, then quickly realized it wasn't trivial at all. :)Yes, I think I'll just stick to the table of integrals. Besides, it's really just the solutions for the integrations we want anyway, not so much finding for the integrals themselves, although I'm thinking about a pet project for trying to find one grand integral that would include the solutions for anything smaller, kind of like a theory of everything for integrals, at least for all of the most common and related ones, since I've noticed some distinct patterns in them, so that if I can just memorize that one (or a few simple ones), then I won't need to refer to a table most of the time.

publius
2007-Nov-12, 12:58 AM
The best I can figure, once I slowed down and thought through it some more over a cup of coffee, is that sqrt(dx^2 + dz^2) would be a small piece of an arc along x and z, so if we extend that piece of arc across y, for a small piece in that direction, the length across that small piece of the arc for y and z would be sqrt(dy^2 + dz^2). We can then find for infinitesimal areas of little sections of the total area with the dimensions sqrt(dx^2 + dz^2) * sqrt(dy^2 + dz^2) = sqrt[(dx^2 +dz^2) (dy^2 + dz^2)] = sqrt[dx^2 dy^2 + dx^2 dz^2 + dy^2 dz^2 + dz^4].

Grav,

You're right on track. sqrt[1 + (dz/dx)^2] and sqrt[1 + (dz/dy)^2] are indeed the arc length along that surface holding y and then x constant. Indeed they are lengths of tangent vectors along the surface at that point. Note for a curve we have a tangent line, but for a surface we have the notion of a tangent *surface*. Both of those tangent vectors lie in and define the tangent plane to the surface at that point.

But think about them. Are they perpendicular? If they were, the area would be a rectangle, and the area would be the direct product of their lengths, call it a and b, for S = ab.

But, they aren't perpendicular, are they? So what is the formula for the area of parallelogram of lengths a and b, and some angle O between them?

-Richard

publius
2007-Nov-13, 11:16 PM
Grav,

Since you haven't replied yet, I'll go ahead with this. The area element,
dS, is that of a parallelogram formed by the two differential lengths of the x and y tangent vectors.

dS = Tx*Ty sinO dx dy = Tx*Ty sinO dA

where dA = dx dy is sort of the "coordinate area element" (hint, hint).

Now, I don't know how familiar you are with vector algebra, but the magnitude of the cross product of two vectors is exactly that,

|A x B| = AB sin O

So we can write:

dS = |Tx x Ty| dA

Where, Tx (vector) is (1, 0, dz/dx) and Ty is (0, 1, dz/dy)

This holds in the general case where we've paramterized the surface in any convienent way we like. Just as we can paramterize a curve by some 't' (or s), we can parameterize a surface by two arbitrary variables, typically denoted by u and v. So we can write:

dS = |Tu x Tv| dA

What is the vector Tu x Tv (where u and v could be x and y)? That is the *normal* vector to the surface at that point, generally denoted by capital 'N'.

So dS = | N | dA. The magnitude of the normal vector is the ratio of the actual area to the "coordinate area", and the direction is perpendicular ("normal") to the tangent plane of the surface at that point.

Now, that's you vector calculus lesson for today. :) More later when I get a round tuit.

-Richard

grav
2007-Nov-14, 02:22 AM
Grav,

Since you haven't replied yet, I'll go ahead with this. The area element,
dS, is that of a parallelogram formed by the two differential lengths of the x and y tangent vectors.

dS = Tx*Ty sinO dx dy = Tx*Ty sinO dA

where dA = dx dy is sort of the "coordinate area element" (hint, hint).

Now, I don't know how familiar you are with vector algebra, but the magnitude of the cross product of two vectors is exactly that,

|A x B| = AB sin O

So we can write:

dS = |Tx x Ty| dA

Where, Tx (vector) is (1, 0, dz/dx) and Ty is (0, 1, dz/dy)

This holds in the general case where we've paramterized the surface in any convienent way we like. Just as we can paramterize a curve by some 't' (or s), we can parameterize a surface by two arbitrary variables, typically denoted by u and v. So we can write:

dS = |Tu x Tv| dA

What is the vector Tu x Tv (where u and v could be x and y)? That is the *normal* vector to the surface at that point, generally denoted by capital 'N'.

So dS = | N | dA. The magnitude of the normal vector is the ratio of the actual area to the "coordinate area", and the direction is perpendicular ("normal") to the tangent plane of the surface at that point.

Now, that's you vector calculus lesson for today. :) More later when I get a round tuit.

-RichardDang. I don't understand that either. I've steadily been trying to find for that area the last few days. I figured the dz factor across x to be dj and across y as dk (partial derivatives or whatever), and all I've been able to come up with so far is that dj + dk = dz, where dz would be the the height of the opposite corner from the "origin", that being the given, if any of that makes any sense to you. Anyway, the area of the square would then be sqrt(dx^2 + dj^2) * sqrt(dy^2 + dk^2), unlike what I had before, but I've tried countless ways to find for that, where dj and dk might ultimately fall out, using dj + dk = dz and Heron's formula for a triangle by splitting the square along the main diagonal, and by finding other similar equalities and trying to relate them in such a way that the factors for dj and dk cancel out and so forth, but all to no use. I've been reading back through what you've posted and those links, and I still don't get it. I'll read back through what you just wrote some more and see if something might eventually click in.

grav
2007-Nov-14, 02:55 AM
Oh, wait. A parallelogram, of course. You mentioned that before but I didn't quite take it in the right way, apparently. So two angles climbing up along x and y from the origin would provide an angle between them that is less than ninety degrees, while I've still been stubbornly thinking in terms of a square. Okay, I'll start all over again.

publius
2007-Nov-14, 04:06 AM
Oh, wait. A parallelogram, of course.

Bingo! I was about to go over it again, but you got it.

The area of that parallelogram is simple Tx Ty sin O, which is the magnitude of their cross product.

The length of Tx is simply sqrt(1 + (dz/dx)^2) dx, and that of Ty is simple sqrt(1 + (dz/dy)^2) dy. Multiply those together and you get:

sqrt[ 1 + (dz/dx)^2 + (dz/dy)^2 + (dz/dx)^2*(dz/dy)^2 ] dx dy.

The effect of the sin O term is to make that last cross term go away.

Again, grav, I don't know if you're too familiar with vector algebra, but it can make dealing with angles like this much easier, more mechanical (except when it doesn't and an angle is easier and more intuitive).

At any rate, the cross product of two vectors A x B has a magnitude equal to AB sin O, where O is the angle between them, and it points normal to both A and B. IOW, it is perpendicular to the plane formed by A and B.

In component form, that cross product goes like this:

i j k

1 0 dz/dx

0 1 dz/dy

Taking the symbolic determinant:

Tx x Ty = -dz/dx i - dz/dy j + 1 k (Oh, i j and k are unit vectors along the x, y, and z directions -- that goes way back to Hamiliton, actually).

That is the components of N, the normal vector, Tx x Ty are
-dz/dx in the x direction, -dz/dy in the y direction, and 1 in the z direction.

The magnitude is simply the square root of the sum of the squares, and so

|N| = sqrt( 1 + (dz/dx)^2 + (dz/dy)^2 ).

-Richard

publius
2007-Nov-14, 06:04 AM
There is some curious about this normal vector, here. Write it in components,

N = (-dz/dx, -dz/dy, 1). You'll note it's dot product with k, the unit vector in the z direction, (0, 0, 1) is identically equal to one:

N dot k = 1.

So, dS = |N| dA = |N| dA/ (N dot k), since that denominator is always 1.

And that is just

dS = dA / ( N/|N| dot k) = dA/(n dot k), where n is the unit normal vector. n dot k is just the cosine of the angle between the normal and the vertical z axis, which is the same angle as the inclination of the tangent plane to the x-y plane.

so dS = dA/(cos O). That is the exact same relation as that of the arc length of a curve in the plane to dx. dx = ds cos O, where O is the angle of the tangent vector to the x-axis. dS cos O is the projection of the area onto the xy plane below.

Now, that is an alternate way to derive the formula for dS, and many texts intially take that route, finding 'n' and noting the cosine of the angle is n dot k. Here, we just started out with N from Tx x Ty directly.

And this sort of odd result, N dot K always equal to one, makes an important point. That only holds for a surface that we've parameterized in terms of z = f(x, y) -- that is, some surface that can be easily expressed in that form. Not all surfaces can, or even if they can, it is very complex.

For example, we've got problems with a simple sphere.

z = +/-sqrt( r^2 - x^2 - y^2). That is very problematic. First of all, it is not single valued. We've got to two z's there for every x and y, and the derivatives blows up in the plane when x^2 + y^2 = r^2.

IOW, N is coordinate dependent. Wait a minute, you say, a normal is a normal. Well, that's true for the *unit normal*, the *DIRECTION* of the normal, but not its magnitude. Its magnitude is indeed a coordinate dependent thing.

A much better coordinatization of a spherical surface is this:

z = r cos u, y = r cos u sin v, x = r cos u cos v

where 0 <= u <= pi, and 0 <= v <= 2pi (sneaky way to use spherical coordinates :lol:, but completely fine).

Now, dS remains |Tu x Tv| du dv, but you'll note that that N dot k is no longer 1. However the unit n of both is the same thing, as it must be.

That does not suffer from blowing up in the plane.

-Richard

publius
2007-Nov-14, 06:12 AM
I'm going to let the cat out of the bag and tell you where we're going. You know me, I love to ramble about General Relativity.

Believe or not, with our surface area sojourn, I'm building something to show how "simple" GR really is. :)

What is going on at the event horizon of a black hole is exactly (conceptually mathematically) as what was going on with our "poor" choice of x-y coordiantes for the spherical surface. :) Indeed, an event horizon (conceptually) exists right there when x^2 + y^2 = 1.

100 points and I'll raise my glass in salute to the first one to 'splain how that be.

-Richard

grav
2007-Nov-14, 06:41 AM
Again, I'm not sure about all you are saying there, but I can identify somewhat with what you were referring to with the sphere, since I was also trying to find for the surface of that, but it seems so far that it can probably only be done in terms of pi and/or sine and cosine. Anyway, it looks like my first assessment for a surface was closest to what you had, but I didn't find it in terms of a parallelogram. I might never have gotten it anyway after I started working on it some more, even if I did associate it with a parallelogram, since it appears I was giving too much flexibility to the surface. I began thinking of it in terms of dj and dk for the height along each axis, whereas the surface could pivot around the diagonal from the origin to the opposite corner, like slicing a cube at some angle between the origin and the opposite corner. But it looks like the cube can only be sliced in a straight cut from the origin to the opposite corner, where the height at symmetrical points on either side is the same (if you can following what I am attempting to describe here :) ) . It would seem the surface should have more flexibility than that, though, but then again, that would be for a cube only, so if the length of dy varies to dx, then then slope along that side of the cube will vary also.

Anyway, so if we have a height of dz on both of the left hand and right hand corners along the angles to the origin, then the opposite corner of the cube (or rectangular box) must be 2 dz. That is about all I managed to figure out in all that time, that dj + dk = the height of the opposite corner, where here, dj = dk = dz, so the height of the opposite corner is 2 dz. But using that then, for this type of sliced cube, we can find the area of the surface using Heron's formula for the area of a triangle knowing the sides (which I discovered independently, by the way, quite a few years back :) ) . So the length of each side of the triangle is a = sqrt(dx^2 + dz^2), b = sqrt(dy^2 + dz^2), and c = sqrt(dx^2 + dy^2 + (2 dz)^2) across the diagonal from the origin to the opposite corner. So we have

A = 2 * (1/4) sqrt(2 a^2 b^2 + 2 a^2 c^2 + 2 b^2 c^2 - a^4 - b^4 - c^4)

(2A)^2 = 2(dx^2 + dz^2)(dy^2 + dz^2) + 2(dx^2 + dz^2)(dx^2 + dy^2 + 4dz^2) + 2(dy^2 + dz^2)(dx^2 + dy^2 + 4dz^2) - (dx^2 + dz^2)^2 - (dy^2 + dz^2)^2 - (dx^2 + dy^2 + 4dz^2)^2

= 2 dx^2 dy^2 + 2 dx^2 dz^2 + 2 dy^2 dz^2 + 2 dz^4 + 2 dx^4 + 2 dx^2 dy^2 + 8 dx^2 dz^2 + 2 dx^2 dz^2 + 2 dy^2 dz^2 + 8 dz^4 + 2 dx^2 dy^2 + 2 dy^4 + 8 dy^2 dz^2 + 2 dx^2 dz^2 + 2 dy^2 dz^2 + 8 dz^4 - dx^4 - 2 dx^2 dz^2 - dz^4 - dy^4 - 2 dy^2 dz^2 - dz^4 - dx^4 - dy^4 - dz^4 - 2 dx^2 dy^2 - 8 dx^2 dz^2 - 8 dy^2 dz^2

which, after cancelling positives and negatives, somewhat surprisingly reduces to just

(2A)^2 = 4 dx^2 dy^2 + 4 dx^2 dz^2 + 4 dy^2 dz^2

so A = sqrt(dx^2 dy^2 + dx^2 dz^2 + dy^2 dz^2)

= sqrt(1 + (dz / dy)^2 + (dz / dx)^2) dx dy

Finally! :)

grav
2007-Nov-14, 10:43 PM
I was thinking about this some more today, and I thought it was sort of strange that we should integrate using the surface of a parallelogram across rectangular boxes of the dimensions dx ,dy, and 2 dz, instead of just dx, dy and dz. I figured that if we were to integrate for, say, a sphere, the dimensions should be symmetrical in all directions, whereas we might just as well integrate for the surface across rectangular boxes of dimensions 2 dx, dy, and dz, or dx, 2 dy, and dz. But that would not make sense either, at least not to me.

But to cut 2 dz down to half the height would also cut the parallelogram in half, producing a triangle, and also bring the integration down to dS = Int sqrt(dx dy + dx dz + dy dz) / 2. However, I'm also thinking that it would be much easier to integrate for a surface using triangles than parallelograms, considering that by connecting any three points in any arbitrary manner, no matter how they are positioned, will produce a triangle, so we can fill the surface in this way in any way we wish, but not so much by using four arbitrary points for a parallelogram. The surface area across the rectangular box also makes more sense to me, since we would simply be integrating across dx and dy, dy and dz, and dx and dz. That is, taking a rectangular box of dimensions dx, dy, dz, and finding for ds = sqrt(dx^2 + dz^2) along one side from the origin (at a lower corner) to one upper corner along the x axis at a length of dx and a height of dz along the z axis, and ds = sqrt(dy^2 + dz^2) along the other, we can also find across dx and dy by cutting from one upper corner to the other upper corner with sqrt(dx^2 + dy^2). The result is a triangle with those lengths for sides, and we can find the area using Heron's formula, the same as I did before, but cut cross-wise across the original parallelogram from the way I did it the first time.

Now, that is my theory anyway, but my thinking might still be a little off. If it is right, though, then if I integrated for the surface of a sphere using dS = sqrt(dx dy + dx dz + dy dz), then I should find the area to be twice what it should be, or 8 pi r^2. But so far, I do not seem to be able to use that formula to integrate for the surface of anything, a cone, a sphere, nothing, because it keeps giving me complicated unknown integrals, rendering them unworkable so far. How come?

grav
2007-Nov-15, 12:13 AM
Now, that is my theory anyway, but my thinking might still be a little off. If it is right, though, then if I integrated for the surface of a sphere using dS = sqrt(dx dy + dx dz + dy dz), then I should find the area to be twice what it should be, or 8 pi r^2.Okay, thinking about it further, if we take a rectangular box and find for the area of a triangular surface from the origin at a lower corner to each of the upper adjacent corners along perpendicular axes, then we are essentially cutting the box in half in the process, diagonally across the upper corners. That is to say that only the volume from origin to corner to corner will be accounted for, and that from each of the adjacent corners to the corner opposite the origin will not. So in order to account for the other half of the box as well in a rudimentary way, but logical, I think, we can just double the area of the triangle to account for the other half of the box in the same way, and it shouldn't affect the integration when dealing with infinitesimals overall, but effectively doubling the area of the triangle, bringing the integration back up to sqrt(dx dy + dx dz + dy dz) = sqrt(1 + (dz / dy)^2 + (dz / dx)^2) dx dy, but figuring it for a box of dimensions of dx, dy, dz all the way.

Also, one could still apply these triangles to a surface, though, by connecting sets of three points across it, but then one would be effectively figuring for half the volume of each box for each triangle, so the integration would come out the same.

grav
2007-Nov-15, 12:35 AM
Not being able to use that formula to integrate for anything, however, is quite frustrating. :wall: Please help.

publius
2007-Nov-15, 01:28 AM
Grav,

Settle down there pardner. :) Grav, I think you're getting hung up on this "2 dz" thing, somehow and I'm not sure. We are integrating a surface area, not a volume. Forget about any little 3D boxes. What we're concerned with is the little patch of surface area formed by those two tangent vectors.

The length of the one on the x side is sqrt(dx^2 + dz^2). Factor the dx out --> ds_x = sqrt( 1 + (dz/dx)^2) dx, and the same for the y side
ds_y = sqrt(1 + (dz/dy)^2) dy.

Those dz's there are *two different ones*. One is the change in z due to varying x and holding y constant, and the other due to varying y while holding x constant. They are two different dz's. We are not forming a 3D box of height 2dz or anything. We're just looking at two tangent vectors in the full 3D space.

Those two line segments form a plane, the tangent plane to the surface at that point. That's all. And we integrate the area of the parallelogram formed by them in that plane:

dS = ds_x*ds_y*sin O, where O is the angle between ds_x and ds_y.

-Richard

grav
2007-Nov-15, 01:49 AM
Yes, I think I've got a grasp on that now, albeit with a slightly different interpretation, but with the same overall result. It's solving for the integration itself I'm having trouble with at this point. How would you perform that? :confused:

grav
2007-Nov-15, 01:04 PM
Oh, whoops. :doh: I'm tremendously sorry I bothered you with all of that "can't perform the integration" stuff yesterday, Richard. It might have helped if I was attempting to perform the right integration. :shifty: It's right there in my posts. I was trying to find for dS = Int sqrt(dx dy + dx dz + dy dz), so I was ending up with square roots within square roots, while the actual integration should have been dS = Int sqrt(dx^2 dy^2 + dx^2 dz^2 + dy^2 dz^2). Looks like I'm starting to trip over myself again. :sad: I'll try again later.

grav
2007-Nov-16, 02:46 AM
Okay, so the integration for the surface of a sphere. I can't actually use Int sqrt[dx^2 dy^2 + dx^2 dz^2 + dy^2 dz^2] directly anyway, I don't suppose, so I must take Int sqrt[1 + (dz/dy)^2 + (dz/dx)^2] dx dy and find for the partial derivatives. I was confused about that at first, but I think I have it now. So the partial derivative of dz/dx is

dz/dx = sqrt(r^2 - x^2 -y^2)' in terms of x, since z = r^2 - x^2 - y^2

F(G(x)) = sqrt(a - x^2), where a = r^2 - y^2, and I'll make F(x) = sqrt(x) and G(x) = a - x^2

F'(x) = (1/2) / sqrt(x) and G'(x) = -2x, so F'(G(x)) G'(x) = -x / sqrt(a - x^2) = -x / sqrt(r^2 - x^2 - y^2)

For dz/dy, that becomes dz/dy = -y / sqrt(r^2 - x^2 - y^2)

So now we get

Int sqrt[1 + (dz/dy)^2 + (dz/dx)^2] dx dy

= Int 2 sqrt[1 + x^2 / (r^2 - x^2 - y^2) + y^2 / (r^2 - x^2 - y^2)] dx dy for -z to z

= Int 2 sqrt[((r^2 - x^2 - y^2) + x^2 + y^2) / (r^2 - x^2 -y^2)] dx dy

= Int 2r sqrt[1 / (r^2 - x^2 - y^2)] dx dy

{Int dy / sqrt(a - y^2) = asin(y / sqrt(a)), a = r^2 - x^2, y = sqrt(r^2 - x^2), so y / sqrt(a) = 1, and asin(1) - asin(-1) = pi/2 - (-pi/2) = pi}

= Int 2 pi r dx

= 4 pi r^2 for x = -r to r

Well, imagine that. Good deal. :)

grav
2007-Nov-16, 03:24 AM
Grav,

Integrating even sqrt( a^2 + x^2) the "long way" is pretty darned involved. I was going to do it, then quickly realized it wasn't trivial at all. :)

Go here:

http://en.wikibooks.org/wiki/Calculus/Further_integration_techniques#Trigonometric_Subst itution

and look at the tangent substitution section to see how that is done. One uses a tangent trig substitution, then integrates by parts *twice*. But one isn't done there, and a little trick happens. One of the integrations by parts returns the original integral, and that allows you to solve for it.

You get something like this:

I = f(x) - k*I + g(x), where I is the desired integral.

You then solve for I.

-RichardOh, yeah. there's also the reverse engineered product rule hh showed me a while back in this thread. I had already forgotten. I'll need to keep going back through the material in this thread periodically to be sure I don't forget anything I can make use of later on. And I thank you both for being so patient as to go through all of this with me.

publius
2007-Nov-16, 03:46 AM
Grav,

Good show -- I though I remembered the surface area of the sphere could be done tractably in cartesians. Tedious, but tractable, and I think you did it. :)

-Richard

hhEb09'1
2007-Nov-16, 09:59 AM
Oh, yeah. there's also the reverse engineered product rule hh showed me a while back in this thread. I had already forgotten. I'll need to keep going back through the material in this thread periodically to be sure I don't forget anything I can make use of later on. And I thank you both for being so patient as to go through all of this with me.I am amazed at how far you've gone in such a short time, and at how long you resisted it! :)
Good showDitto :)

grav
2007-Nov-16, 12:46 PM
Grav,

Well, for the surface area integral, note we have a solid of *revolution*. The cross section is circular. The circumference of a circle is 2pi*r. So the circumference of each circle at y = 1/x is 2pi*y = 2pi/x.

We integrate that from 1 to some x = a. The integral of that is
2pi*(ln(a) - ln(1)). That diverges as a --> infinity. Thus, the surface area of that solid is infinite. Actually, that's not the true surface area formula. :) To get the actual surface area, we need to know the arc length, which is another formula, but the arc length is *greater* than dx (the arc length of a curve is actually sqrt(1 + y'^2)dx). So we know the result is "greater than infinity", which means it's infinite. :)

... [snip] ...

That surface is known as "Gabriel's Horn", and this is a famous "paradox".

-RichardI see what you mean now about using the arc to find the true surface integration. I figured something like that before, but wasn't sure how to think about it. I'll have to be careful of that, and be sure to safeguard against it. So basically, whenever there's a slope along all three dimensions, looks like, we must use dS. Only if one of the dimensions is constant, like with finding for the surface area of a cylinder of radius r and length 2r, minus the ends, as compared to that of finding for the surface of a sphere with the same general length along each of the dimensions, can we use the circumferences in this case like we did for Gabriel's horn, which would then come out to Int 2 pi r dx = 4 pi r^2 for x = -r to r, same as we had for the sphere, and finding for dA would give us Int dx dy = Int 2 sqrt(r^2 - x^2) dx = pi r^2 for x = -r to r, which is just the cross-sectional area, the same as taking circumferences out from the center of a circular area along a plane, but not along the length, so I'll have to watch for that too.

grav
2007-Nov-16, 01:36 PM
I am amazed at how far you've gone in such a short time, and at how long you resisted it! :)Thanks again. I can't believe I've gone this long without learning it either, actually, at this point anyway. I've tried before, but it seemed too involved for me to work through on my own. It definitely helps to have such good mentors as the two of you. You guys have been wonderful! :)

hhEb09'1
2007-Nov-16, 02:07 PM
I've tried before, but it seemed too involved for me to work through on my own. It has been somewhat mystified and definitely jargonized, all sorts of aspects of it are intimidating. But I think that's true of almost any specialized knowledge, probably. Most of the conclusions have been arrived at via common sense, but they're not so obvious out of context. It'd be nice to have a neat little common sense introduction to the topics, but unfortunately not everyone has the same idea of common sense! :)

Have we started the wave equation, yet? I forget. :)

publius
2007-Nov-18, 05:17 PM
Grav,

Now let's look at a more general way to deal with curved surfaces like this. A curve is 1D, it can just curve all around in the larger space in which it is embedded.

In vector lingo, a curve is specified by some R(t), where R is the position vector. While 't' can be thought of as a time, and in many physics cases that's exactly what it is, it is just an "affine" parameter. In the full 3D space, we have three degrees of freedom, but for a curve embedded in it, we've collapsed that down to just one degree of freedom, and that's our 't'.

Another similiar way to do it is parameterize according to the arc length itself, 's'. For example, the coordinate 'x' in the 3D space is just the length along the x axis, and the same for y and z. 's' is the same notion, just that we go along that general curved path which need not be a straight line. But that gets a bit involved.

For a surface, we just go up more dimension. We have two degrees of freedom. So we simply have some R(u, v), where u and v are two independent "affine" parameters the same as 't' for a curve. There's just two of them for a surface, rather than one, and they tell us the position vector, R, of all points on the surface. Continued in a bit..........

-Richard

publius
2007-Nov-19, 03:18 AM
Ok, a surface can be parameterized by some R(u, v). There are then two tangent vectors at a point. dR/du and dR/dv, where the 'd' here needs to be the curly partial derivate symbol. The normal vector at any point is just the cross product of those two N = dR/du x dR/dv.

There is a complication about the orientation of that normal vector. There are generally two sides to a surface and that normal vector could point either way. You want the normal to point "outwards" according to the right-hand sense, and I forget all the details there. For some crazy surfaces, like a Mobius strip, which sort of twists around backwards on itself, there is no clear notion of "side", and defining the proper normal orientation is impossible, or you have to do something else.

However, for calculating the area, all we're concerned about is the magnitude of that normal, so it's direction doesn't matter. But it does for other vector calculus operations, such as finding the flux of some other field through the surface.

Anyway, let's write out that cross product (look up cross product in Wiki to see how this is done in cartersian components).

dR/du = (dx/du) i + (dy/du) v + (dz/du) k

dR/dv = (dx/dv) i +(dy/dv) j + (dz/dv) k ---->

i j k

dx/du dy/du dz/du

dx/dv dy/dv dz/dv

N = (dy/du*dz/dv - dy/dv*dz/du) i +

.... (dz/du*dx/dv - dz/dv*dx/du) j +

.... (dx/du*dy/dv - dx/dv*dy/du) k

The magnitude, |N|, is just the square root of the sum of the squares of components, which we can write using this notation:

|N| = sqrt{ |d(x, y)/d(u, v)|^2 + |d(y, z)/d(u,v)|^2 + |d(z, x)/d(u, v)|^2 }

where {d(x, y)/d(u, v)| is the determinant of the Jacobian matrix of partials for the pair:

... d(x, y)/d(u, v) = dx/du dx/dv
... dy/du dy/dv

This is simply permuting x, y, and z cyclically against u and v for three rank-2 Jacobians. That looks a bit of mess, but if you zet x = u and y = v, you'll see that collapses to the simple version we had above.

This the equivalent of the curve forumula,

ds = |v| dt = sqrt[ (dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2 ] dt

for a surface, dS = |N| du dv = |N| dA

-Richard

publius
2007-Nov-19, 03:30 AM
I used CODE tags on the above to try to preseve my spacing for those matrix forms, but it still sort of spread them out.

Anyway, we can parameterize a sphere of radius r in the following u,v way:

z(u, v) = r cos u

y(u, v) = r sin u sin v

x(u, v) = r sin u cos v

where u and v and now angular variables, and 0 <= u <= pi, and
0 <= v <= 2pi. You can see that specifies all the points on the surface of the sphere.

This a sneaky way to bring in spherical coordinates while still thinking of our R vector in cartesians. :) But seriously, you can sort of see we're actually thinking of R as something of itself, independent of whatever coordiantes we "map" it in.

But you can verify that that Jacobian mess reduces to

dS = |N| du dv = r^2 sin u du dv, and S = 4pi r^2

-Richard

publius
2007-Nov-19, 03:36 AM
Finally, can you find a way to parameterize Gabriel's Horn in this u,v fashion. That comes from revolving the hyperbola y = 1/x around the x-axis. Imagine that in the full x-y-z 3D space.

We start out with y = 1/x lying in the x-y plane, then sweep out a surface by just rotating. The cross section is the y-z plane is just a circle, which suggests a parameter might be a simple angle, a theta, if you will.

Let x = u. v should simply gives us the y and z points on that circle, no? Now what are y and z in terms of u and v?

When you've done that, you've got Gabriel's Horn parameterized in this manner.

-Richard

grav
2007-Nov-21, 05:59 AM
Finally, can you find a way to parameterize Gabriel's Horn in this u,v fashion. That comes from revolving the hyperbola y = 1/x around the x-axis. Imagine that in the full x-y-z 3D space.

We start out with y = 1/x lying in the x-y plane, then sweep out a surface by just rotating. The cross section is the y-z plane is just a circle, which suggests a parameter might be a simple angle, a theta, if you will.

Let x = u. v should simply gives us the y and z points on that circle, no? Now what are y and z in terms of u and v?

When you've done that, you've got Gabriel's Horn parameterized in this manner.

-RichardSorry it's taken me so long to respond. I printed your last few posts and have been going over them, but it'll probably take me a while to take in all of what you are saying. Anyway, from your last post, if we make x(u, v) = u, then would y(u, v) = (1/u) sin v and z(u, v) = (1/u) cos v for Gabriel's horn, or vice versa for sine and cosine, either way, I guess, right?

Anyway, we can parameterize a sphere of radius r in the following u,v way:

z(u, v) = r cos u

y(u, v) = r sin u sin v

x(u, v) = r sin u cos v

where u and v and now angular variables, and 0 <= u <= pi, and
0 <= v <= 2pi. You can see that specifies all the points on the surface of the sphere.

Not that I have worked through this enough to even ponder this yet, but I'm curious. Looking at those parameters you gave for a sphere, and putting that in the context of what you wrote about in another recent thread concerning a hypersphere, could the parameters of a hypersphere, with 'a' as a fourth spatial dimension, become something like

a(u, v) = r cos u sin v

z(u, v) = r cos u cos v

y(u, v) = r sin u sin v

x(u, v) = r sin u cos v

?

publius
2007-Nov-21, 06:38 AM
Sorry it's taken me so long to respond. I printed your last few posts and have been going over them, but it'll probably take me a while to take in all of what you are saying. Anyway, from your last post, if we make x(u, v) = u, then would y(u, v) = (1/u) sin v and z(u, v) = (1/u) cos v for Gabriel's horn, or vice versa for sine and cosine, either way, I guess, right?

Yes indeedy do. Exactly.

Not that I have worked through this enough to even ponder this yet, but I'm curious. Looking at those parameters you gave for a sphere, and putting that in the context of what you wrote about in another recent thread concerning a hypersphere, could the parameters of a hypersphere, with 'a' as a fourth spatial dimension, become something like

a(u, v) = r cos u sin v

z(u, v) = r cos u cos v

y(u, v) = r sin u sin v

x(u, v) = r sin u cos v

?

Let's don't go there, just yet. Please. :) I am not the one to learn you higher dimensional geometry. I can't visualize it. I can see things in no more than 3 dimensions and know if I've got something right or wrong. Going to higher dimensions require you to "fly on instruments" so to speak, completely trust the math on paper because you can't see what you're trying to describe. And that means you have to build up a very rigorous framework so you don't go astray.

But note something above. You're just using two parameters, u and v. You would be trying to describe a 2 dimensional entity that existed in 4 dimensional space. The analogy of the hypersphere is a 3 dimensional "surface volume". 3 curved dimensions on the surface of that hypersphere. So to parameterize a "curved volume", we're going to need 3 parameters, right? We can certainly have a 2D surface in a 4D space, but that sucker can do more things than a surface in 3D dimensions, in the same manner than a curve can "curve more" in 3D than in just 2. :)

But, the way of thinking about this stuff in the 3 dimensions we can visualize helps build the full general notions that allow us to go to higher dimensions, of course.

-Richard

publius
2007-Nov-21, 07:11 AM
Grav,

http://en.wikipedia.org/wiki/Hypersphere

That should sastify your curiosity about hyperspheres (n = 4). The thing can be generalized in as many dimensions as you like. The hypervolume (length^n) enclosed by the n-sphere is given by

V_n = a(n)* r^n, where r is the hyperadius, and a(n) is a somewhat complex function of n. It's just pi for a 2-sphere (area enclosed by a circle), and 4pi/3 for a regular sphere. For a 4-sphere is

pi^2/2

Now, the ratio of the hypervolume to the hypersurface "area" is given by

V_n/S_n = r/n. So S_n = n*V_n/r

The hypervolume of the 4-sphere is (pi^2/2) r^4, so the surface volume of a 4-sphere is

2pi^2 r^3

Note how the "circumference" of a regular sphere remains 2pi*r. That holds all the way up. If were trapped in the closed surface volume of hypersphere, we could travel 2pi*r distance in any direction to get back where we started. We'd be trapped in a finite volume we couldn't escape. We could see the back of our heads so to speak.

In the limit as the number of dimensions goes to infinity, a(n) goes to one! It has maximum at 5 dimensions, then decreases. Now, in something that will really blow your mind, you can think of *fractional* dimensions. Don't ask, but that's what fractals are related to. You can consider, for example, a 3.5 dimensional sphere. Anyway, the maximum of a(n) occurs in between 5 and 6 dimensions. :shifty:

IOW, the volume of a n-sphere approaches that of a n-cube as n gets large.

-Richard

-Richard

grav
2007-Nov-21, 01:01 PM
Grav,

http://en.wikipedia.org/wiki/Hypersphere

That should sastify your curiosity about hyperspheres (n = 4). Thanks for the link. I'll look through that.

The thing can be generalized in as many dimensions as you like. The hypervolume (length^n) enclosed by the n-sphere is given by

V_n = a(n)* r^n, where r is the hyperadius, and a(n) is a somewhat complex function of n. It's just pi for a 2-sphere (area enclosed by a circle), and 4pi/3 for a regular sphere. For a 4-sphere is

pi^2/2a(n) might not be too complex for the surface area of the hypersphere if we use a surface area found with the circumferences across x from -r to r. That would give just

Int 2 pi sqrt(r^2 - x^2) dx

{Int sqrt(a - x^2) dx = (1/2) [x sqrt(a - x^2) - r^2 atan(-x / sqrt(a - x^2)], a = r^2, x = -r to r}

= 2 pi (1/2) [r^2 (pi/2) - r^2 (-pi/2)]

= (pi^2 / 2) r^2

but like you said, it's hard to visualize, so I can't say for sure, but I had worked through this and noticed that the result is the same, so I figured there's a chance it might apply.

Now, the ratio of the hypervolume to the hypersurface "area" is given by

V_n/S_n = r/n. So S_n = n*V_n/r

The hypervolume of the 4-sphere is (pi^2/2) r^4, so the surface volume of a 4-sphere is

2pi^2 r^3

Going through it in the same form as you had before, it looks like the parameters for the "volume" for a hypersphere could be something along the lines of

a(u, v, w) = r cos u

z(u, v, w) = r sin u cos v

y(u, v, w) = r sin u sin v sin w

z(u, v, w) = r sin u sin v cos w

Note how the "circumference" of a regular sphere remains 2pi*r. That holds all the way up. If were trapped in the closed surface volume of hypersphere, we could travel 2pi*r distance in any direction to get back where we started. We'd be trapped in a finite volume we couldn't escape. We could see the back of our heads so to speak.

In the limit as the number of dimensions goes to infinity, a(n) goes to one! It has maximum at 5 dimensions, then decreases. Now, in something that will really blow your mind, you can think of *fractional* dimensions. Don't ask, but that's what fractals are related to. You can consider, for example, a 3.5 dimensional sphere. Anyway, the maximum of a(n) occurs in between 5 and 6 dimensions. :shifty:

IOW, the volume of a n-sphere approaches that of a n-cube as n gets large.That's very interesting. I might look into that some more.

Okay, curiosity satisfied. Not that I was really that curious anyway, yet. :) Just a couple of quick observations I wanted to ask about.

grav
2007-Nov-21, 01:42 PM
Okay, one more quick observation. ;) Looking at the way I wrote that for the surface area of a hypersphere earlier, if it is correct, then it looks like that would be the limit, so that there would be no such thing as surface area for five spatial dimensions or more, at least not with the way we are representing them, is that right? If so, then for volume, it looks like eight dimensions might be the limit for that. But we should still be able to draw a line between two points, straight or curved, and that is only one dimension, not to mention the dimensionless points themselves, so my thinking must be off there. Might it just be a limit for that type of representation for the parameters or something? Can it be expressed differently but in a similar fashion that accounts for higher dimensions?

grav
2007-Nov-21, 02:06 PM
My thinking on that last post is that sine and cosine can be used and switched around in a variety of ways for a small number of dimensions, allowing some freedom for the representation, since those parameters are set at two for each dimension we are actually finding for, but that means that the number of ways they can be used becomes tighter with higher dimensions, so that the number of dimensions they can be used for would depend directly upon how many ways we can express them this way, so it hinges upon this type of representation itself, which only allows it to be used up to 2^m = n dimensions, where m is the number of dimensions of what we are trying to find, for surface area, volume, etc. It still seems like a pretty nifty way to think about it, though. It also appears it might be an easier way to go about it. But then, I haven't actually performed the integration yet. :)

publius
2007-Nov-21, 04:48 PM
Grav,

You can have all the lower dimensional structures in an n dimensional space. In 4D, we can still have curves and surfaces, they can just curve around in even more complex ways. And in 4D, we can have curved volumes as well.

Consider a curve in a plane. That's a curved 1D "surface" in a flat 2D Euclidean space. That curve can curve all around, but note it must be confined to the plane. The notion of area there can only be that enclosed by some closed curve or set of them.

Now, go up to 3D. You can see that gives you more degree of freedom in which way your curve can go. It can now turn out of the plane. And now, we can have curved surfaces, 2D areas that are more than what you can get in the plane.

In 4D, we add yet another way the curve can curve, and yet another way for a 2D surface to bend and curve around. And we gain the notion of a curved volume.

Go to 5 dimensions, and now we've got yet another way for curves, surfaces, and volumes to curve, and now can even have 4D hypervolumes that can curve around as well.

More dimensions are more ways for things to curve around.

In the plane, 2D Euclidean space, it turns out this notion of "how something can curve" can be described by a single number, a scalar. This simple notion is called the "curvature". A circle has constant curvature. If you like we'll look at how that is defined.

Now, add the 3rd dimension. That single number is no longer capable of describing all the ways we can curve that curve. We need another one, which is sometimes called the "torsion" (and this is NOT TO BE CONFUSED with another type of torsion, which has to do with twisting of geodesics.....long story).

A 2D surface can be described completely, I think, by the unit surface normal at every point, which would mean 2 numbers there at each point. So, I think, in general, we need two numbers to describe curvature in 3D.

Going to higher dimensions means more numbers, more functions of position to describe how something can curve, and that requires a *tensor*.

-Richard

grav
2007-Nov-25, 02:11 AM
It also appears it might be an easier way to go about it. But then, I haven't actually performed the integration yet. :)Um, yeah. Did I say easier? I still don't seem to be able to get a grasp on what you're doing there. I printed it and have been going over your posts a little bit each day, but I'm still no further along yet. Looks like I'm even more of a square than I thought. ;) But I figure that if I can learn how you're doing it this way, then I will at least have something to fall back on if I can't get it the other way for some reason, and I can also double check my integration by doing it both ways and comparing the results. That's what I wanted to do before posting what I got the way I was doing it, but I guess I'll go ahead and at least post that for now, anyway.

Int sqrt[1 + (dz/dx)^2 + (dz/dy)^2] dx dy

dz/dx = sqrt(1/x^2 - y^2)'
F(G(x)) = sqrt(1/x^2 - a)
F(x) = sqrt(x), G(x) = 1/x^2 - a
F'(x) = (1/2) / sqrt(x), G'(x) = -2/(x^3)
F'(G(x)) G'(x) = [(1/2) / sqrt(1/x^2 - a)] (-2/(x^3))
= -1 / [x^3 sqrt(1/x^2 - y^2)]

dz/dy = sqrt(a - y^2)'
F(G(x)) = sqrt(a - y^2)
F(x) = sqrt(x), G(x) = a - y^2
F'(x) = (1/2) / sqrt(x), G'(x) = -2y
F'(G(x)) G'(x) = [(1/2) / sqrt(a - y^2)] (-2y)
= -y / sqrt(1/x^2 - y^2)

= Int 2 sqrt[1 + (-1/(x^3 sqrt(1/x^2 - y^2)))^2 + (-y / sqrt(1/x^2 - y^2))^2] dx dy

= Int 2 sqrt[1 + 1/(x^6 (1/x^2 - y^2)) + y^2 / (1/x^2 - y^2)] dx dy

= Int 2 sqrt[(x^6 (1/x^2 - y^2) + 1 + x^6 y^2) / (x^6 (1/x^2 - y^2))] dx dy

= Int 2 sqrt[(x^4 + 1) / (x^4 - x^6 y^2)] dx dy

= Int 2 sqrt[(1/x^2 + 1/x^6) / (1/x^2 - y^2)] dx dy

= Int 2 (1/x) sqrt(1 + 1/x^4) / sqrt(1/x^2 - y^2) dx dy

{ Int 1/sqrt(a^2 - y^2) dy = asin(y/a), a = 1/x, y = -1/x to 1/x}

= Int (2/x) sqrt(1 + 1/x^4) [asin(1) - asin(-1)] dx

= Int (2/x) sqrt(1 + 1/x^4) pi dx

{ Int (sqrt(1 + 1/x^4) / x) dx = (1/2) [asinh(x^2) - sqrt(1 + 1/x^4)], x = j to k}

= pi [ ln(k^2 (1 + sqrt(1 + 1/k^2))) - sqrt(1 + 1/k^4) - ln(j^2 (1 + sqrt(1 + 1/j^4))) + sqrt(1 + 1/j^4)]

publius
2007-Nov-25, 11:11 PM
Grav,

It's been a while and I've sort of lost my train of thought on this thread. What is it you're not grasping? If it's that (u, v) parameterization and the N = Tu x Tv normal thing, I can go over it again.

I don't know if you've been properly exposed to vector algebra, much less calculus, and that may the problem. You just may not be familiar with expressing things in terms of vectors.

And BTW, that reminds me. Back in school, I took a linear algebra class as an elective as a sophomore, I think. That helped me so much in understanding my later quantum courses, the whole eigenvalue/eigenvector, not to mention Fourier/Hilbert space notions, that I can't stress it enough.

It should be a required course for physics students. Heck, it may be, now.

-Richard