Fundamental Theorem of Line Integrals

[MetalManuel]

If someone could link me to a tutorial on how to put in functions into a post, I would appreciate it, thanks. I am going to be putting in screen shots.

Homework Statement

http://img864.imageshack.us/img864/1517/scr1305133657.png"
[PLAIN]http://img864.imageshack.us/img864/1517/scr1305133657.png

Homework Equations

The Attempt at a Solution

I already solved this equation, but I have a dispute with my professor about this problem. I show that the vector is conservative, therefore this is the gradient of a function. When I first integrated this function I got the function to be
http://img703.imageshack.us/img703/8749/scr1305133859.png"
[PLAIN]http://img703.imageshack.us/img703/8749/scr1305133859.png
I thought to myself that has to be wrong. I found the gradient of that function and it was the original vector, multiplied by 2. So that couldn't have been the correct function. So then I realized that arctan (x/y) alone has to be the original function to not get it multiplied by 2, or -arctan (y/x), they must be equal in some domain and range. I found the answer to be pi/12 when my professor said it was pi/6, due to the multiplication of 2. She said it's ok to ignore the 2 when doing it backwards because it's a scalar. I am going to talk to her about it in her office hours in a little while about it, but I was wondering who is correct? She said it's arctan(x/y)-arctan(y/x) I say it's arctan(x/y) alone.

 

[LCKurtz]

You can easily check that the exact differential of -arctan(y/x) gives your integrand, so the general potential function is -arctan(y/x) + C. The additive constant will not affect your answer, which is pi/12.

That being said, I suspect that your original integral to get your answer with the two arctangents was improperly done.

 

[MetalManuel]

You can easily check that the exact differential of -arctan(y/x) gives your integrand, so the general potential function is -arctan(y/x) + C. The additive constant will not affect your answer, which is pi/12.

That being said, I suspect that your original integral to get your answer with the two arctangents was improperly done.

Actually it can be -arctan(y/x) or arctan(x/y), they're both solutions. My integrals were not incorrect, you can do them for yourself and see. I agree with you.

 

[LCKurtz]

Actually it can be -arctan(y/x) or arctan(x/y), they're both solutions. My integrals were not incorrect, you can do them for yourself and see. I agree with you.

OK. I'm just saying if you directly solve for the potential function by properly integrating the exact differential, you will not get a factor of 2.

 

[MetalManuel]

I made a pdf file showing my proof, can any of you look at it? My professor is very stubborn and I want to be very detailed about how I go about to show her that she's wrong. I will also verify with other calculus professors. I honestly don't care about the points, it's just about finding the truth, and not about proving who's wrong or who's right. If I am wrong, let me know.

edit: found a mistake gradf(fx,y)=(fx,fy) not (fx,fx), will edit
edit2:fixed

 

[LCKurtz]

Your teacher's method, which works "most of the time" is absolutely not the way that should be taught. You have worked it correctly. The idea of "adding the two leaving out the repeats" is not well defined and can lead to errors as in this case.

You have solved it correctly. When I was teaching calculus, on an exam I would have docked partial credit on your teacher's "method" of solution even if the answer was correct.

[Edit] In addition, in the first quadrant at least arctan(x/y) = pi/2 - arctan(y/x) so they differ by a constant.

 

[MetalManuel]

Your teacher's method, which works "most of the time" is absolutely not the way that should be taught. You have worked it correctly. The idea of "adding the two leaving out the repeats" is not well defined and can lead to errors as in this case.

You have solved it correctly. When I was teaching calculus, on an exam I would have docked partial credit on your teacher's "method" of solution even if the answer was correct.

[Edit] In addition, in the first quadrant at least arctan(x/y) = pi/2 - arctan(y/x) so they differ by a constant.

Thanks, I know it's not the proper way, but I got to give her credit, it does work most of the time. I tested it out with exact equations in my differential equations class, but I always double check to make sure, in this case her method didn't work.

 

[vela]

Minor nitpick...

After you determine f(y) is a constant, you should say

$$\Phi(x,y) = \arctan \frac{x}{y} + c$$

not

$$\arctan \frac{x}{y} = c$$

But, yeah, your professor is wrong.

 

[MetalManuel]

Minor nitpick...

After you determine f(y) is a constant, you should say

$$\Phi(x,y) = \arctan \frac{x}{y} + c$$

not

$$\arctan \frac{x}{y} = c$$

But, yeah, your professor is wrong.

Oh, I'm sorry, I'm used to solving exact equations setting them equal to a constant. Just looked it up, they do that here too http://tutorial.math.lamar.edu/Classes/DE/Exact.aspx .

 

[vela]

That's because they're starting with the differential equation dΦ=0, from which it follows Φ=constant. In other words, if you were solving

$$\frac{y}{x^2+y^2}dx - \frac{x}{x^2+y^2}dy = 0$$

the solution would be

$$\arctan\frac{x}{y} = c$$.

However, in this problem, you are starting with

$$\frac{\partial\Phi}{\partial x} = \frac{y}{x^2+y^2}$$

so when you integrate, you get

$$\Phi(x,y) = \arctan\frac{x}{y} + f(y) = \arctan\frac{x}{y}+c$$

 

[MetalManuel]

That's because they're starting with the differential equation dΦ=0, from which it follows Φ=constant. In other words, if you were solving

$$\frac{y}{x^2+y^2}dx - \frac{x}{x^2+y^2}dy = 0$$

the solution would be

$$\arctan\frac{x}{y} = c$$.

However, in this problem, you are starting with

$$\frac{\partial\Phi}{\partial x} = \frac{y}{x^2+y^2}$$

so when you integrate, you get

$$\Phi(x,y) = \arctan\frac{x}{y} + f(y) = \arctan\frac{x}{y}+c$$

thanks for the clarification, i appreciate it.