- #1
MetalManuel
- 39
- 0
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.
http://img864.imageshack.us/img864/1517/scr1305133657.png"
[PLAIN]http://img864.imageshack.us/img864/1517/scr1305133657.png
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.
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.
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