Question

I'm a bit lost on the concept of finding the potential of vector F.

EX: vectorF= <2x(y^2), 2y(x^2) + 3(y^2)>

So the gradient of F is <2(y^2), 2(x^2) + 6y>
Where would I go from here?

Answer 1

ohhh maybe i know this give me a sec

Answer 2

YAY. Thank you! :)

Answer 3

@TuringTest @experimentX can u come help

Answer 4

Not quite sure ,,,

Answer 5

Edit: Okay so I believe we integrate each with respect to their corresponding axis. So integration of the gradient would be <2x(y^2) + C, 6xy + (2/3)(x^3) + C>.......And from there on, it's down the rabbit hole...

Answer 6

\[ F = <\frac{\delta V}{\delta x}, \frac{\delta V}{\delta y}> = <2x(y^2), 2y(x^2) + 3(y^2)>\]

I think integrating assuming other variable as constant should give V.

Answer 7

Have a look here :) http://en.wikipedia.org/wiki/Scalar_potential

Answer 8

The same thing what @experimentX has done :) The problem would be the constants arising out of indefinite integration

Answer 9

\[V=<x^2(y^2)+C1, y^2(x^2) + (y^3)+c2>\]

Answer 10

From I
\[ V = \int 2xy^2dx = x^2y^2 +c\]
This constant may also include y terms, since we have considered y as constant

From II
\[ V = \int 2y(x^2) + 3(y^2) dy = x^2y^2+y^3+k\]

Since we have both \( x^2y^2 \) in both \( V = x^2y^2+y^3+k \)

Answer 11

If this is not your answer, don't blame me ... seeya!!

Answer 12

Thank you regardless

Answer 13

@experimentX , Everything is fine until the last statement.
"Since we have both x^2y^2 in both V=x^2y^2+y^3+k"

-->What did you do here ??

Answer 14

V is a scalar quantity, must be same from both integral.

Answer 15

I think the right way must be to take magnitude of V
\[V=<x^2(y^2)+C1, y^2(x^2) + (y^3)+c2>\]
\[V=\sqrt{({x^2(y^2)+C1})^2+({ y^2(x^2) + (y^3)+c2})^2}\]
@mos1635 , can you confirm this please ??

Answer 16

sorry guys i am lost here...

Answer 17

vectorF= <2x(y^2), 2y(x^2) + 3(y^2)>
potentials are just functions such that if you take the gradient of that will give you vector F so:
\[gradient of f=<df/dx,df/dy>=<2xy^2,2yx^2+y^2>vectorF\]
so we'll have:
\[df/dx=2xy^2\] \[df/dy=2yx^2+3y^2\]
that means:
\[f(x,y)=\int\limits 2xy^2dx\]
\[f(x,y)=\int\limits 2yx^2+3y^2dy\]
I'll use the first one 1st:
\[f(x,y)=\int\limits 2xy^2dx=x^2y^2+h(y)\]
h(y) in this case wil be our constant integration since we're integrating wit respect to x, so all other terms without x will be constants. now we need to find h(y):
df/dy=2yx^2+3y^2
so if we differentiate the f(x,y) we got wtih respect to y, then that will be equal to 2yx^2+3y^2
df(x,y)/dy=df/dy
d(x^2y^2+h(y))/dy=2yx^2+3y^2
2yx^2+h'(y)=2yx^2+3y^2
h'(y)=3y^2
integrating this will yield:
h(y)=y^3+c
so the potential for F will be:
f(x,y)=(x^2)(y^2)+y^3<--- Answer
if you want you get the gradient of this and you'll get vector F...

Answer 18

I don't believe the magnitude of that vector fits in at all. If we integrate the gradient of vector F in a way such that <Fxdx, Fydy> (Imagine those being integrated. Importance is with what respect they're being integrated by). If we do that, the constant of integration is dependent on the variable you integrated by. So the C1 of the first component would actually be a g'(x). Or at least it should be something of that nature. From there I am lost even further. Because then we have two equations and how do we put them together to explain the potential? Apologies if I'm rambling. I shall find the answer by tonight and post, I'm sure.

Thank you for any responses thus far, and those coming! :)

Answer 19

you use those two to find the potential :D