# Math Help - Vector Prolem

1. ## Vector Prolem

Let F=(y^2e^4x+2xe^3y)i +G(x,y)j is a gradient field.

Find all G(x,y).

I'm confused all together.

Here's what I've done so far

4y^2e^4x+2e^3y+2ye^4x+6xe^3y

TBH, I don't understand what the question is asking..

Thanks

2. Originally Posted by Khonics89
Let F=(y^2e^4x+2xe^3y)i +G(x,y)j is a gradient field.

Find all G(x,y).

I'm confused all together.

Here's what I've done so far

4y^2e^4x+2e^3y+2ye^4x+6xe^3y

TBH, I don't understand what the question is asking..

Thanks
$grad f(x,y) = \nabla f(x,y) = \frac{ \partial f }{ \partial x } \hat i + \frac{ \partial f }{ \partial y } \hat j$

We are given $\frac{ \partial f }{ \partial x } = y^2e^4x+2xe^3y$ so we need to find a Potential Function such that $\frac{ \partial f }{ \partial y } = G(x,y)$

To do this we will find F(x,y) by integration of the first part. Then find a constant such that our criteria is satisfied

$\frac{ \partial f }{ \partial x } = y^2e^4x+2xe^3y$

$\partial f = y^2e^4x+2xe^3y \partial x$

$f(x,y) = \int y^2e^4x+2xe^3y dx = \frac{ y^2e^4x^2 }{2} + x^2e^3y + Q(y)$ where Q(y) is a function only dependant on y.

Therefore,

$\frac{ \partial f }{ \partial y } = G(x,y) = ye^4x^2 + x^2e^3 + Q(y)$

To get to that step we simply differentiate our potential function with respect to y!

3. Originally Posted by AllanCuz
$grad f(x,y) = \nabla f(x,y) = \frac{ \partial f }{ \partial x } \hat i + \frac{ \partial f }{ \partial y } \hat j$

We are given $\frac{ \partial f }{ \partial x } = y^2e^4x+2xe^3y$ so we need to find a Potential Function such that $\frac{ \partial f }{ \partial y } = G(x,y)$

To do this we will find F(x,y) by integration of the first part. Then find a constant such that our criteria is satisfied

$\frac{ \partial f }{ \partial x } = y^2e^4x+2xe^3y$

$\partial f = y^2e^4x+2xe^3y \partial x$

$f(x,y) = \int y^2e^4x+2xe^3y dx = \frac{ y^2e^4x^2 }{2} + x^2e^3y + Q(y)$ where Q(y) is a function only dependant on y.

Therefore,

$\frac{ \partial f }{ \partial y } = G(x,y) = ye^4x^2 + x^2e^3 + Q(y)$

To get to that step we simply differentiate our potential function with respect to y!

Sorry but I'm still confused

Now I know that F is my gradient field which is "del"theta.. where theta is our potential function.

are we just trying to find the potential function??

4. Originally Posted by Khonics89
Sorry but I'm still confused

Now I know that F is my gradient field which is "del"theta.. where theta is our potential function.

are we just trying to find the potential function??
This is essentially a conservative field problem.

If a vecotor field F is conservative there exists a potential function such that $\nabla f(x,y,z) = \frac { \partial f }{ \partial x } \hat i + \frac { \partial f }{ \partial y } \hat j + \frac { \partial f }{ \partial z } \hat k$ which is the same thing as saying that a gradient exists. Since we are told there is a gradient (which means we have a potential function) we can then find that potential function and find G(x,y).

Like I stated above $G(x.y) = \frac{ \partial f }{ \partial y }$ which is nothing other then the derivitive of our potential function with respect to y. So we find the function and take the derivative. That's all we are looking for here is that component!