Vector Prolem

Apr 2009
108
0
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

I've found gradient function

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

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

Thanks
 
Apr 2010
384
153
Canada
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

I've found gradient function

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

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

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

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

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

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

\(\displaystyle \partial f = y^2e^4x+2xe^3y \partial x \)

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

Therefore,

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

To get to that step we simply differentiate our potential function with respect to y!
 
Apr 2009
108
0
\(\displaystyle grad f(x,y) = \nabla f(x,y) = \frac{ \partial f }{ \partial x } \hat i + \frac{ \partial f }{ \partial y } \hat j \)

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

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

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

\(\displaystyle \partial f = y^2e^4x+2xe^3y \partial x \)

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

Therefore,

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

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??
 
Apr 2010
384
153
Canada
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 \(\displaystyle \nabla f(x,y,z) = \frac { \partial f }{ \partial x } \hat i + \frac { \partial f }{ \partial y } \hat j + \frac { \partial f }{ \partial z } \hat k \) 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 \(\displaystyle G(x.y) = \frac{ \partial f }{ \partial y } \) 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!