• Jan 13th 2013, 09:50 PM
Kvandesterren
find f if grad f = 3xy i + x^3 j
• Jan 14th 2013, 05:49 AM
Barioth
As you know grad f =$\displaystyle \frac{d(f(x,y))}{dx}*i + \frac{d(f(x,y))}{dy}*j$

so I believe you could try integrating both part!

Give that a try and let us know if it worked!

Hope that helped, if not let me know!
• Jan 14th 2013, 06:14 AM
Plato
Quote:

Originally Posted by Kvandesterren
find f if grad f = 3xy i + x^3 j

I think that there is a typo in your post.

Is it not, $\displaystyle \nabla f(x,y) = 3x^2 yi + x^3 k~?$
• Jan 14th 2013, 07:10 AM
HallsofIvy
IF there exist a function f having that gradient, then we must have $\displaystyle \frac{\partial f}{\partial x}= 3xy$ and $\displaystyle \frac{\partial f}{\partial y}= x^3$. If we differentiate the first a second time, but with respect to y, we have $\displaystyle \frac{\partial^2 f}{\partial y\partial x}= 3x$. If we differentiate the second a second time, but with respect to x, we have $\displaystyle \frac{\partial^2 f}{\partial x\partial y}= 3x^2$. As long as all derivatives are continuous, as they are here, the "mixed" second derivatives must be the same. Since this is not true, there is NO function, f, having 3xyi+ x^3j as its gradient.