asked 2014-07-31 11:05:02 -0600
When showing that $\vec F$ is conservative, we take the partial derivatives, just wondering if that is the same as $\nabla f $? I know from there we find the function $f$ to make them equal? But I thought that the gradient was the partial derivatives?
answered 2014-07-31 12:43:04 -0600
I would just like to add that the intuitive reasoning for finding $\frac{\partial P}{\partial y}$ and $\frac{\partial Q}{\partial x}$ checking if they are equal is to show that $f$ is "sufficiently nice," as the book says. The reason that this condition shows whether $f$ exists or not is because of Clairaut’s Theorem. It is Theorem 14.6.2 in the book and it states:
If the mixed partial derivatives are continuous, they are equal.
In other words, if the mixed partials derivatives are not equal, then $f$ does not exist. Note that since $\vec F = \nabla f$, $\langle P, Q \rangle $ = $ \langle f_x, f_y \rangle $. This means that $\frac{\partial P}{\partial y}$ is $f_{xy}$ and $\frac{\partial Q}{\partial x}$ is $f_{yx}$ (i.e. the mixed partials of $f$).
answered 2014-07-31 11:24:46 -0600
It's not the same as $\nabla f$ because you have $\vec F = (P,Q)$ (pretend this is vector notation which isn't showing up for me) and to find if $\vec F$ is conservative, we are taking $\frac{\partial P}{\partial y}$ and $\frac{\partial Q}{\partial x}$. Basically, use the same set up as $\nabla f$ and swap the $\partial x$ with the $\partial y$.
answered 2014-07-31 11:26:17 -0600
OHHHH! Because we're taking with respect to y first and then respect to x second, instead of partial x then partial y!
Thanks!
Asked: 2014-07-31 11:05:02 -0600
Seen: 13 times
Last updated: Jul 31 '14
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