Divergence and Curls

[soggles]

Hey,
Can anyone please help me with this problem?
Thank you,

 

[HallsofIvy]

What do you mean by "help"? Tell you what "divergence" and "curl" of a vector function mean? Okay but I am surprised that your teacher didn't tell you that, or your text book did not have the definitions, before you were given this problem!

Given a "vector valued function", $\displaystyle \vec{F}= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}$, where f, g, and h are differentiable functions of the three independent variables, x, y, and z, the "divergence" of $\displaystyle \vec{F}$ (also denoted by "div $\displaystyle \vec{F}$" or $\displaystyle \nabla\cdot\vec{F}$ is given by $\displaystyle \frac{\partial f}{\partial x}+ \frac{\partial g}{\partial y}+ \frac{\partial h}{\partial z}$. If we think of "$\displaystyle \nabla$" as the (purely symbolic) "vector operator" $\displaystyle \frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial y}\vec{j}+ \frac{\partial}{\partial z}\vec{k}$, then divergence of $\displaystyle \vec{F}$ can be thought of as the "dot product" of $\displaystyle \nabla$ and $\displaystyle \vec{F}$.

In this problem, $\displaystyle f(x,y,z)= x cos(\pi y)$. What is $\displaystyle \frac{\partial f}{\partial x}$? $\displaystyle g(x,y,z)= \frac{2x^2}{\pi}sin(\pi y)- 2y^2e^{-z}$. What is $\displaystyle \frac{\partial g}{\partial y}$? And $\displaystyle h(x,y,z)= ye^{-z}$. What is $\displaystyle \frac{\partial h}{\partial z}$?

The "curl" is a little more complicated! Given the same $\displaystyle \vec{F}$ as above, $\displaystyle curl \vec{F}= \left(\frac{\partial h}{\partial y}- \frac{\partial g}{\partial z}\right)\vec{i}+$$\displaystyle \left(\frac{\partial f}{\partial z}- \frac{\partial h}{\partial x}\right)\vec{j}+$$\displaystyle \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\vec{k}\right)$.

That can be thought of as the "cross product" of the "vector operator", $\displaystyle \nabla$, and the vector function $\displaystyle \vec{F}$, and conveniently remembered as the (symbolic) determinant $\displaystyle \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f(x,y,z) & g(x,y,z) & h(x,y,z) \end{array}\right|$.

 

[soggles]

Thank you for your help. Would this be considered a conservative vector field?
Thank you

 

[HallsofIvy]

Okay, do you know what the definition of "conservative vector field" is? Again, I would expect that you would have at least learned the definition before you were given a problem like this but you write this as if you had no idea!

"conservative" is actually a physics term related to a "conservative force field", like gravity, as opposed to a non-conservative force, like friction. In mathematics the corresponding concept is "exact differential". Given a function of 2 or more variables, say $\displaystyle \phi(x,y,z)$, its gradient is $\displaystyle \nabla \phi= \frac{\partial \phi}{\partial x}\vec{i}+ \frac{\partial \phi}{\partial y}\vec{j}+ \frac{\partial \phi}{\partial z}\vec{k}$.

The question is, given a vector field, $\displaystyle f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}$ that looks like a derivative, is it really? Does there exist a function $\displaystyle \phi$ such that $\displaystyle \nabla \phi$ is equal to $\displaystyle f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}$? (In physics terms, $\displaystyle \phi$ would be the "potential energy field" for this "conservative force field".)

One way to determine that would be to actually try to find that function, $\displaystyle \phi$. But a simpler test depends on the fact that, for a reasonable function, $\displaystyle \phi$, the "mixed second derivatives" do not depend on the order of the derivatives. That is $\displaystyle \frac{\partial \phi}{\partial x\partial y}$ is the same whether you do the derivative with respect to x first, then the derivative with respect to y or vice-versa. If $\displaystyle f= \frac{\partial\phi}{\partial x}$ then $\displaystyle \frac{\partial f}{\partial y}= \frac{\partial \phi^2}{\partial x\partial y}$. If $\displaystyle g= \frac{\partial\phi}{\partial y}$ then $\displaystyle \frac{\partial g}{\partial x}= \frac{\partial \phi}{\partial y\partial x}$ so we must have $\displaystyle \frac{\partial f}{\partial y}= \frac{\partial g}{\partial x}$.

Similarly, we must have $\displaystyle \frac{\partial f}{\partial z}= \frac{\partial h}{\partial x}$ and $\displaystyle \frac{\partial g}{\partial z}= \frac{\partial h}{\partial y}$.
(This is sometimes called the "cross derivative test".)

In this exercise we have $\displaystyle f= x cos(\pi y)$, $\displaystyle g= \frac{2x^2}{\pi}sin(\pi y)- 2y^2e^{-z}$, and $\displaystyle h= ye^{-z}$.

What are $\displaystyle \frac{\partial g}{\partial x}$ and $\displaystyle \frac{\partial f}{\partial y}$?
What are $\displaystyle \frac{\partial h}{\partial x}$ and $\displaystyle \frac{\partial f}{\partial z}$?
What are $\displaystyle \frac{\partial g}{\partial z}$ and $\displaystyle \frac{\partial h}{\partial y}$?