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Divergence and Curls
- Thread starter soggles
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HallsofIvy
Elite Member
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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 \(\displaystyle \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f(x,y,z) & g(x,y,z) & h(x,y,z) \end{array}\right|\).
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 \(\displaystyle \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f(x,y,z) & g(x,y,z) & h(x,y,z) \end{array}\right|\).
HallsofIvy
Elite Member
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- Jan 27, 2012
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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 \(\displaystyle \frac{\partial h}{\partial y}\)?
"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 \(\displaystyle \frac{\partial h}{\partial y}\)?