How Does the Chain Rule Apply in Polar Coordinates?

[JohnTheMan]

Homework Statement

z = ƒ(x,y), x = rcos(θ), y = rsin(θ)
Use the chain rule to show that:
\frac{1}{r^{2}}\frac{\partial ^{2} z}{\partial \theta ^{2}} = sin^{2}(\theta)\frac{\partial ^{2} z}{\partial x^{2}}-2sin(\theta)cos(\theta)\frac{\partial ^{2} z}{\partial x \partial y}+cos^{2}(\theta)\frac{\partial ^{2} z}{\partial y ^{2}} - \frac{1}{r}\frac{\partial z}{\partial r}

Homework Equations

∂z/∂r = (∂z/∂x)(∂x/∂r) + (∂z/∂y)(∂y/∂r)

The Attempt at a Solution

I have worked it through and I have found that:
\frac{\partial z}{\partial \theta } = -rsin(\theta)\frac{\partial z}{\partial x}+rcos(\theta)\frac{\partial z}{\partial y}
\frac{\partial z}{\partial r } = cos(\theta)\frac{\partial z}{\partial x}+sin(\theta)\frac{\partial z}{\partial y}
\frac{\partial ^{2} z}{\partial \theta ^{2}} = r^{2}(sin^{2}(\theta)\frac{\partial ^{2} z}{\partial x^{2}}-2sin(\theta)cos(\theta)\frac{\partial ^{2} z}{\partial x \partial y}+cos^{2}(\theta)\frac{\partial ^{2} z}{\partial y ^{2}})
I have checked these a number of times and they seem right to me. The last one listed is the second partial derivative of z in terms of θ, but it doesn't match that of the problem statement. I'm not really sure what I'm doing wrong. Thanks :)

 

[RUber]

Your second derivative looks like you just squared your first derivative.
You need to actually take one more derivative to get $\frac{\partial^2 z}{\partial \theta^2}$.

 

[JohnTheMan]

Your second derivative looks like you just squared your first derivative.
You need to actually take one more derivative to get $\frac{\partial^2 z}{\partial \theta^2}$.

I didn't square it, I took the second derivative.

 

[RUber]

You have the first part right. But your second derivative is wrong.
$\frac{\partial}{\partial \theta } f(x,y) = \frac{\partial f}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial \theta}$
$\frac{\partial}{\partial \theta }\left( \frac{\partial}{\partial \theta } f(x,y)\right) = \frac{\partial}{\partial \theta }\left( \frac{\partial f}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial \theta}\right)$
Which is
$= \left( \frac{\partial}{\partial \theta }\frac{\partial f}{\partial x}\right) \frac{\partial x}{\partial \theta}+\underline{ \frac{\partial f}{\partial x} \left( \frac{\partial}{\partial \theta }\frac{\partial x}{\partial \theta} \right) } + \left( \frac{\partial}{\partial \theta } \frac{\partial f}{\partial y}\right) \frac{\partial y}{\partial \theta}+ \underline{ \frac{\partial f}{\partial y} \left( \frac{\partial}{\partial \theta }\frac{\partial y}{\partial \theta}\right)}$
In your expansion, you missed the second derivatives with respect to theta on x and y. Which will give you the equivalent parts you need for the partial with respect to r.

 

[JohnTheMan]

You have the first part right. But your second derivative is wrong.
$\frac{\partial}{\partial \theta } f(x,y) = \frac{\partial f}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial \theta}$
$\frac{\partial}{\partial \theta }\left( \frac{\partial}{\partial \theta } f(x,y)\right) = \frac{\partial}{\partial \theta }\left( \frac{\partial f}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial \theta}\right)$
Which is
$= \left( \frac{\partial}{\partial \theta }\frac{\partial f}{\partial x}\right) \frac{\partial x}{\partial \theta}+\underline{ \frac{\partial f}{\partial x} \left( \frac{\partial}{\partial \theta }\frac{\partial x}{\partial \theta} \right) } + \left( \frac{\partial}{\partial \theta } \frac{\partial f}{\partial y}\right) \frac{\partial y}{\partial \theta}+ \underline{ \frac{\partial f}{\partial y} \left( \frac{\partial}{\partial \theta }\frac{\partial y}{\partial \theta}\right)}$
In your expansion, you missed the second derivatives with respect to theta on x and y. Which will give you the equivalent parts you need for the partial with respect to r.

You have the first part right. But your second derivative is wrong.
$\frac{\partial}{\partial \theta } f(x,y) = \frac{\partial f}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial \theta}$
$\frac{\partial}{\partial \theta }\left( \frac{\partial}{\partial \theta } f(x,y)\right) = \frac{\partial}{\partial \theta }\left( \frac{\partial f}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial \theta}\right)$
Which is
$= \left( \frac{\partial}{\partial \theta }\frac{\partial f}{\partial x}\right) \frac{\partial x}{\partial \theta}+\underline{ \frac{\partial f}{\partial x} \left( \frac{\partial}{\partial \theta }\frac{\partial x}{\partial \theta} \right) } + \left( \frac{\partial}{\partial \theta } \frac{\partial f}{\partial y}\right) \frac{\partial y}{\partial \theta}+ \underline{ \frac{\partial f}{\partial y} \left( \frac{\partial}{\partial \theta }\frac{\partial y}{\partial \theta}\right)}$
In your expansion, you missed the second derivatives with respect to theta on x and y. Which will give you the equivalent parts you need for the partial with respect to r.

Ahh. That makes sense. Thanks a lot.