- #1
JohnTheMan
- 3
- 0
Homework Statement
z = ƒ(x,y), x = rcos(θ), y = rsin(θ)
Use the chain rule to show that:
[tex] \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} [/tex]
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:
[tex] \frac{\partial z}{\partial \theta } = -rsin(\theta)\frac{\partial z}{\partial x}+rcos(\theta)\frac{\partial z}{\partial y}[/tex]
[tex] \frac{\partial z}{\partial r } = cos(\theta)\frac{\partial z}{\partial x}+sin(\theta)\frac{\partial z}{\partial y}[/tex]
[tex] \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}}) [/tex]
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 :)