[University Calculus III] Multivariable chain rule
If x=rcos(theta) and y=rsin(theta) and z =f(x,y) find the partial derivative with respect to theta and then find the partial derivative of that with respect to r.
Posted Nov 21, 2016, 4:13 PM UTC
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If x=rcos(theta) and y=rsin(theta) and z =f(x,y) find the partial derivative with respect to theta and then find the partial derivative of that with respect to r.
Posted Nov 21, 2016, 4:13 PM UTC
Okay, I'm using the LaTeX chrome extension to display the maths for this, hopefully it'll go well.
[; \frac{d z}{d\theta} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial \theta} ;]
[; \frac{\partial z}{\partial \theta} = -r\sin(\theta) \frac{\partial z}{\partial x} + r\cos(\theta)\frac{\partial z}{\partial y} ;]
'[; \frac{\partial 2 z}{d\theta dr } = -\sin(\theta) \frac{\partial z}{\partial x} -r\sin(\theta) \frac{\partial}{\partial r}\frac{\partial z}{\partial x} + \cos(\theta)\frac{\partial z}{\partial y} + r\cos(\theta) \frac{\partial}{\partial r}\frac{\partial z}{\partial y} ;]'
Here we've moved the d/dr "through" the d/dx and d/dy. Perfectly legal, though I'm a physicist so I can't tell you why. :p
[; \frac{\partial 2 z}{\partial \theta \partial r } = -\sin(\theta) \frac{\partial z}{\partial x} -r\sin(\theta) \frac{\partial}{\partial x}\left[ \frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r} \right] + \cos(\theta)\frac{\partial z}{\partial y} + r\cos(\theta) \frac{\partial}{\partial y}\left[ \frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r} \right] ;]
Here, just like in the first part, we're using the chain rule to partially differentiate z with respect to r this time instead of θ.
[; \frac{\partial 2 z}{\partial \theta dr } = -\sin(\theta) \frac{\partial z}{\partial x} -r\sin(\theta) \left[ \frac{\partial 2 z}{\partial x2}\frac{\partial x}{\partial r} + \frac{\partial 2 z}{\partial x\partial y }\frac{\partial y}{\partial r} \right] + \cos(\theta)\frac{\partial z}{\partial y} + r\cos(\theta) \left[ \frac{\partial 2 z}{\partial x \partial y}\frac{\partial x}{\partial r} + \frac{\partial 2 z}{\partial y2}\frac{\partial y}{\partial r} \right] ;]
There, after some wrangling the only concession I've had to make is d^2 looks like d2. So yeah, we've made liberal use of the chain rule and in the later steps moved d/dr through d/dx so we could use the chain rule again.