- #1

Master1022

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- Homework Statement:
- Show that the result obtained can be done directly via the chain rule for partial derivatives.

- Relevant Equations:
- Chain rule for partial derivatives

For context, we have an equation [itex] f(x,y) = \frac{x}{y} [/itex] and we had used the substitutions [itex] x = r \cos\theta [/itex] and [itex] y = r \sin\theta [/itex]. In the previous parts of the question, we have shown the following result:

[tex] \frac{\partial f}{\partial x} = \cos\theta \Big(\frac{\partial f}{\partial r}\Big) - \frac{\sin\theta}{r} \Big(\frac{\partial f}{\partial \theta}\Big) [/tex]

Now we have been asked to get to this result directly using the chain rule for partial derivatives.

My attempt: the chain rule states that:

[tex] \frac{\partial f}{\partial x} = \Big(\frac{\partial f}{\partial r}\Big) \Big(\frac{\partial r}{\partial x}\Big) + \Big(\frac{\partial f}{\partial \theta}\Big) \Big(\frac{\partial \theta}{\partial x}\Big) [/tex]

This is where I seem to have gone wrong, and I would appreciate any insight as to why that may have been the case. The answers have used [itex] r^2 = x^2 + y^2 [/itex] and [itex] \tan\theta = \frac{y}{x} [/itex] and have gone on to the answer. I, however, just used the substitution above [itex] x = r\cos\theta [/itex], but this doesn't seem to lead us in the right direction. Showing my method, we get [itex] \frac{\partial r}{\partial x} = \frac{1}{cos\theta} [/itex] and [itex] \frac{\partial \theta}{\partial x} = \frac{-1}{r\sin\theta} [/itex]. Substituting into the above chain rule yields:

[tex] \frac{\partial f}{\partial x} = \Big(\frac{\partial f}{\partial r}\Big) \Big(\frac{1}{cos\theta} \Big) + \Big(\frac{\partial f}{\partial \theta}\Big) \Big(\frac{-1}{r\sin\theta}\Big) [/tex]. I am not sure I can see any clever way to get us to the right answer from here.

Looking back, I think the fact that I had two expressions for [itex] r [/itex] in terms of [itex] x [/itex] and [itex] y[/itex] could have been part of the problem. However, I cannot seem to put an exact reason as to why that would be the case.

Thanks in advance.

[tex] \frac{\partial f}{\partial x} = \cos\theta \Big(\frac{\partial f}{\partial r}\Big) - \frac{\sin\theta}{r} \Big(\frac{\partial f}{\partial \theta}\Big) [/tex]

Now we have been asked to get to this result directly using the chain rule for partial derivatives.

My attempt: the chain rule states that:

[tex] \frac{\partial f}{\partial x} = \Big(\frac{\partial f}{\partial r}\Big) \Big(\frac{\partial r}{\partial x}\Big) + \Big(\frac{\partial f}{\partial \theta}\Big) \Big(\frac{\partial \theta}{\partial x}\Big) [/tex]

This is where I seem to have gone wrong, and I would appreciate any insight as to why that may have been the case. The answers have used [itex] r^2 = x^2 + y^2 [/itex] and [itex] \tan\theta = \frac{y}{x} [/itex] and have gone on to the answer. I, however, just used the substitution above [itex] x = r\cos\theta [/itex], but this doesn't seem to lead us in the right direction. Showing my method, we get [itex] \frac{\partial r}{\partial x} = \frac{1}{cos\theta} [/itex] and [itex] \frac{\partial \theta}{\partial x} = \frac{-1}{r\sin\theta} [/itex]. Substituting into the above chain rule yields:

[tex] \frac{\partial f}{\partial x} = \Big(\frac{\partial f}{\partial r}\Big) \Big(\frac{1}{cos\theta} \Big) + \Big(\frac{\partial f}{\partial \theta}\Big) \Big(\frac{-1}{r\sin\theta}\Big) [/tex]. I am not sure I can see any clever way to get us to the right answer from here.

Looking back, I think the fact that I had two expressions for [itex] r [/itex] in terms of [itex] x [/itex] and [itex] y[/itex] could have been part of the problem. However, I cannot seem to put an exact reason as to why that would be the case.

Thanks in advance.