The Chain Rule - finding differentials

Question

I have been given:

$V=f(x,y,z)$, with $x= r\cos\theta$, $y= r\sin\theta$ , and $z=t$.
And asked, find $dV/dr$, $dV/d\theta$ and $dV/dt$

Would $$ \frac{dV}{dr} = \frac{dV}{dx} * \frac{dx}{dr} + \frac{dV}{dy} * \frac{dy}{dr} $$ ? If so, what is $dV/dx$ if I have just been given $V=f(x,y,z)$?

I've obtained:

$$V_{r}=V_{x}\cos(\theta )+V_{y}\sin(\theta )$$

I am unsure how to display my final answer.

Do you know about partial derivatives? You really want $$\frac{\partial V}{\partial x}=\frac{\partial f}{\partial x}$$ — saulspatz, Apr 13 '18 at 14:43
Yes I'm well aware of partial derivatives and I think I've used them correctly in my most recent edit — QuestionAnswer13, Apr 13 '18 at 14:47
I've used partial differentiation correctly and obtained my edit — QuestionAnswer13, Apr 13 '18 at 14:48

score 1 · Answer 1 · answered Apr 13 '18 at 16:51

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First: using the same name for different things is a bad habit that causes confusion. Use different names for different things. In your case: $$V(r,\theta,t) = f(r\cos\theta,r\sin\theta,t).$$ Then: $$ \frac{\partial V}{\partial r} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial r} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial r} = \cdots $$

answered Apr 13 '18 at 16:51

Martín-Blas Pérez Pinilla

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I'm confused as to what V is defined as? What function is it? – QuestionAnswer13 Apr 15 '18 at 20:25
@JonnyWhite13, read again the definition: $V(r,\theta,t) = f(r\cos\theta,r\sin\theta,t)$. – Martín-Blas Pérez Pinilla Apr 15 '18 at 20:30

The Chain Rule - finding differentials

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