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This is an exam practice question.

Suppose that $f(x, y)$ is a function of two variables with $f_x(0, 2) = 2$ and $f_y(0, 2) = -1$. Using the chain rule, compute the numerical value of $f_\theta(r\cos\theta, r\sin\theta)$ at $r=2$, $\theta=\pi/2$.

So for this question, I have $x = r\cos\theta$ and $y = r\sin\theta$.

Using the chain rule, I have: \[ f_\theta(r\cos\theta, r\sin\theta) = f_x(x,y)\frac d{d\theta}(r\cos\theta) + f_y(x,y)\frac d{d\theta}(r\sin\theta) \] \[ f_\theta(2\cos\pi/2, 2\sin\pi/2) = (2)(-2\sin\pi/2) - 1(2\cos\pi/2) = -4 \] Do I have the right idea here? Or am I totally off?

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@mathstudent Yes. Your computations are good! – user17762 Jun 13 '12 at 1:36
@Marvis - Thanks for your feedback! – JackReacher Jun 13 '12 at 1:46
Your working is correct. – user22805 Jun 13 '12 at 2:31
You are totally on. – copper.hat Jun 13 '12 at 2:35

Yes. Your computations are good! – Marvis Jun 13 at 1:36

Your working is correct. – David Wallace Jun 13 at 2:31

You are totally on. – copper.hat Jun 13 at 2:35

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merged comment-answers into a CW. – user31373 Jun 19 '12 at 18:50

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