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I'm a graduate student and I'm currently teaching multivariable calculus. I gave my students a question about a bug traveling along a circle of radius $200$cm in the $xy$-plane. We suppose also that the speed of the bug is $3$ cm/s. Now let $T(x,y)$ be the temperature at any time $t$. The question is, what is the change in temperature at $t = \frac{\pi}{3}$.

If we parametrize the circle as $\gamma(t) = \langle 200 \cos t, 200 \sin t \rangle$ and since speed at time $t$ is $\|\gamma'(t)\| = 200$, I was thinking that I would need to alter this parametrization so that $\|\gamma'(t)\| = 3$ for all $t$ to agree with the speed of the bug. Doing so I get the parametrization;

$$g(s) = \left\langle 200 \cos\left( \frac{3s}{200}\right), 200 \sin\left( \frac{3s}{200} \right)\right\rangle $$

Now we just gave to compute;

$$\frac{d}{dt} \left(T(g(s)) \right|_{t = \pi/3} = \nabla T (g(\pi/3)) \cdot g'(\pi/3)$$

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  • $\begingroup$ Have them draw a little picture. $\endgroup$
    – copper.hat
    Jun 30, 2016 at 18:14

2 Answers 2

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You have $\tau = T \circ \gamma$, and you want $D \tau( { \pi \over 3}) = D T(\gamma( { \pi \over 3})) D \gamma ({ \pi \over 3})$.

The value $D T(\gamma( { \pi \over 3}))$ is independent of the parameterisation and $D \gamma ({ \pi \over 3})$ is given by the fact that the angle is ${ \pi \over 3}$ and the speed.

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  • $\begingroup$ So do I just use $\nabla T(\gamma(\pi/3)) \cdot \gamma'(\pi/3) =\|\nabla T(\gamma(\pi/3))\| \cdot 3 \cdot \cos (\pi/3)$ ? $\endgroup$ Jun 30, 2016 at 19:05
  • $\begingroup$ No. It would be $\nabla T((\cos{ \pi \over 3}, \sin{ \pi \over 3})) \cdot (3 ( -\sin { \pi \over 3} , \cos{ \pi \over 3}))$. $\endgroup$
    – copper.hat
    Jun 30, 2016 at 19:08
  • $\begingroup$ That makes sense. I was confusing that fact that given a parametrization, $\|\gamma'(t)\|$ gives the speed for some particle at time $t$. In this case, the speed of the particle does not coincide with that of the curve, the curve is just used for position. Does that make sense? That was the reason why I tried to alter the curve so that that speed for the parametrization agreed with that of the bug. $\endgroup$ Jun 30, 2016 at 19:17
  • $\begingroup$ As Ted mentioned below the only thing that really matters is the bug's velocity at the point in question. Also, in this case, there is no alternative parameterisation if the bug is stuck on the circle and the speed is fixed. So, the parameterisation you chose is the only one. $\endgroup$
    – copper.hat
    Jun 30, 2016 at 21:20
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You don't need a specific parametrization to apply the chain rule. Just dot the gradient vector of $T$ at the point with the velocity vector (which you know because you know length — the speed — and direction).

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  • $\begingroup$ Why is this not dependent on parametrization? The function T(x,y) is dependent on (x,y), so position matters. $\endgroup$ Jun 30, 2016 at 17:56
  • $\begingroup$ So the velocity vector should be $3 \langle \cos \pi/3, \sin \pi/3 \rangle$. $\endgroup$ Jun 30, 2016 at 18:02
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    $\begingroup$ You need to know the point $\mathbf g(t_0)$ and the velocity vector $\mathbf g'(t_0)$. You do not need to know $\mathbf g(t)$!! (This is discussed you-know-where. :) ) Note that you could just plug in the parametrization and never use the multivariable chain rule at all; why not teach your students to think and apply the new material thoughtfully? $\endgroup$ Jun 30, 2016 at 18:05
  • $\begingroup$ I wanted this to be done specifically using this chain rule for path. I mean sometimes it is unnecessary to use chain rule if you have simple functions but to get used to the concept we employ it in simple cases and say it will be really used for harder cases. I know this could have been easily done by plugging in the coordinates $x(t), y(t)$ into $T$ and taking a derivative but that wasn't the point. $\endgroup$ Jun 30, 2016 at 19:20

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