Application of Chain Rule for Paths

Question

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;

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

Answer 1

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.

Answer 2

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).