Answer
The rate of change of the bug's temperature at the angle $\theta = \frac{\pi }{3}$ is $ - 7.33$ degrees Celsius per second.
Work Step by Step
We are given $T\left( {x,y} \right) = 20 + 0.1\left( {{x^2} - xy} \right)$, the radius of the circle $R=200$ cm and the speed of the bug, $v=3$ cm/s. So, the gradient of $T$ is
$\nabla T\left( {x,y} \right) = 0.1\left( {2x - y, - x} \right)$
Recall from Example 5 of Section 12.2 on page 609, the angular velocity of the bug is given by $\omega = \frac{v}{R}$. So, $\omega = \frac{3}{{200}} = 0.015$ radian per second. Thus, the path of the bug can be parametrized by
${\bf{r}}\left( t \right) = 200\left( {\cos 0.015t,\sin 0.015t} \right) + {\bf{c}}$ ${\ \ }$ for $0 \le t \le 2\pi $,
where ${\bf{c}}$ is a constant determined by the initial condition.
At time $t=0$, the bug is located at $\left( {200,0} \right)$, so
${\bf{r}}\left( 0 \right) = 200\left( {1,0} \right) + {\bf{c}} = \left( {200,0} \right)$
Thus, ${\bf{c}} = 0$. Therefore, the path is given by
${\bf{r}}\left( t \right) = \left( {200\cos 0.015t,200\sin 0.015t} \right)$ ${\ \ }$ for $0 \le t \le 2\pi $
The velocity is
${\bf{r}}'\left( t \right) = \left( { - 3\sin 0.015t,3\cos 0.015t} \right)$
At the angle $\theta = \frac{\pi }{3}$, we find the corresponding time $t$:
$0.015t = \frac{\pi }{3}$
$t = \frac{{200\pi }}{9}$
So, the velocity at $\theta = \frac{\pi }{3}$ is
${\bf{r}}'\left( {\frac{{200\pi }}{9}} \right) = \left( { - 3\sin \left( {0.015\cdot\frac{{200\pi }}{9}} \right),3\cos \left( {0.015\cdot\frac{{200\pi }}{9}} \right)} \right)$
${\bf{r}}'\left( {\frac{{200\pi }}{9}} \right) = \left( { - 2.6,1.5} \right)$
The unit velocity vector ${\bf{u}}$ is
${\bf{u}} = \frac{{\left( { - 2.6,1.5} \right)}}{{||\left( { - 2.6,1.5} \right)||}} = \left( { - 0.87,0.5} \right)$
Recall from our previous result:
$\nabla T\left( {x,y} \right) = 0.1\left( {2x - y, - x} \right)$
So,
$\nabla T\left( {{\bf{r}}\left( t \right)} \right) = 0.1\left( {400\cos \left( {0.015t} \right) - 200\sin \left( {0.015t} \right), - 200\cos \left( {0.015t} \right)} \right)$
At $t = \frac{{200\pi }}{9}$, we get
$\nabla T\left( {{\bf{r}}\left( {\frac{{200\pi }}{9}} \right)} \right) = \left( {2.68, - 10} \right)$
Using Theorem 3, the rate of change of temperature at $t = \frac{{200\pi }}{9}$ is
${D_{\bf{u}}}T{|_{t = \frac{{200\pi }}{9}}} = \nabla T\left( {{\bf{r}}\left( {\frac{{200\pi }}{9}} \right)} \right)\cdot{\bf{u}}$
${D_{\bf{u}}}T{|_{t = \frac{{200\pi }}{9}}} = \left( {2.68, - 10} \right)\cdot\left( { - 0.87,0.5} \right) \simeq - 7.33$
So, the rate of change of the bug's temperature at the angle $\theta = \frac{\pi }{3}$ is $ - 7.33$ degrees Celsius per second.
