Answer
At $t = \pi $, the rate of change of the difference $D$ in the temperatures the two spacecraft experience is
$\frac{{dD}}{{dt}}{|_{t = \pi }} = 1 + 3{\pi ^2} - 4{\pi ^3} \simeq - 93.42$
Work Step by Step
We are given the paths of two spacecraft ${{\bf{r}}_1}\left( t \right) = \left( {\sin t,t,{t^2}} \right)$ and ${{\bf{r}}_2}\left( t \right) = \left( {\cos t,1 - t,{t^3}} \right)$.
If the temperature for points in space are given by $T\left( {x,y,z} \right) = {x^2}y\left( {1 - z} \right)$, then the temperatures the two spacecraft experience are
$T\left( {{{\bf{r}}_1}\left( t \right)} \right) = t\left( {1 - {t^2}} \right){\sin ^2}t$ ${\ \ }$
and
$T\left( {{{\bf{r}}_2}\left( t \right)} \right) = \left( {1 - t} \right)\left( {1 - {t^3}} \right){\cos ^2}t$,
respectively
Thus, the difference $D$ in temperatures is
$D = T\left( {{{\bf{r}}_1}\left( t \right)} \right) - T\left( {{{\bf{r}}_2}\left( t \right)} \right) = t\left( {1 - {t^2}} \right){\sin ^2}t - \left( {1 - t} \right)\left( {1 - {t^3}} \right){\cos ^2}t$
$D = t{\sin ^2}t - {t^3}{\sin ^2}t - {\cos ^2}t + t{\cos ^2}t + {t^3}{\cos ^2}t - {t^4}{\cos ^2}t$
Since ${\cos ^2}t + {\sin ^2}t = 1$, so
$D = t - {t^3}{\sin ^2}t - {\cos ^2}t + {t^3}{\cos ^2}t - {t^4}{\cos ^2}t$
$D = t - {t^3}{\sin ^2}t - \left( {1 - {t^3} + {t^4}} \right){\cos ^2}t$
So, the rate of change of the difference $D$ in the temperatures the two spacecraft experience is
$\frac{{dD}}{{dt}} = 1 - 3{t^2}{\sin ^2}t - 2{t^3}\sin t\cos t - \left( { - 3{t^2} + 4{t^3}} \right){\cos ^2}t + 2\left( {1 - {t^3} + {t^4}} \right)\cos t\sin t$
At $t = \pi $, we get
$\frac{{dD}}{{dt}}{|_{t = \pi }} = 1 + 3{\pi ^2} - 4{\pi ^3} \simeq - 93.42$
