If $\mathit{P}\mathbf{=}\mathit{r}\mathit{c}\mathit{o}\mathit{s}\mathit{t}$ and role="math" localid="1664266535265" $\mathit{r}\mathit{s}\mathit{i}\mathit{n}\mathit{t}\mathbf{-}\mathbf{2}\mathit{t}{\mathit{e}}^{\mathbf{r}}\mathbf{=}\mathbf{0}$, find $\frac{\mathbf{d}\mathbf{P}}{\mathbf{d}\mathbf{t}}$ .

If any function $v=v(x,y)$has independent variables which are also functions of some independent variables such that $x=x(s,t)andy=y(s,t)$, then chain rule for the function is given by:

$$\begin{gathered} \frac{\partial v}{\partial s}=\frac{\partial v}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial v}{\partial y}\frac{\partial y}{\partial s} \\ \frac{\partial v}{\partial t}=\frac{\partial v}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial v}{\partial y}\frac{\partial y}{\partial t} \end{gathered}$$

The given functions are: $P=r\cos tandr=\sin t-2t{e}^r=0$.

Differentiate$P=r\cos t$as follows:

localid="1664271551113" $$\begin{gathered} P=r\cos t \\ dp=\cos tdr-r\sin tdt........\left(1\right) \end{gathered}$$

Similarly, differentiate$r\sin t-2t{e}^r=0$as follows:

localid="1664271586869" $$\begin{gathered} r\sin t=2t{e}^r=0 \\ (r\cos t-2e^r)dt+(\sin t-2t{e}^r)dr=0 \\ dr=\frac{r\cos t-2e^r}{2t{e}^r-\sin t}dt.........\left(2\right) \end{gathered}$$

From equations (1) and (2):

$$\begin{gathered} dP=\cos t\left[\frac{r\cos t-2e^r}{2t{e}^r-\sin t}dt\right]-r\sin tdt \\ =\left[\frac{r{\cos }^{2}t-2\cos t{e}^r}{2t{e}^r-\sin t}-r\sin t\right]dt \\ dP=\left[\frac{r-2tr{e}^2\sin t-2e^r\cos t}{2t{e}^r-\sin t}\right]dt \\ \frac{dP}{dt}=\frac{r-2tr{e}^r\sin t-2e^r\cos t}{2t{e}^r-\sin t} \end{gathered}$$

Hence, the required answer is:

$\frac{dP}{dt}=\frac{r-2tr{e}^r\sin t-2e^r\cos t}{2t{e}^r-\sin t}$