91Ó°ÊÓ

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:

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

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:

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