91Ó°ÊÓ

Given:

$$\begin{gathered} z=x^2{y}^3, \\ x=t\sin sand \\ y=s\cos t \end{gathered}$$

We have to find the indicated derivatives and express your answers as functions of a single variable.

By Theorem 12.33, we have

$\frac{∂z}{∂s}=\frac{∂z}{∂x}·\frac{∂x}{∂s}+\frac{∂z}{∂y}·\frac{∂y}{∂s}---\left(1\right)$

So first we find $\frac{∂z}{∂x},\frac{∂x}{∂s},\frac{∂z}{∂y}·\frac{∂y}{∂s}$

So we have

$$\begin{gathered} \frac{∂z}{∂x}=2x{y}^3 \\ \frac{∂z}{∂y}=3x^2{y}^2 \\ \frac{∂x}{∂s}=t\cos s \\ \frac{∂y}{∂s}=\cos t \end{gathered}$$

Use these values in (1) we get

localid="1649693736225" $\frac{∂z}{∂s}=2x{y}^3·t·\cos t+3x^2{y}^2·cost$

This result is correct, but it is preferable to write the function as a function of just s and t.

We use $x=t\sin sandy=s\cos t$to do so:

localid="1649695745634" role="math" $$\begin{gathered} \frac{∂z}{∂s}=2(t\sin s){(s\cos t)}^3·t·\cos t+3{(t\sin s)}^2{(s\cos t)}^2·cost \\ \frac{∂z}{∂s}=2s^3{t}^2·\sin s·{\cos }^{4}t+3s^2{t}^2{\sin }^{2}s·{\cos }^{3}t \\ \frac{∂z}{∂s}=s^2{t}^2·\sin s·{\cos }^{3}t(2s\cos t+3\sin s) \end{gathered}$$