Q. 27 - Calculus

Question

Use Theorem 12.33 to find the indicated derivatives in Exercises 27–30. Express your answers as functions of two variables.

$\frac{\partial z}{\partial s} when z=x^2{y}^3, x=t \sin s, and y=s \cos t$

Step-by-Step Solution

Answer

The value is $\frac{\partial z}{\partial s}=s^2{t}^2·\sin s·{\cos }^3 t (2s \cos t+3 \sin s)$

Step 1. Given Information:

Given:

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

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

Step 2. Solution:

By Theorem 12.33, we have

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

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

So we have

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

Use these values in (1) we get

$\frac{\partial z}{\partial s}=2x{y}^3·t·\cos t+3x^2{y}^2·cos t$

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

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

$$\begin{gathered} \frac{\partial z}{\partial s}=2(t \sin s){(s \cos t)}^3·t·\cos t+3{(t \sin s)}^2{(s \cos t)}^2·cos t \\ \frac{\partial z}{\partial s}=2s^3{t}^2·\sin s·{\cos }^4 t+3s^2{t}^2 {\sin }^2 s·{\cos }^3 t \\ \frac{\partial z}{\partial s}=s^2{t}^2·\sin s·{\cos }^3 t (2s \cos t+3 \sin s) \end{gathered}$$