Q. 28 - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

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

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

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

So we have

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

Use these values in (1) we get

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

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:

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