Q. 29 - Calculus

Question

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

$\frac{\partial z}{\partial r} when z=(x^2+ xy)e^y, x=r \cos \theta and y=r \sin \theta$

Step-by-Step Solution

Answer

The value is $\frac{\partial z}{\partial r}=r \cos \theta e^{r \sin \theta }\left[(2+ \sin \theta +\sin \theta (r \cos \theta +r \sin \theta +1)\right]$

Step 1. Given Information:

Given:

$$\begin{gathered} z=(x^2+ xy)e^y, \\ x=r \cos \theta and \\ y=r \sin \theta \end{gathered}$$

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

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

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

So we have

$$\begin{gathered} \frac{\partial z}{\partial x}=e^y(2x+y) \\ \frac{\partial z}{\partial y}=e^y·x^2+x{e}^y+xy{e}^y \\ \frac{\partial x}{\partial r}=\cos \theta \\ \frac{\partial y}{\partial r}=\sin \theta \end{gathered}$$

Use these values in (1) we get

$$\begin{gathered} \frac{\partial z}{\partial r}=e^y(2x+y)·\cos \theta +x{e}^y(x+1+y)·\sin \theta \\ \frac{\partial z}{\partial r}=e^y\left[(2x+y)·\cos \theta +x \sin \theta (x+y+1)\right] \end{gathered}$$

This result is correct, but it is preferable to write the function as a function of just $r and \theta$.

We use $x=r \cos \theta and y=r \sin \theta$ to do so:

$$\begin{gathered} \frac{\partial z}{\partial r}=e^{r \sin \theta }\left[\left(2\right(r \cos \theta )+r \sin \theta )·\cos \theta +r \cos \theta \sin \theta (r \cos \theta +r \sin \theta +1)\right] \\ \frac{\partial z}{\partial r}=r \cos \theta e^{r \sin \theta }\left[(2+ \sin \theta +\sin \theta (r \cos \theta +r \sin \theta +1)\right] \end{gathered}$$