Chapter 12: Q. 30 (page 964)

Use Theorem 12.33 to find the indicated derivatives in Exercises 27–30. Express your answers as functions of two variables.
$\frac{\partial z}{\partial θ} w h e n z = (x^{2} + x y) e^{y}, x = r \cos θ a n d y = r \sin θ$

Short Answer

Expert verified

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

Step by step solution

Step 1. Given Information:

Given:

$z = (x^{2} + x y) e^{y}, x = r \cos θ a n d y = r \sin θ$

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

$\frac{\partial z}{\partial θ} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial θ} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial θ} - - - (1)$

So first we find $\frac{\partial z}{\partial x}, \frac{\partial x}{\partial θ}, \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial θ}$

So we have

$\frac{\partial z}{\partial x} = e^{y} (2 x + y) \frac{\partial z}{\partial y} = e^{y} \cdot x^{2} + x e^{y} + x y e^{y} \frac{\partial x}{\partial θ} = - r s i n θ \frac{\partial y}{\partial θ} = r \cos θ$

Use these values in (1) we get

$\frac{\partial z}{\partial r} = e^{y} (2 x + y) \cdot - r s i n θ + x e^{y} (x + 1 + y) \cdot r \cos θ \frac{\partial z}{\partial r} = r e^{y} [(2 x + y) \cdot s i n θ + x \cos θ (x + y + 1)]$

This result is correct, but it is preferable to write the function as a function of just $r a n d θ$ .

We use $x = r \cos θ a n d y = r \sin θ$ to do so:

$\frac{\partial z}{\partial r} = r e^{r \sin θ} [(2 (r \cos θ) + r \sin θ) \cdot \sin θ + r \cos θ \cos θ (r \cos θ + r \sin θ + 1)] \frac{\partial z}{\partial r} = r^{2} e^{r \sin θ} [(2 \sin θ \cos θ + \sin^{2} θ + \cos^{2} θ (r \cos θ + r \sin θ + 1)] \frac{\partial z}{\partial r} = r^{2} e^{r \sin θ} [(\sin 2 θ + \sin^{2} θ + \cos^{2} θ (r \cos θ + r \sin θ + 1)]$

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Use Theorem 12.33 to find the indicated derivatives in Exercises 27–30. Express your answers as functions of two variables.
$\frac{\partial z}{\partial θ} w h e n z = (x^{2} + x y) e^{y}, x = r \cos θ a n d y = r \sin θ$

Short Answer

Step by step solution

Step 1. Given Information:

Step 2. Solution:

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Use Theorem 12.33 to find the indicated derivatives in Exercises 27–30. Express your answers as functions of two variables.∂z∂θwhenz=(x2+xy)ey,x=rcosθandy=rsinθ

Short Answer

Step by step solution

Step 1. Given Information:

Step 2. Solution:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

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Study anywhere. Anytime. Across all devices.

Use Theorem 12.33 to find the indicated derivatives in Exercises 27–30. Express your answers as functions of two variables.
$\frac{\partial z}{\partial θ} w h e n z = (x^{2} + x y) e^{y}, x = r \cos θ a n d y = r \sin θ$