Chapter 12: Q. 28 (page 964)

Use Theorem 12.33 to find the indicated derivatives in Exercises 27–30. Express your answers as functions of two variables.
$\frac{\partial z}{\partial t} w h e n z = x^{2} y^{3}, x = t \sin s, a n d y = s \cos t$

Short Answer

Expert verified

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

Step by step solution

Step 1. Given Information:

Given:

$z = x^{2} y^{3}, x = t \sin s a n d y = s \cos t$

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

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

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

So we have

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

Use these values in (1) we get

role="math" localid="1649695474446" $\frac{\partial z}{\partial t} = 2 x y^{3} \cdot \sin s + 3 x^{2} y^{2} \cdot (- s \sin t) \frac{\partial z}{\partial t} = 2 x y^{3} \cdot \sin s - 3 x^{2} y^{2} s \cdot \sin t$

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

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

$\frac{\partial z}{\partial t} = 2 (t \sin s) {(s \cos t)}^{3} \cdot \sin s + 3 {(t \sin s)}^{2} {(s \cos t)}^{2} s \cdot \sin t \frac{\partial z}{\partial t} = 2 s^{3} t \cdot \sin^{2} s \cdot \cos^{3} t + 3 s^{2} t^{2} \sin^{2} s \cdot \cos^{2} t \cdot \sin t \frac{\partial z}{\partial t} = s^{2} t \cdot \sin^{2} s \cdot \cos^{2} t (2 s \cos t + 3 t \sin t)$

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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.
$\frac{\partial z}{\partial t} w h e n z = x^{2} y^{3}, x = t \sin s, a n d y = s \cos t$

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∂twhenz=x2y3,x=tsins,andy=scost

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

Statistics

Pure Maths

Geometry

Applied Mathematics

Discrete Mathematics

Theoretical and Mathematical Physics

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.
$\frac{\partial z}{\partial t} w h e n z = x^{2} y^{3}, x = t \sin s, a n d y = s \cos t$