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Answers without the blur. Sign up and see all textbooks for free! Q. 28

Expert-verified Found in: Page 964 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Use Theorem 12.33 to find the indicated derivatives in Exercises 27–30. Express your answers as functions of two variables.$\frac{\partial z}{\partial t}whenz={x}^{2}{y}^{3},x=t\mathrm{sin}s,andy=s\mathrm{cos}t$

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

See the step by step solution

## Step 1. Given Information:

Given:

$z={x}^{2}{y}^{3},\phantom{\rule{0ex}{0ex}}x=t\mathrm{sin}sand\phantom{\rule{0ex}{0ex}}y=s\mathrm{cos}t$

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

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

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

$\frac{\partial z}{\partial x}=2x{y}^{3}\phantom{\rule{0ex}{0ex}}\frac{\partial z}{\partial y}=3{x}^{2}{y}^{2}\phantom{\rule{0ex}{0ex}}\frac{\partial x}{\partial t}=\mathrm{sin}s\phantom{\rule{0ex}{0ex}}\frac{\partial y}{\partial t}=-s\mathrm{sin}t$

Use these values in (1) we get

role="math" localid="1649695474446" $\frac{\partial z}{\partial t}=2x{y}^{3}·\mathrm{sin}s+3{x}^{2}{y}^{2}·\left(-s\mathrm{sin}t\right)\phantom{\rule{0ex}{0ex}}\frac{\partial z}{\partial t}=2x{y}^{3}·\mathrm{sin}s-3{x}^{2}{y}^{2}s·\mathrm{sin}t$

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

We use $x=t\mathrm{sin}sandy=s\mathrm{cos}t$ to do so:

$\frac{\partial z}{\partial t}=2\left(t\mathrm{sin}s\right){\left(s\mathrm{cos}t\right)}^{3}·\mathrm{sin}s+3{\left(t\mathrm{sin}s\right)}^{2}{\left(s\mathrm{cos}t\right)}^{2}s·\mathrm{sin}t\phantom{\rule{0ex}{0ex}}\frac{\partial z}{\partial t}=2{s}^{3}t·{\mathrm{sin}}^{2}s·{\mathrm{cos}}^{3}t+3{s}^{2}{t}^{2}{\mathrm{sin}}^{2}s·{\mathrm{cos}}^{2}t·\mathrm{sin}t\phantom{\rule{0ex}{0ex}}\frac{\partial z}{\partial t}={s}^{2}t·{\mathrm{sin}}^{2}s·{\mathrm{cos}}^{2}t\left(2s\mathrm{cos}t+3t\mathrm{sin}t\right)$ ### Want to see more solutions like these? 