Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

Find the partial derivative \(\frac{{\partial z}}{{\partial s}}\)and \(\frac{{\partial z}}{{\partial t}}\) with the help of chain rule. The functions are \(z = {x^2}{y^3},x = s\cos t\) and \(y = s\sin t.\)

The partial derivative \(\frac{{\partial z}}{{\partial s}}\) and \(\frac{{\partial z}}{{\partial t}}\) with the help of chain rule are \(2x{y^3}\cos t + 3{x^2}{y^2}\sin t\) and \( - 2sx{y^3}\sin t + 3s{x^2}{y^2}\cos t\) respectively.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The chain rule.

With the chain rule, the partial derivative of function \({\rm{f(x, y)}}\) with respect to \({\rm{t}}\) becomes \(\frac{{\partial f}}{{\partial t}} = \frac{{\partial f}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial t}} + \frac{{\partial f}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial t}}.\)

Step 2: Use the chain rule for calculation.

Differentiate \(z\) partially with respect to \(s\) as follows.

\(\frac{{\partial z}}{{\partial s}} = \frac{{\partial z}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial s}} + \frac{{\partial z}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial s}}\)

Differentiate \(z\) partially with respect to \(x\) as follows.

\(\begin{aligned}{l}\frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {{x^2}{y^3}} \right)\\ = {y^3}\frac{\partial }{{\partial x}}\left( {{x^2}} \right)\\ = {y^3}(2x)\\ = 2x{y^3}\end{aligned}\)

Differentiate \(z\) partially with respect to \(y\) as follows.

\(\begin{aligned}{l}\frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {{x^2}{y^3}} \right)\\ = {x^2}\frac{\partial }{{\partial y}}\left( {{y^3}} \right)\\ = {x^2}\left( {3{y^2}} \right)\\ = 3{x^2}{y^2}\end{aligned}\)

Differentiate \(x\) partially with respect to \(s\) as follows.

\(\frac{{\partial x}}{{\partial s}} = \frac{\partial }{{\partial s}}(s\cos t) = \cos t\)

Differentiate \(y\) partially with respect to \(s\) as follows.

\(\frac{{\partial y}}{{\partial s}} = \frac{\partial }{{\partial s}}(s\sin t) = \sin t\)

Then, the partial derivative of\(z\) with respect to\(s\) becomes.

\(\begin{aligned}{l}\frac{{\partial z}}{{\partial s}} = \left( {2x{y^3}} \right)(\cos t) + \left( {3{x^2}{y^2}} \right)(\sin t)\\ = 2x{y^3}\cos t + 3{x^2}{y^2}\end{aligned}\)

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Essential Calculus: Early Transcendentals

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Short Answer

Find the partial derivative \(\frac{{\partial z}}{{\partial s}}\)and \(\frac{{\partial z}}{{\partial t}}\) with the help of chain rule. The functions are \(z = {x^2}{y^3},x = s\cos t\) and \(y = s\sin t.\)

Step by Step Solution

Step 1: The chain rule.

Step 2: Use the chain rule for calculation.

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Find the partial derivative \(\frac{{\partial z}}{{\partial s}}\)and \(\frac{{\partial z}}{{\partial t}}\) with the help of chain rule. The functions are \(z = {x^2}{y^3},x = s\cos t\) and \(y = s\sin t.\)

Step by Step Solution

Step 1: The chain rule.

Step 2: Use the chain rule for calculation.

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