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Q38E

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Essential Calculus: Early Transcendentals
Found in: Page 657
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Show the equation \({\left( {\frac{{\partial u}}{{\partial x}}} \right)^2} + {\left( {\frac{{\partial u}}{{\partial y}}} \right)^2} = {e^{ - 2s}}\left( {{{\left( {\frac{{\partial u}}{{\partial s}}} \right)}^2} + {{\left( {\frac{{\partial u}}{{\partial t}}} \right)}^2}} \right)\) if\(u = f(x,y)\), where \(x = {e^s}\cos t\) and \(y = {e^s}\sin t.\)

The equation is \({\left( {\frac{{\partial u}}{{\partial x}}} \right)^2} + {\left( {\frac{{\partial u}}{{\partial y}}} \right)^2} = {e^{ - 2s}}\left( {{{\left( {\frac{{\partial u}}{{\partial s}}} \right)}^2} + {{\left( {\frac{{\partial u}}{{\partial t}}} \right)}^2}} \right)\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Chain rule

"Assume that \(z = f(x,y)\) is a differentiable function of \(x\) and \(y\), where \(x = g(t)\) and \(y = h(t)\) are both differentiable functions of \(t\). Then, \(z\) is differentiable function of \(t\) and\(\frac{{dz}}{{dt}} = \frac{{dz}}{{dx}} \cdot \frac{{dx}}{{dt}} + \frac{{dz}}{{dy}} \cdot \frac{{dy}}{{dt}}\)”

Step 2: Find the value of \(\partial u/\partial s\)and \(\partial u/\partial t\)

As given, the function is \(z = u(x,y)\)

\(\begin{aligned}{l}\frac{{\partial u}}{{\partial s}} = \frac{{\partial u}}{{\partial x}}\frac{{\partial x}}{{\partial s}} + \frac{{\partial u}}{{\partial y}}\frac{{\partial y}}{{\partial s}}\\\frac{{\partial u}}{{\partial s}} = {e^s}\cos t\frac{{\partial u}}{{\partial x}} + {e^s}\sin t\frac{{\partial u}}{{\partial y}} \ldots \ldots \ldots (1)\end{aligned}\)

Similarly,

\(\begin{aligned}{l}\frac{{\partial u}}{{\partial t}} = \frac{{\partial u}}{{\partial x}}\frac{{\partial x}}{{\partial t}} + \frac{{\partial u}}{{\partial y}}\frac{{\partial y}}{{\partial t}}\\\frac{{\partial u}}{{\partial t}} = - {e^s}\sin t\frac{{\partial u}}{{\partial x}} + {e^s}\cos t\frac{{\partial u}}{{\partial y}} \ldots \ldots \ldots (2)\end{aligned}\)

Step 3: Calculate the value of \({\left( {\frac{{\partial u}}{{\partial x}}} \right)^2} + {\left( {\frac{{\partial u}}{{\partial y}}} \right)^2}\)

By using equation (1) and (2)

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

Multiply both sides of the equation by \({e^{ - 2s}}\) to isolate \({(\partial u/\partial x)^2} + {(\partial u/\partial y)^2}\)

\({\left( {\frac{{\partial u}}{{\partial x}}} \right)^2} + {\left( {\frac{{\partial u}}{{\partial y}}} \right)^2} = {e^{ - 2s}}\left( {{{\left( {\frac{{\partial u}}{{\partial s}}} \right)}^2} + {{\left( {\frac{{\partial u}}{{\partial t}}} \right)}^2}} \right)\)

Hence, \({\left( {\frac{{\partial u}}{{\partial x}}} \right)^2} + {\left( {\frac{{\partial u}}{{\partial y}}} \right)^2} = {e^{ - 2s}}\left( {{{\left( {\frac{{\partial u}}{{\partial s}}} \right)}^2} + {{\left( {\frac{{\partial u}}{{\partial t}}} \right)}^2}} \right)\).

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