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Q38E

Expert-verifiedFound in: Page 657

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

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

**"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**\(\frac{{dz}}{{dt}} = \frac{{dz}}{{dx}} \cdot \frac{{dx}}{{dt}} + \frac{{dz}}{{dy}} \cdot \frac{{dy}}{{dt}}\)**”**

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,

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

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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