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Found in: Page 657

### Essential Calculus: Early Transcendentals

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

See the step by step solution

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