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### Essential Calculus: Early Transcendentals

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

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

See the step by step solution

## Step 1: Obtain the partial derivative.

The function is$${\rm{z = u(x, y)}}.$$

Obtain the partial derivative$$\frac{{\partial u}}{{\partial s}}.$$

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

Thus, the partial derivative$$\frac{{\partial u}}{{\partial s}} = \frac{{\partial u}}{{\partial x}} \cdot {e^s}\cos t + \frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t.$$

Obtain the partial derivative$$\frac{{du}}{{dt}}.$$

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

Thus, the partial derivative$$\frac{{\partial u}}{{\partial t}} = \frac{{\partial u}}{{\partial x}}\left( { - {e^s}\sin t} \right) + \frac{{\partial u}}{{\partial y}}\left( {{e^s}\cos t} \right).$$

Obtain$$\frac{{{\partial ^2}u}}{{\partial {s^2}}}.$$

## Step 2: Partial derivative with respect to$$s$$and$$t.$$

Take partial derivative of$$\frac{{\partial u}}{{\partial s}}.$$with respect to$$s.$$

\begin{aligned}{l}\frac{{{\partial ^2}u}}{{\partial {s^2}}} = \frac{{\partial u}}{{\partial x}} \cdot {e^s}\cos t + {e^s}\cos t \cdot \frac{\partial }{{\partial s}}\left( {\frac{{\partial u}}{{\partial x}}} \right) + \frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t + {e^s}\sin t \cdot \frac{\partial }{{\partial s}}\left( {\frac{{\partial u}}{{\partial y}}} \right)\\ = \left( {\begin{aligned}{*{20}{l}}{\frac{{\partial u}}{{\partial x}} \cdot {e^s}\cos t + {e^s}\cos t\left( {\frac{{{\partial ^2}u}}{{\partial {x^2}}} \cdot \frac{{\partial x}}{{\partial s}} + \frac{{{\partial ^2}u}}{{\partial y\partial x}} \cdot \frac{{\partial y}}{{\partial s}}} \right) + }\\{\frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t + {e^s}\sin t\left( {\frac{{{\partial ^2}u}}{{\partial {y^2}}} \cdot \frac{{\partial y}}{{\partial s}} + \frac{{{\partial ^2}u}}{{\partial x\partial y}} \cdot \frac{{\partial x}}{{\partial s}}} \right)}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{\frac{{\partial u}}{{\partial x}} \cdot {e^s}\cos t + {e^s}\cos t\left( {\frac{{{\partial ^2}u}}{{\partial {x^2}}} \cdot {e^s}\cos t + \frac{{{\partial ^2}u}}{{\partial y\partial x}} \cdot {e^s}\sin t} \right) + }\\{\frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t + {e^s}\sin t\left( {\frac{{{\partial ^2}u}}{{\partial {y^2}}} \cdot {e^s}\sin t + \frac{{{\partial ^2}u}}{{\partial x\partial y}} \cdot {e^s}\cos t} \right)}\end{aligned}} \right)\end{aligned}

The partial derivatives$$\frac{{{\partial ^2}u}}{{\partial y\partial x}} = \frac{{{\partial ^2}u}}{{\partial x\partial y}}$$, as functions are continuous.

Simplify$$\frac{{{\partial ^2}u}}{{\partial {s^2}}}$$as follows.

\begin{aligned}{l}\frac{{{\partial ^2}u}}{{\partial {s^2}}} = \left( {\begin{aligned}{*{20}{l}}{\frac{{\partial u}}{{\partial x}} \cdot {e^s}\cos t + {e^s}\cos t\left( {\frac{{{\partial ^2}u}}{{\partial {x^2}}} \cdot {e^s}\cos t + \frac{{{\partial ^2}u}}{{\partial x\partial y}} \cdot {e^s}\sin t} \right) + }\\{\frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t + {e^s}\sin t\left( {\frac{{{\partial ^2}u}}{{\partial {y^2}}} \cdot {e^s}\sin t + \frac{{{\partial ^2}u}}{{\partial x\partial y}} \cdot {e^s}\cos t} \right)}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{l}}{\frac{{\partial u}}{{\partial x}} \cdot {e^s}\cos t + \frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t + {e^{2s}}{{\cos }^2}t \cdot \frac{{{\partial ^2}u}}{{\partial {x^2}}} + }\\{2{e^{2s}}\cos t\sin t \cdot \frac{{{\partial ^2}u}}{{\partial x\partial y}} + {e^{2s}}{{\sin }^2}t \cdot \frac{{{\partial ^2}u}}{{\partial {y^2}}}}\end{aligned}} \right)\end{aligned}

Similarly, find the value of$$\frac{{{\partial ^2}u}}{{\partial {t^2}}}$$ as shown.

Obtain$$\frac{{{\partial ^2}u}}{{\partial {t^2}}}$$

Take partial derivative on both sides of$$\frac{{\partial u}}{{\partial t}}$$ with respect to$$t.$$

\begin{aligned}{l}\frac{{{\partial ^2}u}}{{\partial {t^2}}} = \frac{{\partial u}}{{\partial x}} \cdot \left( { - {e^s}\cos t} \right) - {e^s}\sin t \cdot \frac{\partial }{{\partial t}}\left( {\frac{{\partial u}}{{\partial x}}} \right) + \frac{{\partial u}}{{\partial y}} \cdot \left( { - {e^s}\sin t} \right) + {e^s}\cos t \cdot \frac{\partial }{{\partial t}}\left( {\frac{{\partial u}}{{\partial y}}} \right)\\ = \left( {\begin{aligned}{*{20}{l}}{\left. {\frac{{\partial u}}{{\partial x}} \cdot \left( { - {e^s}\cos t} \right) - {e^s}\sin t\left( {\frac{{{\partial ^2}u}}{{\partial {x^2}}} \cdot \frac{{\partial x}}{{\partial t}} + \frac{{{\partial ^2}u}}{{\partial y\partial x}} \cdot \frac{{\partial y}}{{\partial t}}} \right) - } \right)}\\{\frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t + {e^s}\cos t\left( {\frac{{{\partial ^2}u}}{{\partial {y^2}}} \cdot \frac{{\partial y}}{{\partial s}} + \frac{{{\partial ^2}u}}{{\partial x\partial y}} \cdot \frac{{\partial x}}{{\partial s}}} \right)}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{l}}{\frac{{\partial u}}{{\partial x}} \cdot \left( { - {e^s}\cos t} \right) - {e^s}\sin t\left( {\frac{{{\partial ^2}u}}{{\partial {x^2}}} \cdot {e^s}\cos t + \frac{{{\partial ^2}u}}{{\partial y\partial x}} \cdot {e^s}\sin t} \right) - }\\{\frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t + {e^s}\cos t\left( {\frac{{{\partial ^2}u}}{{\partial {y^2}}} \cdot {e^s}\sin t + \frac{{{\partial ^2}u}}{{\partial x\partial y}} \cdot {e^s}\cos t} \right)}\end{aligned}} \right)\end{aligned}

The partial derivatives$$\frac{{{\partial ^2}u}}{{\partial y\partial x}} = \frac{{{\partial ^2}u}}{{\partial x\partial y}}$$, as the functions are continuous.

Thus, it can be simplified as follows.

\begin{aligned}{l}\frac{{{\partial ^2}u}}{{\partial {t^2}}} = \left( {\begin{aligned}{*{20}{l}}{\frac{{\partial u}}{{\partial x}} \cdot \left( { - {e^s}\cos t} \right) - {e^s}\sin t\left( {\frac{{{\partial ^2}u}}{{\partial {x^2}}} \cdot {e^s}\cos t + \frac{{{\partial ^2}u}}{{\partial y\partial x}} \cdot {e^s}\sin t} \right) - }\\{\frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t + {e^s}\cos t\left( {\frac{{{\partial ^2}u}}{{\partial {y^2}}} \cdot {e^s}\sin t + \frac{{{\partial ^2}u}}{{\partial x\partial y}} \cdot {e^s}\cos t} \right)}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{l}}{\frac{{\partial u}}{{\partial x}} \cdot \left( { - {e^s}\cos t} \right) - \frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t + {e^{2s}}{{\sin }^2}t\frac{{{\partial ^2}u}}{{\partial {x^2}}} - }\\{2{e^{2s}}\cos t\sin t\frac{{{\partial ^2}u}}{{\partial x\partial y}} + {e^{2s}}{{\cos }^2}t\frac{{{\partial ^2}u}}{{\partial {y^2}}}}\end{aligned}} \right)\end{aligned}

Find the value of the Right-Hand side of the equation, $${e^{ - 2s}}\left( {\frac{{{\partial ^2}u}}{{\partial {s^2}}} + \frac{{{\partial ^2}u}}{{\partial {t^2}}}} \right).$$

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

Thus, the Left-Hand side is obtained.

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