Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 15 - Differentiation in Several Variables - 15.3 Partial Derivatives - Exercises - Page 782: 78

Answer

We show that when the bug is sitting on the rod at distance $x$ from the origin she feels the maximum temperature at time $t = \frac{1}{2}{x^2}$.

Work Step by Step

Write $f\left( {x,t} \right) = \frac{1}{{2\sqrt \pi }}{t^{ - 1/2}}{{\rm{e}}^{ - {x^2}/4t}}$ The derivative with respect to $t$ is (see Example 11) $\frac{{\partial f}}{{\partial t}} = \frac{\partial }{{\partial t}}\left( {\frac{1}{{2\sqrt \pi }}{t^{ - 1/2}}{{\rm{e}}^{ - {x^2}/4t}}} \right)$ $ = - \frac{1}{{4\sqrt \pi }}{t^{ - 3/2}}{{\rm{e}}^{ - {x^2}/4t}} + \frac{1}{{8\sqrt \pi }}{x^2}{t^{ - 5/2}}{{\rm{e}}^{ - {x^2}/4t}}$ From Figure 6, we see that there is maximum for $f$. We can find the maximum by solving the equation $\frac{{\partial f}}{{\partial t}} = 0$: $0 = - \frac{1}{{4\sqrt \pi }}{t^{ - 3/2}}{{\rm{e}}^{ - {x^2}/4t}} + \frac{1}{{8\sqrt \pi }}{x^2}{t^{ - 5/2}}{{\rm{e}}^{ - {x^2}/4t}}$ Multiplying both sides by $4\sqrt \pi {t^{3/2}}{{\rm{e}}^{{x^2}/4t}}$ gives $0 = - 1 + \frac{1}{2}{x^2}{t^{ - 1}}$ $\frac{1}{2}\frac{{{x^2}}}{t} = 1$, ${\ \ \ \ }$ $t = \frac{1}{2}{x^2}$ Hence, when the bug is sitting on the rod at distance $x$ from the origin she feels the maximum temperature at time $t = \frac{1}{2}{x^2}$.
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