Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.3 - Partial Derivatives - 14.3 Exercise - Page 927: 95

Answer

$\dfrac{\partial K}{\partial m}\dfrac{\partial^{2}K}{\partial v^{2}}=K$

Work Step by Step

For $ \dfrac{\partial K}{\partial m}$, here all variables aside from $m$ to be treated as constants. $ \dfrac{\partial K}{\partial m}=\dfrac{\partial}{\partial m}[\dfrac{1}{2}mv^{2}]=\dfrac{1}{2}v^{2}$ $\dfrac{\partial K}{\partial v}=\dfrac{\partial}{\partial v}[(\dfrac{1}{2})mv^{2}]=mv$ This gives: $ \dfrac{\partial^{2}K}{\partial v^{2}}=\dfrac{\partial}{\partial v}[mv]=m$ and $\dfrac{\partial K}{\partial m}\times\dfrac{\partial^{2}K}{\partial v^{2}}=\dfrac{1}{2}v^{2}m=K$ Hence, we have $\dfrac{\partial K}{\partial m}\dfrac{\partial^{2}K}{\partial v^{2}}=K$
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