Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.3 - Partial Derivatives - Exercises 14.3 - Page 807: 39


$W_P=V$ and $W_V=P+\dfrac{\delta v^2}{ 2g}$ $W_{\delta}=\dfrac{V v^2}{ 2g}$; $W_{v}=\dfrac{V \delta v}{g}$; $W_{g}=-\dfrac{ V \delta v^2}{ 2g^2}$;

Work Step by Step

We need to take the first partial derivatives of the given function. In order to find the partial derivative, we will differentiate with respect to one of its variables, by keeping all as a constant. $W_P=V$ and $W_V=P+\dfrac{\delta v^2}{ 2g}$ $W_{\delta}=\dfrac{V v^2}{ 2g}$; $W_{v}=\dfrac{2 V \delta v}{ 2g}=\dfrac{V \delta v}{g}$; $W_{g}=-\dfrac{ V \delta v^2}{ 2g^2}$
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