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 14.4 - The Chain Rule - Exercises 14.4 - Page 816: 4


a. $\dfrac{16}{1+16t}$ and b. $\dfrac{16}{49}$

Work Step by Step

a. Using chain rule: $\dfrac{dw}{dt}=\dfrac{\partial w}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial w}{\partial y}\dfrac{dy}{dt}+\dfrac{\partial w}{\partial z}\dfrac{dz}{dt}$ and $(\dfrac{-2x \sin t}{x^2+y^2+z^2})+(\dfrac{-2y \cos t}{x^2+y^2+z^2})+(\dfrac{4zt^{-1/2}}{x^2+y^2+z^2})=\dfrac{16}{1+16t}$ Now, by using direct differentiation. We have $w=\ln (x^2+y^2+z^2)$ or, $w=\ln (cos^2 t+\sin ^2 t+16t)$ Also, $w=\ln (1+16t)$; $\dfrac{dw}{dt}=\dfrac{16}{(1+16t)}$ b. $\dfrac{dw}{dt}(3)=\dfrac{16}{(1+16t)}=\dfrac{16}{1+16(3)}=\dfrac{16}{49}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.