Calculus (3rd Edition)

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

Chapter 15 - Differentiation in Several Variables - 15.4 Differentiability and Tangent Planes - Exercises - Page 789: 6


$$z=\frac{1}{2}+ \frac{\sqrt{3}}{2} (u-\pi/6)+ \frac{\sqrt{3}}{12}(w-1)$$

Work Step by Step

Given $$G(u, w)=\sin (u w), \quad\left(\frac{\pi}{6}, 1\right)$$ Since \begin{align*} G(u, w)&=\sin (u w)\ \ \ \ &G\left(\frac{\pi}{6}, 1\right)&=\frac{1}{2} \\ G_u(u, w)&=w\cos(uw) \ \ \ \ &G_u \left(\frac{\pi}{6}, 1\right)&= \frac{\sqrt{3}}{2}\\ G_w(u, w)&= u \cos(uw)\ \ \ \ & G_w\left(\frac{\pi}{6}, 1\right)&= \frac{\sqrt{3}\pi}{12} \\ \end{align*} Then, the tangent plane at $(1,0.8)$ is given by \begin{align*} z&=G\left(\frac{\pi}{6}, 1\right)+G_u \left(\frac{\pi}{6}, 1\right)(u-\pi/6)+G_v \left(\frac{\pi}{6}, 1\right)(w-1)\\ &=\frac{1}{2}+ \frac{\sqrt{3}}{2} (u-\pi/6)+ \frac{\sqrt{3}}{12}(w-1) \end{align*}
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.