Introduction to Electrodynamics 4e

Published by Pearson Education
ISBN 10: 9332550441
ISBN 13: 978-9-33255-044-5

Chapter 1 - Section 2.2 - Differential Calculus - Problem - Page 15: 12

Answer

a) The top of the hill is 3 miles north, 2 miles west, of South Hadley. b)Height of the hill, h=720ft. c)The slope at the point is 200$\sqrt{2}$ ft/mile in the northwest direction.

Work Step by Step

a)$\nabla{h}=(\large\hat{x}\frac{\partial}{\partial{x}}+\hat{y}\frac{\partial}{\partial{y}}+\hat{z}\frac{\partial}{\partial{z}})\normalsize(h(x,y))$ $\hspace{1cm}=10(2y-6x-18)\hat{x}+(2x-8y+28)\hat{y}$ At the top of hill $\nabla{h}=0$, so (by comparison) $2y-6x-18=0\hspace{1cm}$ --(1) $2x-8y+28=0\hspace{1cm}$ --(2) eq(1)+3$\times$eq(2) gives $-22y+66=0$ ; $\Rightarrow y=3$ Putting y=3 in eq(1) , $\Rightarrow x=-2$ Since x is the distance(in miles) east, and y the distance north, of South Hadley, thus the top is located at 3 miles north, 2 miles west, of South Hadley. b)Height of the hill, $h(-2,3)=10(-12-12-36+36+84+12)=720$ feet or ft. c)The slope at x=1 and y=1 (slope at a point 1 mile north and 1 mile east of South Hadley) is given by, $\hspace{4cm}\nabla{h}\Big|_{x=1,y=1}=10(2-6-18)\hat{x}+(2-8+28)\hat{y}$ $\hspace{6.1cm}=220(-\hat{x}+\hat{y})$ $|\nabla{h}|=200\sqrt{2}$ The direction of this steepest slope at the point (x=1,y=1) is $(-\hat{x}+\hat{y})$, which corresponds to northwest.
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.