Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 16 - Vector Calculus - 16.1 Vector Fields - 16.1 Exercises - Page 1114: 22

Answer

$\frac{1}{\sqrt{2s+3t}}i + \frac{3}{2\sqrt{2s+3t}}j$

Work Step by Step

Derive $f(s, t) $with respect to $s$, which is: $f_{s}(s, t) = (\frac{1}{2})\cdot(2s + 3t)^\frac{-1}{2}\cdot 2 = \frac{1}{\sqrt{2s+3t}}$ Derive $f(s, t) $with respect to $s$, which is: $f_{t}(s, t) = (\frac{1}{2})\cdot(2s + 3t)^\frac{-1}{2}\cdot 3 = \frac{3}{2\sqrt{2s+3t}}$ Thus the gradient of $f$ is calculated: $\nabla f(s,t) = f_{s}(s, t)i + f_{t}(s, t)j$ $ = \frac{1}{\sqrt{2s+3t}}i + \frac{3}{2\sqrt{2s+3t}}j$
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