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Q. 33

Expert-verified

Calculus

Found in: Page 989

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Find the gradient of the given function, and find the direction in which the function increases most rapidly at the specified point P.
$f (x, y) = x \sin y, P = (3, \frac{π}{2})$

$p = (3, \frac{π}{4}) is \nabla f (3, \frac{π}{4}) = ⟨ 1, 0 ⟩$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given Information

$The gradient of a function f (x, y) is the vector function defined by \nabla f (x, y) = \frac{\partial f}{\partial x} i + \frac{\partial f}{\partial y} j, Here f (x, y) = x \sin y .$

Step 2. Solution

Gradient is given by,

$\begin{array}{l} \nabla f (x, y) = \frac{\partial}{\partial x} (x \sin y) i + \frac{\partial}{\partial y} (x \sin y) j \\ = \sin y i + x \cos y j \end{array}$

Hence the gradient is, $\nabla f (x, y) = ⟨ \sin y, x \cos y ⟩$

$As the gradient of a function f at a point p points in the direction in which f increases most rapidly and$

at $p = (3, \frac{π}{2})$

$\begin{array}{l} \nabla f (3, \frac{π}{2}) = \sin \frac{π}{2} i + 3 \cos \frac{π}{2} j \\ = (i + 0) \end{array}$

So the direction in which f increases most rapidly at $p = (3, \frac{π}{4}) is \nabla f (3, \frac{π}{4}) = ⟨ 1, 0 ⟩$ .

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