Explanation Given information: f ( x , y ) = x y 3 − x 2 , v = i − j , P = ( 2 , − 1 ) We have to use a unit vector in directional derivative computation. Calculation: Since the gradient of function is denoted by ∇ f = 〈 ∂ f ∂ x , ∂ f ∂ y 〉 The given equation f ( x , y ) = x y 3 − x 2 , Differentiating f with respect to x we get, ∂ f ∂ x = ∂ ∂ x x y 3 − x 2 ⇒ ∂ f ∂ x = ∂ ∂ x x y 3 − ∂ ∂ x x 2 ⇒ ∂ f ∂ x = y 3 ∂ ∂ x x − 2 x = y 3 − 2 x Differentiating f with respect to y we get, ∂ f ∂ y = ∂ ∂ y x y 3 − x 2 ⇒ ∂ f ∂ y = ∂ ∂ y x y 3 − ∂ ∂ y x 2 = 3 x y 2 − 0 = 3 x y 2 Hence ∇ f = 〈 ∂ f ∂ x , ∂ f ∂ y 〉 = 〈 y 3 − 2 x , 3 x y 2 〉 At P = ( 2 , − 1 ) ∇ f ( − 2 , 1 ) = 〈 ( 1 ) 3 − 2 ( − 2 ) , 3 ( − 2 ) ( 1 ) 2 〉 ⇒ ∇ f ( − 2 , 1 ) = 〈 1 + 4 , 3 ( − 2 ) 〉 = 〈 5 , − 6 〉 We will find the derivative with respect to v = i − j i

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable

4th Edition

ISBN: 9781319312886

Author: Rogawski, Jon, Adams, Colin, FRANZOSA, Robert

Publisher: W. H. Freeman

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Chapter 14.5, Problem 22E

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the directional derivative in the direction of v at the given point.

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Chapter 14.5, Problem 21E

Chapter 14.5, Problem 23E

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Chapter 14 Solutions

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E

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Tocalculate: the directional derivative in the direction of v at the given point.

Solution Summary: The author calculates the directional derivative in the direction of v at the given point P by using a unit vector.

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Tocalculate: the directional derivative in the direction of v at the given point.

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Chapter 14 Solutions

Tocalculate:

the directional derivative in the direction of v at the given point.