Explanation Given: The given surface is 3 z 3 + x 2 y − y 2 x = 1 . Given point P = ( 1 , − 1 , 1 ) Formula used: Normal vectorto a surface f ( x , y , z ) = k at point P ( x 0 , y 0 , z 0 ) is given by ∇ f ( x 0 , y 0 , z 0 ) Here, ∇ f ( x , y , z ) = f x ( x , y , z ) i + f y ( x , y , z ) j + f z ( x , y , z ) k Calculation: Consider the given surface 3 z 3 + x 2 y − y 2 x = 1 . f ( x , y , z ) = 3 z 3 + x 2 y − y 2 x − 1 ...... (1) Consider the point P = ( 1 , − 1 , 1 ) x 0 = 1 , y 0 = − 1 , z 0 = 1 Differentiate equation (1) partially with respect to x . f x ( x , y , z ) = 2 x y − y 2 So, f x ( 1 , − 1 , 1 ) = 2 ( 1 ) ( − 1 ) − ( − 1 ) 2 = − 3 Differentiate equation (1) partially with respect to y

CALCULUS EARLY TRANSCENDENTALS W/ACHIEVE

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ISBN: 9781319057664

Author: Rogawski

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Chapter 14.5, Problem 42E

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To find:

A vector normal to thegiven surface at the point P(1,−1,1).

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Chapter 14.5, Problem 41E

Chapter 14.5, Problem 43E

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Chapter 14 Solutions

CALCULUS EARLY TRANSCENDENTALS W/ACHIEVE

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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A vector normal to thegiven surface at the point P ( 1 , − 1 , 1 ) .

Solution Summary: The author calculates a vector normal to the given surface at the point P(1,-1,1).

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To find:

A vector normal to thegiven surface at the point P(1,−1,1).

Want to see the full answer?

Chapter 14 Solutions

A vector normal to thegiven surface at the point P ( 1 , − 1 , 1 ) .

Solution Summary: The author calculates a vector normal to the given surface at the point P(1,-1,1).

Videos

To find: A vector normal to thegiven surface at the point P(1,−1,1).

Want to see the full answer?

Chapter 14 Solutions

To find:

A vector normal to thegiven surface at the point P(1,−1,1).