Explanation Given information: Thesurface z 2 − 2 x 4 − y 4 = 16 , and point P = ( 2 , 2 , 8 ) Formula used: The normal vector to a surface f ( x , y , z ) is n ^ = ∇ f ‖ ∇ f ‖ , where ∇ f = ( f x , f y , f z ) is gradient of f ( x , y , z ) Calculation: As surface is given z 2 − 2 x 4 − y 4 = 16 , thus f ( x , y , z ) = z 2 − 2 x 4 − y 4 − 16 = 0 , thus to find unit normal vector on it First find its gradient, thus ∇ f = ( f x , f y , f z ) = ( ∂ f ∂ x , ∂ f ∂ y , ∂ f ∂ z ) = ( ∂ ( z 2 − 2 x 4 − y 4 − 16 ) ∂ x , ∂ ∂ y ( z 2 −

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ISBN: 9781319055905

Author: FRANZOSA

Publisher: VST

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Chapter 14.5, Problem 61E

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Find unit n normal vector to the surface z2−2x4−y4=16, at P=(2,2,8) in direction of xy- plane

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Chapter 14.5, Problem 60E

Chapter 14.5, Problem 62E

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Chapter 14 Solutions

EBK CALCULUS: EARLY TRANSCENDENTALS

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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Find unit n normal vector to the surface z 2 − 2 x 4 − y 4 = 16 , at P = ( 2 , 2 , 8 ) in direction of xy- plane

Solution Summary: The author calculates that the normal vector to a surface is f(x,y,z).

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Find unit n normal vector to the surface z2−2x4−y4=16, at P=(2,2,8) in direction of xy- plane

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Chapter 14 Solutions