Explanation Given: The given surface is x z + 2 x 2 y + y 2 z 3 = 11 . Given point P = ( 2 , 1 , 1 ) Formula used: Equation of tangent plane of a surface f ( x , y , z ) = k at point P ( x 0 , y 0 , z 0 ) is given by f x ( x 0 , y 0 , z 0 ) ( x − x 0 ) + f y ( x 0 , y 0 , z 0 ) ( y − y 0 ) + f z ( x 0 , y 0 , z 0 ) ( z − z 0 ) = 0 Calculation: Consider the given surface x z + 2 x 2 y + y 2 z 3 = 11 . f ( x , y , z ) = x z + 2 x 2 y + y 2 z 3 − 11 ...... (1) Consider the point P = ( 2 , 1 , 1 ) x 0 = 2 , y 0 = 1 , z 0 = 1 Differentiate equation (1) partially with respect to x . f x ( x , y , z ) = z + 4 x y So, f x ( 2 , 1 , 1 ) = 1 + 4 × 2 × 1 = 9 Differentiate equation (1) partially with respect to y

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ISBN: 9781319308858

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 14.5, Problem 47E

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To find:

An equation of the tangent plane to the given surface at the point P(2,1,1).

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Chapter 14.5, Problem 46E

Chapter 14.5, Problem 48E

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Chapter 14 Solutions

CALCULUS:EARLY TRANS.(LL)-W/ACCESS

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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An equation of the tangent plane to the given surface at the point P ( 2 , 1 , 1 ) .

Solution Summary: The author calculates an equation of the tangent plane to the given surface at the point P(2,1,1).

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To find:

An equation of the tangent plane to the given surface at the point P(2,1,1).

Want to see the full answer?

Chapter 14 Solutions

An equation of the tangent plane to the given surface at the point P ( 2 , 1 , 1 ) .

Solution Summary: The author calculates an equation of the tangent plane to the given surface at the point P(2,1,1).

Videos

To find: An equation of the tangent plane to the given surface at the point P(2,1,1).

Want to see the full answer?

Chapter 14 Solutions

To find:

An equation of the tangent plane to the given surface at the point P(2,1,1).