Explanation Given: The given surface is ln ( 1 + 4 x 2 + 9 y 4 ) − 0.1 z 2 = 0 . Given point P = ( 3 , 1 , 6.1876 ) 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 ln ( 1 + 4 x 2 + 9 y 4 ) − 0.1 z 2 = 0 . f ( x , y , z ) = ln ( 1 + 4 x 2 + 9 y 4 ) − 0.1 z 2 ...... (1) Consider the point P = ( 3 , 1 , 6.1876 ) x 0 = 3 , y 0 = 1 , z 0 = 6.1876 Differentiate equation (1) partially with respect to x . f x ( x , y , z ) = 8 x 1 + 4 x 2 + 9 y 4 So, f x ( 3 , 1 , 6.1876 ) = 8 ( 3 ) 1 + 4 ( 3 ) 2 + 9 ( 1 ) 4 = 24 1 + 36 + 9 = 12 23 Differentiate equation (1) partially with respect to y

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Author: Rogawski

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Chapter 14.5, Problem 49E

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An equation of the tangent plane to the given surface at the point P(3,1,6.1876).

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Chapter 14.5, Problem 48E

Chapter 14.5, Problem 50E

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Chapter 14 Solutions

CALCULUS:EARLY TRANS.-SAPLINGPLUS

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 ( 3 , 1 , 6.1876 ) .

Solution Summary: The author calculates an equation of the tangent plane to the given surface at the point P=(3,1,6.1876).

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To find:

An equation of the tangent plane to the given surface at the point P(3,1,6.1876).

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Chapter 14 Solutions

An equation of the tangent plane to the given surface at the point P ( 3 , 1 , 6.1876 ) .

Solution Summary: The author calculates an equation of the tangent plane to the given surface at the point P=(3,1,6.1876).

Videos

To find: An equation of the tangent plane to the given surface at the point P(3,1,6.1876).

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(3,1,6.1876).