(a) Explanation Given: The surface is, sin ( x y z ) = x + 2 y + 3 z . Result used: “The tangent plane to the level surface at the point P ( x 0 , y 0 , z 0 ) is defined as 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: Let the surface function be, F ( x , y , z ) = x + 2 y + 3 z − sin ( x y z ) (1) The equation of the tangent plane to the given surface at the point ( 2 , − 1 , 0 ) is defined by, F x ( 2 , − 1 , 0 ) ( x − 2 ) + F y ( 2 , − 1 , 0 ) ( y + 1 ) + F z ( 2 , − 1 , 0 ) ( z − 0 ) = 0 (2) Take partial derivative with respect to x at the point ( 2 , − 1 , 0 ) in the equation (1), F x ( x , y , z ) = ∂ ∂ x ( x + 2 y + 3 z − sin ( x y z ) ) = 1 + 0 + 0 − y z sin ( x y z ) = 1 − y z cos ( x y z ) The value of F x ( x , y , z ) at the point ( 2 , − 1 , 0 ) is, F x ( 2 , − 1 , 0 ) = 1 − ( − 1 ) ( 0 ) cos ( 2 ( − 1 ) 0 ) = 1 − 0 = 1 Thus, the value of F x ( 2 , − 1 , 0 ) = 1 (b) To determine To find: The equation of the normal line to the surface sin ( x y z ) = x + 2 y + 3 z at the point ( 2 , − 1 , 0 ) .

Essential Calculus: Early Transcendentals

2nd Edition

ISBN: 9781133112280

Author: James Stewart

Publisher: Cengage Learning

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Chapter 11, Problem 27RE

(a)

To determine

To find: The equation of the tangent planeto the surface sin(xyz)=x+2y+3z at the point (2,−1,0) .

(b)

To determine

To find: The equation of the normal line to the surface sin(xyz)=x+2y+3z at the point (2,−1,0) .

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Chapter 11, Problem 26RE

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Chapter 11 Solutions

Essential Calculus: Early Transcendentals

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The equation of the tangent planeto the surface sin ( x y z ) = x + 2 y + 3 z at the point ( 2 , − 1 , 0 ) .

Solution Summary: The author calculates the equation of the tangent planeto the surface mathrmsin(xyz)=x+2y+3z at the point

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To find: The equation of the tangent planeto the surface sin(xyz)=x+2y+3z at the point (2,−1,0) .

To find: The equation of the normal line to the surface sin(xyz)=x+2y+3z at the point (2,−1,0) .

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Chapter 11 Solutions