Explanation Given: Thesurface x 3 + 2 x y 2 − 7 x 3 + 3 y + 1 = 0 and a point P 0 ( 1 , 1 , 1 ) Concept Used: For a function that defines a surface f ( x , y , z ) differentiated at a point ( x 0 , y 0 , z 0 ) . The tangent plane to f ( x , y , z ) at ( x 0 , y 0 , z 0 ) is the plane with normal vector ∇ f ( x 0 , y 0 , z 0 ) that passes through the point ( x 0 , y 0 , z 0 ) . The equation of the tangent plane is given by ∇ f ( x 0 , y 0 , z 0 ) · 〈 x − x 0 , y − y 0 , z − z 0 〉 = 0 Where ∇ f is the gradient of the function f which is ∇ f = 〈 f x , f y , f z 〉 . And, the equation of the normal line is given by x ( t ) = x 0 + f x ( x 0 , y 0 , z 0 ) t y ( t ) = y 0 + f y ( x 0 , y 0 , z 0 ) t z ( t ) = z 0 + f z ( x 0 , y 0 , z 0 ) t Calculation: For the givensurface x 3 + 2 x y 2 − 7 x 3 + 3 y + 1 = 0 and a point P 0 ( 1 , 1 , 1 ) Consider f ( x , y , z ) = x 3 + 2 x y 2 − 7 x 3 + 3 y + 1 f ( x , y , z ) = 2 x y 2 − 6 x 3 + 3 y + 1 The gradient of the function f is ∇ f = 〈 f x , f y , f z 〉 that is ∇ f = 〈 2 y 2 − 18 x 2 , 4 x y + 3 , 0 〉 At P 0 ( 1 , 1 , 1 ) ∇ f ( 1 , 1

Student's Solution and Survival Manual for Calculus

7th Edition

ISBN: 9781524934040

Author: STRAUSS MONTY J, TODA MAGDALENA DANIELE, SMITH KARL J

Publisher: Kendall Hunt Publishing

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Chapter 11, Problem 37SP

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To find:The equation for the tangent plane and normal line to the surface x3+2xy2−7x3+3y+1=0 at P0(1,1,1) .

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Chapter 11, Problem 36SP

Chapter 11, Problem 38SP

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Student's Solution and Survival Manual for Calculus

Ch. 11.1 - Prob. 1PS Ch. 11.1 - Prob. 2PS Ch. 11.1 - Prob. 3PS Ch. 11.1 - Prob. 4PS Ch. 11.1 - Prob. 5PS Ch. 11.1 - Prob. 6PS Ch. 11.1 - Prob. 7PS Ch. 11.1 - Prob. 8PS Ch. 11.1 - Prob. 9PS Ch. 11.1 - Prob. 10PS

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The equation for the tangent plane and normal line to the surface x 3 + 2 x y 2 − 7 x 3 + 3 y + 1 = 0 at P 0 ( 1 , 1 , 1 ) .

Solution Summary: The author explains the equation for the tangent plane and normal line to the surface p_0(1,1,1).

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To find:The equation for the tangent plane and normal line to the surface x3+2xy2−7x3+3y+1=0 at P0(1,1,1) .

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Chapter 11 Solutions