Explanation Given information: z = x y 3 + 8 y − 1 ,tangent plane is parallel to 2 x + 7 y + 2 z = 0 Calculation: Consider, f ( x , y , z ) = x y 3 + 8 y − 1 − z We will find the normal vector to the plane. The normal vector to the plane is ∇ f = 〈 ∂ f ∂ x , ∂ f ∂ y , ∂ f ∂ z 〉 f ( x , y , z ) = x y 3 + 8 y − 1 − z Differentiating f ( x , y , z ) with respect to x we get, ∂ f ∂ x = ∂ ∂ x x y 3 + ∂ ∂ x 8 y − 1 − ∂ ∂ x z ⇒ ∂ f ∂ x = y 3 ∂ ∂ x x + ∂ ∂ x 8 y − 1 − ∂ ∂ x z ⇒ ∂ f ∂ x = y 3 ( 1 ) + ( 0 ) − ( 0 ) = y 3 So, ∂ f ∂ x = y 3 Differentiating f ( x , y , z ) with respect to y we get, ∂ f ∂ y = ∂ ∂ y x y 3 + ∂ ∂ y 8 y − 1 − ∂ ∂ y z ⇒ ∂ f ∂ y = x ∂ ∂ y y 3 + ∂ ∂ y 8 y − 1 − ∂ ∂ y z ⇒ ∂ f ∂ y = 3 x y 2 + ( 8 ) ( − 1 y − 2 ) − ( 0 ) ⇒ ∂ f ∂ y = 3 x y 2 − 8 y − 2 So, ∂ f ∂ y = 3 x y 2 − 8 y − 2

Loose-leaf Version For Calculus: Early Transcendentals

4th Edition

ISBN: 9781319055912

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Chapter 14.4, Problem 12E

To determine

Tofind:

the points on the graph of z=xy3+8y−1 where the tangent plane is parallel to 2x+7y+2z=0

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Chapter 14.4, Problem 11E

Chapter 14.4, Problem 13E

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Chapter 14 Solutions

Loose-leaf Version For 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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Tofind: the points on the graph of z = x y 3 + 8 y − 1 where the tangent plane is parallel to 2 x + 7 y + 2 z = 0

Solution Summary: The author explains that the tangent plane is parallel to 2x+7y+2z.

Videos

Tofind:

the points on the graph of z=xy3+8y−1 where the tangent plane is parallel to 2x+7y+2z=0

Want to see the full answer?

Chapter 14 Solutions

Tofind: the points on the graph of z = x y 3 + 8 y − 1 where the tangent plane is parallel to 2 x + 7 y + 2 z = 0

Solution Summary: The author explains that the tangent plane is parallel to 2x+7y+2z.

Videos

Tofind: the points on the graph of z=xy3+8y−1 where the tangent plane is parallel to 2x+7y+2z=0

Want to see the full answer?

Chapter 14 Solutions

Tofind:

the points on the graph of z=xy3+8y−1 where the tangent plane is parallel to 2x+7y+2z=0