Explanation Given: The paraboloid z = 6 − x − x 2 − 2 y 2 intersects the plane x = 1 at the point ( 1 , 2 , − 4 ) . Result used: “Suppose f has continuous partial derivatives. An equation of the tangent plane to the surface z = f ( x , y ) at the point P ( x 0 , y 0 , z 0 ) is, z − z 0 = f x ( x 0 , y 0 ) ( x − x 0 ) + f y ( x 0 , y o ) ( y − y 0 ) ”. Calculation: The equation of paraboloid is, z = 6 − x − x 2 − 2 y 2 . Substitute x = 1 in the above equation, z = 6 − ( 1 ) − ( 1 ) 2 − 2 y 2 = 6 − 1 − 1 − 2 y 2 = 6 − 2 − 2 y 2 = 4 − 2 y 2 Take partial derivative for the above equation z with respect to y , ∂ z ∂ y = ∂ ∂ y ( 4 − 2 y 2 ) = ∂ ∂ y ( 4 ) − ∂ ∂ y ( 2 y 2 ) = 0 − 2 ( 2 y ) = − 4 y Substitute y = 2 in the above equation to obtain the slope value, ∂ z ∂ y = − 4 ( 2 ) = − 8 Thus, the slope of the line is − 8

Bundle: Multivariable Calculus, 8th + WebAssign Printed Access Card for Stewart's Calculus, 8th Edition, Single-Term

8th Edition

ISBN: 9781305779198

Author: James Stewart

Publisher: Cengage Learning

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Chapter 14.3, Problem 98E

To determine

To find: The parametric equations for the tangent line to the parabola at the point (1,2,−4) if the paraboloid z=6−x−x2−2y2 intersects the plane x=1 in a parabola.

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Chapter 14.3, Problem 97E

Chapter 14.3, Problem 99E

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Chapter 14 Solutions

Bundle: Multivariable Calculus, 8th + WebAssign Printed Access Card for Stewart's Calculus, 8th Edition, Single-Term

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The parametric equations for the tangent line to the parabola at the point ( 1 , 2 , − 4 ) if the paraboloid z = 6 − x − x 2 − 2 y 2 intersects the plane x = 1 in a parabola.

Solution Summary: The author calculates the parametric equations for the tangent line to the parabola at the point (1,2,-4).

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To find: The parametric equations for the tangent line to the parabola at the point (1,2,−4) if the paraboloid z=6−x−x2−2y2 intersects the plane x=1 in a parabola.

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Chapter 14 Solutions