Find an equation of the tangent plane to the surface z=−1x^2+1y^2+3x−1y+3 at the point (5, 5, 13). z=

Trigonometry (MindTap Course List)
10th Edition
ISBN:9781337278461
Author:Ron Larson
Publisher:Ron Larson
Chapter6: Topics In Analytic Geometry
Section6.2: Introduction To Conics: parabolas
Problem 4ECP: Find an equation of the tangent line to the parabola y=3x2 at the point 1,3.
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Find an equation of the tangent plane to the surface z=−1x^2+1y^2+3x−1y+3 at the point (5, 5, 13).
z=

Expert Solution
Step 1

Given the equation of surface is:

                              z=-x2+y2+3x-y+3

We nee to find the equation of tangent plane at the point 5,5,13.

Let fx,y=-x2+y2+3x-y+3

Partially differentiating f with respect to x and y respectively, we get

                                               fx=-2x+3fy=2y-1

Therefore at x=5, y=5

                                             fx5,5=-25+3=-7fy5,5=25-1=9

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