Check that the point (1, –1, 2) lies on the given surface. Then, viewing the surface as a level surface for a function f(x, y, z), find a vector normal to the surface and an equation for the tangent plane to the surface at (1, –1, 2). 2а2 — 3у? + 422 %3D15 - 3y + 422 = 15

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter6: The Trigonometric Functions
Section6.6: Additional Trigonometric Graphs
Problem 78E
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11.6: Problem 6
Check that the point (1, –1, 2) lies on the given surface. Then,
viewing the surface as a level surface for a function f(x, y, z),
find a vector normal to the surface and an equation for the
tangent plane to the surface at (1, –1, 2).
2æ² – 3y? + 422 = 15
vector normal =
tangent plane:
Transcribed Image Text:11.6: Problem 6 Check that the point (1, –1, 2) lies on the given surface. Then, viewing the surface as a level surface for a function f(x, y, z), find a vector normal to the surface and an equation for the tangent plane to the surface at (1, –1, 2). 2æ² – 3y? + 422 = 15 vector normal = tangent plane:
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