Find an equation of the plane tangent to the surface xy z2 = 2 at the point (2, 1, -1). О А.х+ бу + 4z%3D 4 О В.х + бу - 4z %3D 0 O C.x = (2 + t) i + (1 + 6t) j + (-1 - 4t) k, t e R O D. x + 5y - 4z = 11 О Еx + 6у - 4z %3D 12

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Find an equation of the plane tangent to the surface xy z2 = 2 at the point (2, 1, -1).
О А.х+ бу + 4z%3D 4
О В.х + бу - 4z %3D 0
O C.x = (2 + t) i + (1 + 6t) j + (-1 - 4t) k, t e R
O D. x + 5y - 4z = 11
О Еx + 6у - 4z %3D 12
Transcribed Image Text:Find an equation of the plane tangent to the surface xy z2 = 2 at the point (2, 1, -1). О А.х+ бу + 4z%3D 4 О В.х + бу - 4z %3D 0 O C.x = (2 + t) i + (1 + 6t) j + (-1 - 4t) k, t e R O D. x + 5y - 4z = 11 О Еx + 6у - 4z %3D 12
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