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1. The equation r²y³ +2ln(yz) + ye" - 1 determines a surface. Note that you cannot easily write z as a function of a and y. (a) Use implicit differentiation to find at the point (0,1,2). (b) Use implicit differentiation to find at the point (0,1,2).

1. The equation r²y³ +2ln(yz) + ye" - 1 determines a surface. Note that you cannot easily write z as a function of a and y. (a) Use implicit differentiation to find at the point (0,1,2). (b) Use implicit differentiation to find at the point (0,1,2).

Algebra and Trigonometry (MindTap Course List)
Algebra and Trigonometry (MindTap Course List)
4th Edition
ISBN:9781305071742
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter8: Polar Coordinates And Parametric Equations
Section8.CT: Chapter Test
Problem 8CT
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1. The equation r²y³ +2ln(yz) + ye™² - 1 determines a surface. Note that you cannot easily write z as a function of a and y. (a) Use implicit differentiation to find at the point (0,1,2). (b) Use implicit differentiation to find at the point (0,1,2). 2. A right triangle has sides of length a abd b and a hypotenuse of length c. Suppose the triangle initially has a 5 and 6-12 but then the smaller side length increases to 5.1 while the longer side decreases to 11.8. By how much does the hypotenuse's length change? 3. Find equation of the tangent plane to f(x,y) - e²+² at the point (1,1) and use it to approximate the value of ee¹² +2²
Transcribed Image Text:1. The equation r²y³ +2ln(yz) + ye™² - 1 determines a surface. Note that you cannot easily write z as a function of a and y. (a) Use implicit differentiation to find at the point (0,1,2). (b) Use implicit differentiation to find at the point (0,1,2). 2. A right triangle has sides of length a abd b and a hypotenuse of length c. Suppose the triangle initially has a 5 and 6-12 but then the smaller side length increases to 5.1 while the longer side decreases to 11.8. By how much does the hypotenuse's length change? 3. Find equation of the tangent plane to f(x,y) - e²+² at the point (1,1) and use it to approximate the value of ee¹² +2²
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