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(c) The tangent line is horizontal if y=✔ Using the expression for y' from part (a), we see that y'=0 when ~-{|}x²)'«u{ |fv²_) which simplifies tox-16x3 Since x 0 in the first quadrant, we have x³.Ifx ✓ Looking at the figure below, we see that our answer is reasonable. = 0 (provided that y² - 2x0). Substituting in the equation of the curve, we get ←x-(0)² - 0√O: then y .Thus the tangent is horizontal at (0, 0) and at (x, y) =

(c) The tangent line is horizontal if y=✔ Using the expression for y' from part (a), we see that y'=0 when ~-{|}x²)'«u{ |fv²_) which simplifies tox-16x3 Since x 0 in the first quadrant, we have x³.Ifx ✓ Looking at the figure below, we see that our answer is reasonable. = 0 (provided that y² - 2x0). Substituting in the equation of the curve, we get ←x-(0)² - 0√O: then y .Thus the tangent is horizontal at (0, 0) and at (x, y) =

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 92E
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Hi i need help with the answers underlined in red please. i included what i have so far if that helps

Example 2 Video Example (a) Find y' if x² + y² = 6xy. (b) Find the tangent to the folium of Descartes x³ + y = 6xy at the point (3, 3). (c) At what point in the first quadrant the tangent line horizontal? Solution (a) Differentiating both sides of x³ + y³ = 6xy with respect to x, regarding y as a function of x, and using the Chain Rule on the y³ term and the Product Rule on the 6xy term, we get 3x² + ₂2 )y₁ y² + 6( y (y² + + 2 y or +². We now solve for y'. 1² - 2x -3 2.2 (b) When xy = 3, y²y' - 2xy' = 2y-x² )y² = 2y = x² and a glance at the figure below confirms that this is a reasonable value for the slope at (3, 3). y=-x+6 (c) The tangent line 1/4+² 2y=x2 1² - 2x Ⓡ So an equation of the tangent to the folium at (3, 3) is y-(3✔ ) = -1(x - (3✓ or (3, 3) which simplifies to x6 = 16x³ horizontal if y' = 0 x )²-√3x² ✔. Using the expression for y' from part (a), we see that y'= 0 when . Since * # 0 in the first quadrant, we have x³ = Looking at the figure below, we see that our answer reasonable. If x 2-x² = 0 (provided that y² - 2x + 0). Substituting y = x² in the equation of the curve, we get then y = . Thus the tangent is horizontal at (0, 0) and at (x, y) =
Transcribed Image Text:Example 2 Video Example (a) Find y' if x² + y² = 6xy. (b) Find the tangent to the folium of Descartes x³ + y = 6xy at the point (3, 3). (c) At what point in the first quadrant the tangent line horizontal? Solution (a) Differentiating both sides of x³ + y³ = 6xy with respect to x, regarding y as a function of x, and using the Chain Rule on the y³ term and the Product Rule on the 6xy term, we get 3x² + ₂2 )y₁ y² + 6( y (y² + + 2 y or +². We now solve for y'. 1² - 2x -3 2.2 (b) When xy = 3, y²y' - 2xy' = 2y-x² )y² = 2y = x² and a glance at the figure below confirms that this is a reasonable value for the slope at (3, 3). y=-x+6 (c) The tangent line 1/4+² 2y=x2 1² - 2x Ⓡ So an equation of the tangent to the folium at (3, 3) is y-(3✔ ) = -1(x - (3✓ or (3, 3) which simplifies to x6 = 16x³ horizontal if y' = 0 x )²-√3x² ✔. Using the expression for y' from part (a), we see that y'= 0 when . Since * # 0 in the first quadrant, we have x³ = Looking at the figure below, we see that our answer reasonable. If x 2-x² = 0 (provided that y² - 2x + 0). Substituting y = x² in the equation of the curve, we get then y = . Thus the tangent is horizontal at (0, 0) and at (x, y) =
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