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(a) If x2 + y2 = 100, find dx (b) Find an equation of the tangent to the circle x² + y² = 100 at the point (6, 8). SOLUTION 1 (a) Differentiating both sides of the equation x? + y? = 100: -(100) dx dx + = 0. Remembering that y is a function of x and using the Chain Rule, we have -(v²) dy dy dx dy dx Thus dy 2x + = 0. dx Now we solve this equation for dy/dx: dy dx (b) At the point (6, 8) we have x = 6 and y = 8, so dy dx An equation of the tangent to the circle at (6, 8) is therefore y (x or

(a) If x2 + y2 = 100, find dx (b) Find an equation of the tangent to the circle x² + y² = 100 at the point (6, 8). SOLUTION 1 (a) Differentiating both sides of the equation x? + y? = 100: -(100) dx dx + = 0. Remembering that y is a function of x and using the Chain Rule, we have -(v²) dy dy dx dy dx Thus dy 2x + = 0. dx Now we solve this equation for dy/dx: dy dx (b) At the point (6, 8) we have x = 6 and y = 8, so dy dx An equation of the tangent to the circle at (6, 8) is therefore y (x or

Trigonometry (MindTap Course List)
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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(a) If x2 + y2 = 100, find dx (b) Find an equation of the tangent to the circle x² + y² = 100 at the point (6, 8). SOLUTION 1 (a) Differentiating both sides of the equation x? + y? = 100: -(100) dx dx + = 0. Remembering that y is a function of x and using the Chain Rule, we have -(v²) dy dy dx dy dx Thus dy 2x + = 0. dx Now we solve this equation for dy/dx: dy dx (b) At the point (6, 8) we have x = 6 and y = 8, so dy dx An equation of the tangent to the circle at (6, 8) is therefore y (x or
Transcribed Image Text:(a) If x2 + y2 = 100, find dx (b) Find an equation of the tangent to the circle x² + y² = 100 at the point (6, 8). SOLUTION 1 (a) Differentiating both sides of the equation x? + y? = 100: -(100) dx dx + = 0. Remembering that y is a function of x and using the Chain Rule, we have -(v²) dy dy dx dy dx Thus dy 2x + = 0. dx Now we solve this equation for dy/dx: dy dx (b) At the point (6, 8) we have x = 6 and y = 8, so dy dx An equation of the tangent to the circle at (6, 8) is therefore y (x or
SOLUTION 2 (b) Solving the equation x2 + y? = 100, we get y = ±/100 – x2. The point (6, 8) lies on the upper semicircle y = V 100 – x and so we consider the function f(x) = V100 – x2. Differentiating f using the %3D Chain Rule, we have 1) dx x2)-1/2 So f'(6) and, as in Solution 1, an equation of the tangent is
Transcribed Image Text:SOLUTION 2 (b) Solving the equation x2 + y? = 100, we get y = ±/100 – x2. The point (6, 8) lies on the upper semicircle y = V 100 – x and so we consider the function f(x) = V100 – x2. Differentiating f using the %3D Chain Rule, we have 1) dx x2)-1/2 So f'(6) and, as in Solution 1, an equation of the tangent is
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