(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

Calculus: Early Transcendentals
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Author:James Stewart
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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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Question
(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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