15. Let r be a positive real number. The equation for a circle of radius r whose center is the origin is x² + y² = r². dy dx (a) Use implicit differentiation to determine (b) Let (a, b) be a point on the circle with a ‡ 0 and b ‡ 0. Determine the slope of the line tangent to the circle at the point (a, b). (c) Prove that the radius of the circle to the point (a, b) is perpendicular to the line tangent to the circle at the point (a, b). Hint: Two lines (neither of which is horizontal) are perpendicular if and only if the products of their slopes is equal to −1.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Question
15. Let r be a positive real number. The equation for a circle of radius r whose
center is the origin is x² + y² = r².
dy
dx
(a) Use implicit differentiation to determine
(b) Let (a, b) be a point on the circle with a ‡ 0 and b ‡ 0. Determine
the slope of the line tangent to the circle at the point (a, b).
(c) Prove that the radius of the circle to the point (a, b) is perpendicular to
the line tangent to the circle at the point (a, b). Hint: Two lines (neither
of which is horizontal) are perpendicular if and only if the products of
their slopes is equal to −1.
Transcribed Image Text:15. Let r be a positive real number. The equation for a circle of radius r whose center is the origin is x² + y² = r². dy dx (a) Use implicit differentiation to determine (b) Let (a, b) be a point on the circle with a ‡ 0 and b ‡ 0. Determine the slope of the line tangent to the circle at the point (a, b). (c) Prove that the radius of the circle to the point (a, b) is perpendicular to the line tangent to the circle at the point (a, b). Hint: Two lines (neither of which is horizontal) are perpendicular if and only if the products of their slopes is equal to −1.
Expert Solution
Step 1: Solution

Given equation:

x squared plus y squared equals r squared                     . . . (1)

(a) Differentiating (1) with respect to x,

    fraction numerator d over denominator d x end fraction open parentheses x squared plus y squared close parentheses equals fraction numerator d over denominator d x end fraction open parentheses r squared close parentheses

fraction numerator d over denominator d x end fraction x squared plus fraction numerator d over denominator d x end fraction y squared equals 0

     2 x plus 2 y fraction numerator d y over denominator d x end fraction equals 0

             2 y fraction numerator d y over denominator d x end fraction equals negative 2 x

Solving further,

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