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Multivariable Calculus

8th Edition
James Stewart
ISBN: 9781305266643
Textbook Problem

Find the points on the given curve where the tangent line is horizontal or vertical.

62. r = 1 − sin θ

To determine

To find: The tangent line is horizontal or vertical and the point for the curve r=1sinθ .

Explanation

Given:

The polar curve is as below.

r=1sinθ

Calculation:

Horizontal tangents is obtained when the condition dydθ=0 is true and the vertical tangents is obtained by equation dxdθ=0 .

Differentiate the curve equation (r=1sinθ) with respect to θ .

r=1sinθdrdθ=cosθ

Write the chain rule for dydx .

dydx=dydθdxdθ = drdθsinθ+rcosθdrdθcosθrsinθ

For horizontal tangents, use the condition dydθ=0 .

Substitute (cosθ) for (drdθ) and (1sinθ) for (r) in equation (drdθcosθrsinθ) .

dxdθ=drdθcosθrsinθ=(cosθ)cosθ(1sinθ)sinθ=cos2θsinθ+sin2θ=2sin2θsinθ1=(2sinθ+1)(sinθ1)

Substitute (cosθ) for (drdθ) and (1sinθ) for (r) in equation (drdθsinθ+rcosθ) .

dydθ=drdθsinθ+rcosθ

dydθ=(cosθ)sinθ+(1sinθ)cosθ=cosθ(2sinθ1)

Horizontal tangents is obtained by the equation dydθ=0 .

Substitute (cosθ(2sinθ1)) for (dydθ) .

dydθ=0(cosθ(2sinθ1))=0θ=π6,π2,5π6,3π2

Substitute (π6) for θ in equation r=1sinθ .

r=1sinθ=1sin(π6)=12

Substitute (π2) for θ in equation r=1sinθ .

r=1sinθ=1sin(π2)=11=0

Substitute (5π6) for θ in equation   r=1sinθ .

r=1sinθ=1sin(5π6)=12

Substitute (3π6) for θ in equation r=1sinθ .

r=1sinθ=1sin(3π6)=1(1)=2

The points are (r,θ)

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