Find the slope of the tangent line to the polar curve r = 2 - sin(θ) at the point specified by θ = π/3.

Solution:

Given, r = 2 - sinθ is the polar equation.

Then, dr/dθ = -cosθ

x= r cosθ and y = r sin θ

Thus x = (2 - sinθ) cos θ 

y = (2 - sin θ) sin θ

dy/dx= \(\dfrac{dy/dθ}{dx/dθ}\)

= [(d.(2 - sinθ) sin θ)/dθ]/ [(d.(2 - sinθ) cos θ)/dθ]

= [-cosθ sinθ + (2-sinθ )cosθ] / [-cosθ cosθ - (2-sinθ)sinθ ]

= [-cosθ sinθ + 2cosθ - sinθ cosθ] / [-cos²θ - 2sinθ + sin²θ]

= (2cosθ - 2sinθ cosθ) / (-(cos²θ - sin²θ) - 2sinθ)

= (2cosθ - sin2θ ) / (-cos2θ - 2sinθ )

The slope of the tangent at the point where θ = 𝜋/3 is

dy/dx at (θ =𝜋/3) = [2cos( 𝜋/3) - sin(2 𝜋/3)] / [-cos(2 𝜋/3) - 2sin( 𝜋/3)]

= [(2×1/2) - (√3/2)] / [-(-√3/2) - (2√3/2)]

=[(2- √3)/2] / [(1- 2√3)/2]

= (2-√3) / (1-2√3)

Therefore, the slope of the tangent at θ = 𝜋/3 is (2-√3) / (1-2√3).

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Find the slope of the tangent line to the polar curve r = 2 - sin(θ) at the point specified by θ = π/3.

Summary:

The slope of the tangent line to the polar curve r = 2 - sin(θ) at the point specified by θ = π/3 is (2-√3) / (1-2√3).