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

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First recall that

x = rcosθ and y = rsinθ

Thus

x = (2 - sinθ)(cosθ) and y = (2 - sinθ)(sinθ)

Now

dx/dθ = 2(sinθ)^2 - 2sinθ - 1 and dy/dθ = 2cosθ - 2sinθcosθ

Now evaluating each derivative above @ θ = π/3, we obtain

dx/dθ = (1 - 2√3)/2 and dy/dθ = (2 - √3)/2

Now

dy/dx = (dy/d
θ)(dθ/dx)

Thus @ θ = π/3,

dy/dx = [(2 -
√3)/2][2/(1 - 2√3)]

dy/dx = (2 -
√3)/(1 - 2√3) is the slope of the required tangent

User Napkinsterror
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