Tangents with polar coordinates

Question

Find the slope of the tangent line to the given polar curve at the point specified by the value of theta.

r = 5-sin(theta)

theta = pi/3

Answer 1

Derivative with Polar Coordinates

dy/dx = [ [dr/dθ]sinθ+rcosθ ] / [ dr/dθ]cosθ-rsinΘ ]

OUR PROBLEM: r = 5 - sinθ and θ=π/3

dr/dθ = -cosθ Plug it into the formula for the "Derivative with Polar Coordinates"

dy/dx = [ (-cosθ)sinθ + (5-sinθ)cosθ ] / [ (-cosθ)cosθ - (5-sinθ)sinθ ]

This simplifies to

dy/dx = [ 5cosθ - sinθcosθ - sinθcosθ ] / [ -cos²θ - 5sinθ + sin²θ ] or

dy/dx = [ 5cosθ - 2sinθcosθ] / [sin²θ - cos²θ - 5sinθ]

we were given that θ=π/3 and cos(pi/3) = 1/2 and sin(pi/3) = √3/2

Plug those in and you will have the slope of the tangent line.

I get [ 5-√3 ] / [1 - 5√3 ]