Question

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

Answer 1

The given curve has equation

r(θ) = 9 + 8 cos(θ)

and its derivative is

dr/dθ = -8 sin(θ)

When θ = π/3, we have r (π/3) = 13, and dr/dθ (π/3) = -4√3.

Differentiate these with respect to θ :

dy/dθ = dr/dθ sin(θ) + r(θ) cos(θ)

dx/dθ = dr/dθ cos(θ) - r(θ) sin(θ)

In polar coordinates, we have

y(θ) = r(θ) sin(θ)

x(θ) = r(θ) cos(θ)

and when θ = π/3, we have y (π/3) = 13√3/2 and x (π/3) = 13/2.

The slope of the tangent line to the curve is dy/dx. By the chain rule,

dy/dx = dy/dθ • dθ/dx = (dy/dθ) / (dx/dθ)

dy/dx = (dr/dθ sin(θ) + r(θ) cos(θ)) / (dr/dθ cos(θ) - r(θ) sin(θ))

When θ = π/3, the slope is

dy/dx = (-4√3 sin(π/3) + 13 cos(π/3)) / (-4√3 cos(π/3) - 13 sin(π/3))

dy/dx = (-4√3 (√3/2) + 13 (1/2)) / (-4√3 (1/2) - 13 (√3/2))

dy/dx = - 1/(17√3)

So, the tangent line has slope -1/(17√3) and passes through (13/2, 13√3/2). Using the point-slope formula, its equation is

y - 13√3/2 = -1/(17√3) (x - 13/2)

y = -(x - 338)/(17√3)