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)
