Find the slope of the tangent line of the curve r = cos (3theta) at theta = pi / 3

The slope of the tangent line to the curve at a point (x, y) is dy/dx. By the chain rule, this is equivalent to

dy/dθ × dθ/dx = (dy/dθ) / (dx/dθ)

where y = r(θ) sin(θ) and x = r(θ) cos(θ). Then

dy/dθ = dr/dθ sin(θ) + r(θ) cos(θ)

dx/dθ = dr/dθ cos(θ) - r(θ) sin(θ)

Given r(θ) = cos(3θ), we have

dr/dθ = -3 sin(3θ)

and so

dy/dx = (-3 sin(3θ) sin(θ) + cos(3θ) cos(θ)) / (-3 sin(3θ) cos(θ) - cos(3θ) sin(θ))

When θ = π/3, we end up with a slope of

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

dy/dx = -cos(π/3) / sin(π/3)

dy/dx = -cot(π/3) = -1/√3

29 is the correct answer

step-by-step explanation: