Tangent to polar curves quick clarification?

Question

Find the slope of the tangent line to the curve r = 4 cosθ at the point where θ = pi/4.

Then the instructor had this formula: dy/dx = (dr/dθ)/(dr/dθ) =
(sinθ + r cosθ)/cosθ - r sinθ), which for some reason resulted (using r above): (-4sin^2θ + 4cos^2θ)/(4sinθcosθ - 4cosθsinθ). I know that dr/dθ = -4sin, however, I just don't see the correlation between the formula, the derivative of r, and the result she had. Any help is appreciated, thanks!

Answer 1

The formula is a little different from what you wrote.

r = f(θ)
x = rcosθ = f(θ)cosθ
y = rsinθ = f(θ)sinθ

dx/dθ = f'(θ)cosθ - f(θ)sinθ
dy/dθ = f'(θ)sinθ + f(θ)cosθ

dy/dx = (dy/dθ) / (dx/dθ)
___________

r = 4cosθ

x = rcosθ = (4cosθ)(cosθ) = 4cos²θ
y = rsinθ = (4cosθ)(sinθ) = 2(sin 2θ)

dx/dθ = -8(cosθ)(sinθ) = -4(sin 2θ)
dy/dθ = 4(cos 2θ)

dy/dx = (dy/dθ) / (dx/dθ) = 4(cos 2θ) / [-4(sin 2θ)] = -(cot 2θ)

Plug in θ = π/4

dy/dx = -(cot 2θ) = -[cot 2(π/4)] = -(cot π/2) = 0

The slope of the tangent line at the point (r, π/4) is zero.

Answer 2

Maybe you got confused somewhere.
If you divide (sinθ + r cosθ)/cosθ - r sinθ) by the cos θ, you get this formula,
m = slope of tangent line =
(r + (tan θ)(dr/dθ))/[-r(tan θ)+(dr/dθ)]

Here is how to solve your problem:

(dr/dθ) = -4sin θ
(dr/dθ)(pi/4) = -2sqrt(2)
r = 4 cosθ
r(pi/4) = 2sqrt(2)

m = [2sqrt(2) +(1)(-2sqrt(2))]/[-2sqrt(2)(1) + (-2sqrt(2))] = 0
So the slope of the tangent line is 0 which means that the line is horizontal.