calculus help needed

Question

Find the exact value of the slope of the line which is tangent to the curve given by the equation

r = 1 - 2cos θ at  . You must show your work.

I missed today's class so could someone please help me out with this? thanks

Answer 1

Slope m = dy/dx at θ = π/2;

y = rsinθ = sinθ - 2cosθsinθ = sinθ - sin2θ; x = rcosθ = cosθ - 2cos²θ = cosθ - 1 - cos2θ

dy/dx = (dy/dθ)/(dx/dθ) = (cosθ -2cos2θ)/(- sinθ + 2sin2θ);\Slope of tangent line dy/dx(π/2) =

(cosπ/2 - 2cosπ)/(-sinπ/2 + 2sinπ) = 2/(-1) = - 2

Answer 2

Nic,,,,,,I am going to use t instead of θ.

the goal is to find dy/dx in terms of t.

both x and y are functions of t,

x=r(t)*cos(t)=(1-2cos(t))*cos(t)

y=r(t)*sin(t)=(1-2cos(t))*sin(t))

because dy/dx=dy/dt*dt*dx (chain rule)

and dy/dt*dt/dx=[dy/dt]/[dx/dt] ;;;;;;;;;;;;;;;(minor correction made with divide sign)

dy/dx=dy/dt divided by dx/dt (the formula for d²y/dx² is more complicated)

here dy/dt =2*sin²(t)+(1-2cos(t))*cos(t)

dx/dt=2cos(t)sin(t)+(1-2cos(t))*-sin(t)

at t=π/2 dy/dx=-2