Answer to Question #266347 in Calculus for Kooper

Question

The graph of the parametric equations is called a cycloid.

x=θ-sinθ and y=1-cosθ for 0≤θ≤2π

(a) find dy/dx

(b) find an equation of the tangent to the cycloid at the point where θ=π/3

(c) at what point is the tangent horizontal?

(d) graph the cycloid and the tangent lines in parts (b) and (c)

Answer 1

(a) $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

$y=1-cos\theta ,\ \frac{dy}{d\theta}=sin\theta$

$x=\theta-sin\theta,\$ $\frac{dx}{d\theta}=1-cos\theta$

$therefore \ \frac{dy}{dx}=\frac{sin\theta}{1-cos\theta}$

(b)Equation of tangent line to the cycloid

$when \ \theta=\frac{\pi}{3}\ we \ have$

$x=r(\frac{\pi}{3}-sin\frac{\pi}{3})=r(\frac{\pi}{3}-\frac{\sqrt{3}}{2})$

$y=r(1-cos\frac{\pi}{3})=\frac{r}{2}$

and$\frac{dy}{dx}=\frac{sin(\frac{\pi}{3})}{1-cos\frac{\pi}{3}}=\frac{\frac{\sqrt{3}}{2}}{1-\frac{1}{2}}=\sqrt{3}$

The slope of the tangent is $\sqrt{3}$

its equation is $y-\frac{r}{2}=\sqrt{3}(x-\frac{r\pi}{3}+\frac{r\sqrt{3}}{3})\ or \sqrt{3}x-y=r(\frac{\pi}{\sqrt{3}}-2)$

(c)The tangent is horizontal when $\frac{dy}{dx}=0$

This occurs when sin θ = 0 and 1 – cos θ ≠ 0,

that is, θ = (2n – 1)π, n an integer.

The corresponding point on the cycloid is

((2n – 1)π, 2).

When θ = 2nπ, both dx/dθ and dy/dθ are 0.

(d)Graph