Q. 73 - Calculus

Question

In Example 5 we saw that the cycloid

$x=r\theta -r\sin \theta ,y=r-r\cos \theta ,\theta \in \mathrm{ℝ}$

has a horizontal tangent line at each odd multiple of$\pi$. Show that the cycloid has a vertical tangent at each even multiple of $\pi$ by showing that $\underset{\theta \to 2\mathrm{li}}{\lim }\frac{dy}{dx}$does not exist wherever k is an integer.

Step-by-Step Solution

Answer

At an even multiple of $\theta$, the cycloid has a vertical tangent line.

Step 1: Given Information

The cycloid is $x=r\theta -r\sin \theta ,y=r-r\cos \theta ,\theta \in \mathrm{ℝ}$

Step 2: Simplification

Consider the cycloid, $x=r\theta -r\sin \theta ,y=r-r\cos \theta \theta \in \mathrm{ℝ}$.

The objective is to prove that the cycloid has a vertical tangent at each even multiple of $\mathrm{\pi }$.

First find the derivatives of the parametric equations and equate them to zero to get the points where it is vertical or horizontal.

A vertical tangent line occurs when $\frac{dx}{dt}=0$w.

To find the vertical tangent line of the cycloid the denominator is zero and the numerator is not zero.

Take the equation, $x=r\theta -r\sin \theta$.

Differentiate with respect to$\theta$

$$\begin{gathered} \frac{dx}{d\theta }=\frac{d}{d\theta }(r\theta -r\sin \theta ) \\ \frac{dx}{d\theta }=\frac{d}{d\theta }r\theta -r\frac{d}{d\theta }\sin \theta \\ \frac{dx}{d\theta }=r-r\cos \theta \end{gathered}$$

Step 3: Further simplification

Now take the equation, $y=r-r\cos \theta$.

Differentiate with respect to$\theta$.

$$\begin{gathered} \frac{dy}{d\theta }=\frac{d}{d\theta }(r-r\cos \theta ) \\ \frac{dy}{d\theta }=\frac{d}{d\theta }r-r\frac{d}{d\theta }\cos \theta \\ \frac{dy}{d\theta }=r\sin \theta \end{gathered}$$

Now the derivative is, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}$

$$\begin{gathered} \frac{dy}{dx}=\frac{r\sin \theta }{r-r\cos \theta }\left[since\frac{dx}{d\theta }=r-r\cos \theta ,\frac{dy}{d\theta }=r\sin \theta \right] \\ \frac{dy}{dx}=\frac{r\sin \theta }{r(1-\cos \theta )} \end{gathered}$$

Then,

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

Now the limit of the derivative at $2k\pi$.

$\begin{array}{rcl}\underset{\theta \to 2k{\pi }^{*}}{\lim }\frac{dy}{dx}& =& \underset{\theta \to 2k\pi }{\lim }\frac{\sin \theta }{(1-\cos \theta )}\\ & =& \underset{\theta \to 2kz}{\lim }\frac{\cos \theta }{\sin \theta }\text{ By using L'Hospitals }\\ & =& \infty \end{array}$

Step 4: Prove that the cycloid has a vertical tangent at each even multiple of π

Now the limit of the derivative at $2k{\pi }^{-}$.

For the vertical tangent line the denominator is zero.

$\begin{array}{rcl}\underset{\theta \to 2k{\pi }^{-}}{\lim }\frac{dy}{dx}& =& \underset{\theta \to 2k{\pi }^{-}}{\lim }\frac{\sin \theta }{(1-\cos \theta )}\\ & =& \underset{\theta \to 2k{\pi }^{-}}{\lim }\frac{\cos \theta }{\sin \theta }\text{ By using L'Hospitals }\\ & =& -\infty \end{array}$

Thus,

$\underset{\theta \to 2kx}{\lim }\frac{dy}{dx}$ does not exists.

The cycloid has a vertical tangent line at even multiple of $\theta$.

Hence Proved.