Chapter 6.3, Problem 53E

Solution

a)

To find: The maximum value of $y=1-\cos t$ .

The equation $y=1-\cos t$ has a maximum value at the points $t=\pi ,3\pi ,5\pi ,\mathrm{...}$

Given information:

The cycloid: $x=t-\sin t$ , $y=1-\cos t$

Formula used:

Let $y=f\left(x\right)$ be a function and it is maximum at $x_0$ if

  $\frac{dy}{dx}=0$ and $\frac{d^{2}y}{d{x}^2}<0$ at $x=x$ .

Calculation:

  $\frac{dy}{dt}=\sin t$

Solve: $\frac{dy}{dx}=0$

  $\begin{array}{l}\sin t=0\\ t=\pi ,2\pi ,\mathrm{...}\end{array}$

  $\begin{array}{c}\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)\\ =\frac{d}{dx}\left(-\sin t\right)\\ =-\cos t\end{array}$

Substitute the values of $t$ in the equation $\begin{array}{c}\frac{d^{2}y}{d{x}^2}=-\cos t\\ \end{array}$ :

At $t=\pi ,3\pi ,5\pi ,\mathrm{...}$

  $\begin{array}{c}\frac{d^{2}y}{d{x}^2}<0\\ \end{array}$

And

At $t=2\pi ,4\pi ,6\pi ,\mathrm{...}$

  $\begin{array}{c}\frac{d^{2}y}{d{x}^2}>0\\ \end{array}$

Thus, the equation $y=1-\cos t$ has a maximum value at the points $t=\pi ,3\pi ,5\pi ,\mathrm{...}$ and substitute any of these values, the value of $y$ is 2.

The value $y=2$ is the highest value corresponding to the points $t=\pi ,3\pi ,5\pi ,\mathrm{...}$ in the graph.

  

Figure (1)

b)

To find:the distance between neighboring x-intercepts in Cycloid.

The distance between the neighboring x-intercept is $2\pi$ .

Given information:

Given the parametric equations of the cycloid: $x=t-\sin t,$

  $y=1-\cos t$

Calculation:

The intercept of the cycloid can be computed by assuming $y=0$ .

  $\begin{array}{l}y=0\\ 1-\cos t=0\\ \cos t=1\\ t=0,2\pi ,4\pi ,\mathrm{...}\end{array}$

The cycloid intersect x-axis at $x=0,2\pi ,4\pi ,\mathrm{...}$

Thus, the distance between the neighboring x-intercept is $2\pi$ .

  

Figure (1)