Chapter 1, Problem 34RE

a.

To determine

Find parametric equations for the set of all points $P$ .

Answer to Problem 34RE

  $\left(a-a\cos 2\theta ,2a\tan \theta -a\sin 2\theta \right)$

Explanation of Solution

Given information:

Find parametric equations for the set of all points $P$ determined as shown in the figure such that $\left|OP\right|=\left|AB\right|$ .(This curve is called the cissoid of Diocles after the Greek scholar Diocles, who introduced the cissode as a graphical method for constructing the edge of a cube whose volume is twice that of a given cube.)

  

Calculation:

Consider the figure

  

Whereas $OG\&GA$ are radii of the circle radius $a$ . So $OG\&GA$ are congruent and so their opposite angles $\overline{)OAG}\&\overline{)AOG}$ are congruent.

Hence $\overline{)OAG}=\overline{)AOG}=\theta$ .

For the triangles $ABE\&ODP$ , the sides $\left|OP\right|=\left|AB\right|$ and lines $BE\Vert PD$ (these lines are perpendicular to the $x-axis$ ) so the angles $\overline{)BAE}\&\overline{)POD}$ are congruent.

Hence $\overline{)BAE}=\overline{)POD}=\theta$ .

Consider the triangle $OBC$ to find the coordinates of the point $B$ .

Hence $OC=2a$ (diameter of the circle) and $\overline{)BOC}=\theta$ , we have

  $\begin{array}{l}\tan \theta =\frac{BC}{2a}\\ BC=2a\tan \theta \end{array}$

The $x$ -coordinate of the point $B$ is $2a$ and the $y-$ coordinates is $2a\tan \theta$ .

Hence the point $B=\left(2a,2a\tan \theta \right)$ .

To find the point $A$ , we need to determine the distances $AF\&OF$ .

In the triangle $OAG$ , sum of the angles are

  $\begin{array}{l}\theta +\theta +\overline{)AGO}=180°\\ \overline{)AGO}=180°-2\theta \end{array}$

We know that

  $\begin{array}{l}\overline{)AGO}+\overline{)AGF}=180°\\ 180°-2\theta +\overline{)AGF}=180°\\ \overline{)AGF}=2\theta \end{array}$

Now consider the triangle $AGF,$ get

  $\begin{array}{l}\sin 2\theta =\frac{AF}{AG}\\ =\frac{AF}{a}\\ AF=a\sin 2\theta \end{array}$

And

  $\begin{array}{l}\cos 2\theta =\frac{GF}{AG}\\ =\frac{GF}{a}\\ GF=a\cos 2\theta \end{array}$

The $x-$ coordinate of the point $A$ is

  $OG+GF=a+a\cos 2\theta$

The $y-$ coordinate of the point $A$ is $AF=a\sin 2\theta$ .

Hence, the point $A=\left(a+\cos 2\theta ,a\sin 2\theta \right)$ .

Find the distance $AE\&BE:$

To get the distance $AE$ , we need to subtract the $x$ -coordinate of $A$ from the $x$ -coordinate of $B$ .

  $\begin{array}{l}AE=2a-\left(a+a\cos 2\theta \right)\\ =a-a\cos 2\theta \end{array}$

To get the distance $BE$ , we need to subtract the $y$ -coordinate of $A$ from the $y$ -coordinate of $B$ .

  $BE=2a\tan \theta -a\sin 2\theta$ .

The triangles $ABE\&OPD$ are congruent whereas the sides $OP,AB$ are equal and the angles $\overline{)BAE}\&\overline{)POD}$ are equal.

Hence, in these two triangles $ABE\&OPD$ the corresponding sides $AE,OD$ are equal and $BE,PD$ are equal.

  $AE=OD=a-a\cos 2\theta ,BE=PD=2a\tan \theta -a\sin 2\theta$

Hence, the required coordinates of the point $P$ is $\left(a-a\cos 2\theta ,2a\tan \theta -a\sin 2\theta \right)$ .

b.

To determine

Use the geometric description of the curve to draw a rough of the curve by hand.

Answer to Problem 34RE

  

Explanation of Solution

Given information:

Use the geometric description of the curve to draw a rough of the curve by hand. Check your work by using the parametric equations to graph the curve.

  

Calculation:

Draw the curve represented by the parametric equations

  $x\left(\theta \right)=a-a\cos 2\theta ,y\left(\theta \right)=2a\tan \theta -a\sin 2\theta$ .

This curve is known as cissoids of Diocles.

The graph of the curve is shown below:

  

Hence, the result is plotted.