Chapter 0.4, Problem 4E

To determine

To match: the parametric equation $x=12\sin t-3\sin \left(6t\right)$ , $y=12\cos t+3\cos \left(6t\right)$ with their graph. Also state the approximate dimension of the viewing window and given a parameter interval that traces the curve exactly once.

Answer to Problem 4E

The graph of the given parametric equation is the graph (b).

The viewing window is: $\left[-15,15\right]$ by $\left[-15,15\right]$

The parameter interval is: $0\le t\le 2\pi$

Explanation of Solution

Given information:

The given graphs are:

  

Formula Used:

  $\sin \left(0\right)=0$

  $\cos \left(0\right)=1$

Period of $\sin \left(x\right)=2\pi$

Period of $\cos \left(x\right)=2\pi$

Consider the given parametric equation,

  $x=12\sin t-3\sin \left(6t\right)$ , $y=12\cos t+3\cos \left(6t\right)$

Substitute $t=0$, then the corresponding values of x and y are:

  $\begin{array}{c}x=12\sin \left(0\right)-3\sin \left(6\left(0\right)\right)\\ =12\left(0\right)-3\left(0\right)\\ =0-0\\ =0\end{array}$

  $\begin{array}{c}y=12\cos \left(0\right)+3\cos \left(6\left(0\right)\right)\\ =12\left(1\right)+3\left(1\right)\\ =12+3\\ =15\end{array}$

This implies that the graph of this parametric equation must contain the point $\left(0,15\right)$ . This eliminates the choices (a), (c) and (d).

So, the correct answer is graph (b).

  

Now, on observing the graph (b), it can be noticed that the point $\left(0,15\right)$ lie on the third mark on the positive y-axis. So, it can be implied that each mark denotes 3 units. So, the viewing window is $\left[-15,15\right]$ by $\left[-15,15\right]$ .

Now, it can also be observed that this graph is a closed curve in the window. This is possible when the parameter of t is such that gives one full cycle for both x and y .

Now, the period of both $\sin t$ and $\cos t$ is $2\pi$ .

Thus, for the parameter t to be able to complete one full cycle it must lie between 0 and $2\pi$ .

That is, $0\le t\le 2\pi$