Chapter 0.4, Problem 1E

To determine

To match: the parametric equation $x=3\sin \left(2t\right)$ , $y=1.5\cos t$ 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 1E

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

The viewing window is: $\left[-4,4\right]$ by $\left[-3,3\right]$

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

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

Consider the given parametric equation,

  $x=3\sin \left(2t\right)$ , $y=1.5\cos t$

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

  $\begin{array}{c}x=3\sin \left(2\left(0\right)\right)\\ =3\sin \left(0\right)\\ =0\end{array}$

  $\begin{array}{c}y=1.5\cos \left(0\right)\\ =1.5\left(1\right)\\ =1.5\end{array}$

This implies that the graph of this parametric equation must contain the point $\left(0,1.5\right)$ . This eliminates the choices (a), (b) and (d).

So, the correct answer is graph (c).

  

Now, on observing the graph (c), it can be noticed that the point $\left(0,1.5\right)$ lie almost in the mid of the first and the second mark on the positive y-axis. So, it can be implied that each mark denotes 1 unit. So, the viewing window is $\left[-4,4\right]$ by $\left[-3,3\right]$ .

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 $3\sin \left(2t\right)$ is $\pi$ and the period of $1.5\cos t$ 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, $0\le t\le 2\pi$