Chapter 1.4, Problem 4E

To determine

To match: The parametric equations with its appropriate graph, write the approximate dimensions of the viewing window and a parameter interval that traces the curve exactly once.

Answer to Problem 4E

The correct match of the graph for given parametric equation is (b), the dimension of the window is $\left[-15,15\right]$ by $\left[-15,15\right]$ and parameter interval that traces the curve exactly once is $0\le t\le 2\pi$.

Explanation of Solution

Given information: The parametric equations are $x=12\sin t-3\sin \left(6t\right)$ and $y=12\cos t+3\cos \left(6t\right)$.

Calculation:

Use the following step to graph the parametric equations by graphing calculator.

Step 1: Press the $\boxed{\text{MODE}}$ button, under FUNC and choose PAR option.

Step 2: Press $\boxed{\text{Y=}}$ and set the parametric equations under ${\text{X}}_{1\text{T}}$ and ${\text{Y}}_{1\text{T}}$.

Step 3: Press $\boxed{\text{GRAPH}}$ to obtain the graph of the parametric equations and the window can be set to $\left[-15,15\right]$ by $\left[-15,15\right]$ as shown below.

  

Figure (1)

First find the starting point at $t=0$ to find the period of the function as shown below.

Substitute 0 for $t$ in the parametric equation $x=12\sin t-3\sin \left(6t\right)$.

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

Substitute 0 for $t$ in the parametric equation $y=12\cos t+3\cos \left(6t\right)$.

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

Find the next value of $t$ for which the coordinates of the curve are $\left(0,15\right)$ . After $t=0$ , the next value of $t$ for which $y=15$ is $t=2\pi$.

Substitute $2\pi$ for $t$ in the parametric equation $y=12\cos t+3\cos \left(6t\right)$.

  $\begin{array}{c}y=12\cos \left(2\pi \right)+3\cos \left(12\pi \right)\\ =12+3\\ =15\end{array}$

Substitute $2\pi$ for $t$ in the parametric equation $x=12\sin t-3\sin \left(6t\right)$.

  $\begin{array}{c}x=12\sin \left(2\pi \right)-3\sin \left(12\pi \right)\\ =0\end{array}$

The interval is $0\le t\le 2\pi$.

Therefore, the correct match of the graph for given parametric equation is (b), the dimension of the window is $\left[-15,15\right]$ by $\left[-15,15\right]$ and parameter interval that traces the curve exactly once is $0\le t\le 2\pi$.