Chapter 1, Problem 46RE

(a)

To determine

To find: The graph of the parametric equations, the initial and terminal points. Also, indicate the direction in which the curve is traced.

Answer to Problem 46RE

The graph of the parametric equations with the direction is shown in figure (1), the initial point is $\left(0,4\right)$ and no terminal point.

Explanation of Solution

Given information:The equations are $x=4\cos t$ and $y=4\sin t$ such that $\frac{\pi }{2}\le t\le 3\pi$ .

Calculation:

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

Step 1: First press the “ON” button graphical calculator.

Step 2: Press the $\boxed{\text{MODE}}$ button, under FUNC and choose PAR option.

Step 3: Press $\boxed{\text{Y=}}$ , and set the parametric equations under ${\text{X}}_{\text{1T}}$ and ${\text{Y}}_{\text{1T}}$ .

Step 4: Press $\boxed{\text{GRAPH}}$ to obtain the graph of the parametric equations and the window can be set to $\left[-4,1\right]$ by $\left[-4,4\right]$ as shown below.

  

Figure (1)

Substitute $\frac{\pi }{2}$ for $t$ in the equation $x=4\cos t$ .

  $\begin{array}{c}x=4\cos \left(\frac{\pi }{2}\right)\\ =0\end{array}$

Substitute 0 for $t$ in the equation $y=4\sin t$ .

  $\begin{array}{c}y=4\sin \left(\frac{\pi }{2}\right)\\ =4\end{array}$

The initial point is $\left(0,4\right)$ that lies in the interval $\frac{\pi }{2}\le t\le 3\pi$ . There is no terminal point as the interval is $\frac{\pi }{2}\le t\le 3\pi$ .

Therefore, the graph of the parametric equations with the direction is shown in figure (1), the initial point is $\left(0,4\right)$ and no terminal point.

(b)

To determine

To find:The Cartesian equation and the portion of the graph of the Cartesian equation that is traced by the parameterized curve.

Answer to Problem 46RE

The Cartesian equation is $y^2+x^2=16$ and the parameterized curve traces the left half of the circle.

Explanation of Solution

Given information:The equations are $x=4\cos t$ and $y=4\sin t$ such that $\frac{\pi }{2}\le t\le 3\pi$ .

Calculation:

Square both the given equations and add them.

  $\begin{array}{c}{x}^2+y^2={\left(4\sin t\right)}^2+{\left(4\cos t\right)}^2\\ {\left(\frac{y}{4}\right)}^2+{\left(\frac{x}{4}\right)}^2={\sin }^{2}t+{\cos }^{2}t\\ \frac{x^2}{16}+\frac{y^2}{16}=1\\ x^2+y^2=16\end{array}$

This is the equation of a circle. The parameterization traces the half part of the circle.

Therefore, the Cartesian equation is $y^2+x^2=16$ and the parameterized curve traces the left half of the circle.