Chapter 1.4, Problem 6E

(a)

To determine

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

Answer to Problem 6E

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

Explanation of Solution

Given information: The equations are $x=-\sqrt{t}$ and $y=t$ , $t\ge 0$.

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

Step 3: Press $\boxed{\text{GRAPH}}$ to obtain the graph of the parametric equations and the window can be set to $\left[-3,3\right]$ by $\left[-1,3\right]$ as shown below.

  

Figure (1)

If $t=0$ , then $x=0$ and $y=0$ . So, the initial point is $\left(0,0\right)$ . There is no terminal point as $t$ approaches to infinity when $x$ approaches to $-\infty$ and $y$ approaches to $\infty$.

Therefore, the graph of the parametric equations with the direction is shown in figure (1), the initial point is $\left(0,0\right)$ , 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 6E

The Cartesian equation is $y=x^2$ and the left half of the graph of the Cartesian equation is traced by the parameterized curve.

Explanation of Solution

Given information: The equations are $x=-\sqrt{t}$ and $y=t$ , $t\ge 0$.

Calculation:

Square both sides of the equation $x=-\sqrt{t}$.

  $\begin{array}{c}{x}^2={\left(-\sqrt{t}\right)}^2\\ =t\end{array}$

Substitute $x^2$ for $t$ in the equation $y=t$.

  $y=x^2$

The above Cartesian form is an equation of parabola. From the graph shown in part (a), the left half of the graph of the Cartesian equation is traced by the parameterized curve.

Therefore, the Cartesian equation is $y=x^2$ and the left half of the graph of the Cartesian equation is traced by the parameterized curve.