Chapter 0.4, Problem 6E

a.

To determine

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

Answer to Problem 6E

  

Initial point is $\left(0,0\right)$ and no terminal point.

Explanation of Solution

Given information:

The parametric equations:

  $x=-\sqrt{t},y=t,t\ge 0$

Use the graphing calculator to graph the given curve.

First step is to set the mode to parametric.

For that press the MODE key.

Then Scroll down to FUNC.

Then move it to right till PAR

Press ENTER key.

  

Now, go back to main window by quitting (to do that press and then ).

Now, press the key

Now enter the given parametric equations as shown below:

  

Press the WINDOW key.

Enter values

  

Press GRAPH (here observe the direction in which the graph is being traced.)

  

This is the required graph and the red arrow gives the direction in which it is traced.

To find the initial point, substitute $t=0$ in the given parametric equation.

  $x=-\sqrt{0}=0$

  $y=0$

So, the initial point is $\left(0,0\right)$ .

Since t goes to infinity, so the graph has no terminal point.

b.

To determine

To find: Cartesian equation for the curve that contains the parameterized curve. Also explain what portion of the Cartesian equation is traced by the parameterized curve.

Answer to Problem 6E

  $y=x^2$

Only left half of the graph of the Cartesian equation is traced by the parameterized curve.

Explanation of Solution

Given information:

The parametric equations:

  $x=-\sqrt{t},y=t,t\ge 0$

Squaring both sides of the parametric equation $x=-\sqrt{t}$ ,

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

Thus, the required Cartesian equation is $y=x^2$ .

The Cartesian equation $y=x^2$ represents a parabola and from the graph in part (a), it is clearly just the left half of the parabola.