Chapter 1.4, Problem 29E

To determine

To find: The value of $t$ that produces the graph in the given quadrant.

Answer to Problem 29E

The value of $t$ is $1<t<3$ such that the graph is in the Quadrant I.

Explanation of Solution

Given information: The given parametric equations are $x=3-\left|t\right|$ , $y=t-1$ at the interval $-5\le t\le 5$ . The graph is given below:

  

Calculation:

From the given graph, it can be observed that Quadrant I consists of the line segment for the interval $0<y<2$ or points $\left(0,2\right)$ and $\left(2,0\right)$ .

For the point $\left(0,2\right)$ , $y=2$ . Substitute 2 for $y$ in the equation $y=t-1$ to find the value of $t$ .

  $\begin{array}{c}2=t-1\\ t=2+1\\ =3\end{array}$

For the point $\left(2,0\right)$ , $y=0$ . Substitute 0 for $y$ in the equation $y=t-1$ to find the value of $t$ .

  $\begin{array}{c}0=t-1\\ t=1\end{array}$

To check the values, substitute 1 for t in the equation $x=3-\left|t\right|$ .

  $\begin{array}{c}x=3-\left|1\right|\\ =2\end{array}$

To check the values, substitute 3 for t in the equation $x=3-\left|t\right|$ .

  $\begin{array}{c}x=3-\left|3\right|\\ =0\end{array}$

Therefore, the value of $t$ is $1<t<3$ such that the graph is in the Quadrant I.