Chapter 1.4, Problem 32E

To determine

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

Answer to Problem 32E

The value of $t$ is $3<t\le 5$ such that the graph is in the Quadrant IV.

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 IV consists of the line segment for the interval $-4<y<0$ or points $\left(-2,-4\right)$ and $\left(2,0\right)$ .

For the point $\left(-2,-4\right)$ , $y=-4$ . Substitute $-4$ for $y$ in the equation $y=t-1$ to find the value of $t$ .

  $\begin{array}{c}-4=t-1\\ t=-4+1\\ =-3\end{array}$

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 $-3$ for t in the equation $x=3-\left|t\right|$ .

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

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

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

Therefore, the value of $t$ is $3<t\le 5$ such that the graph is in the Quadrant IV.