Chapter 1.4, Problem 31E

To determine

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

Answer to Problem 31E

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

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 III consists of the line segment for the interval $-6\le y<-4$ or point $\left(-2,-6\right)$ and $\left(0,-4\right)$ .

For the point $\left(-2,-6\right)$ , $y=-6$ . Substitute $-6$ for $y$ in the equation $y=t-1$ to find the value of $t$ .

  $\begin{array}{c}-6=t-1\\ t=-6+1\\ =-5\end{array}$

For the point $\left(0,-4\right)$ , $y=-4$ . Substitute $-4$ for $y$ in the equation $y=t-1$ to find the value of $t$ .

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

To check the values, substitute $-5$ for t in the equation $x=3-\left|t\right|$ .

  $\begin{array}{c}x=3-\left|-5\right|\\ =-2\end{array}$

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 $-5\le t<-3$ such that the graph is in the Quadrant III.