Chapter 1.4, Problem 30E

To determine

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

Answer to Problem 30E

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

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

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

  $\begin{array}{c}4=t-1\\ t=4+1\\ =5\end{array}$

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

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

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