Chapter 6.3, Problem 36E

Solution

To find: the values of the parameter $t$ that produces the graph in the fourth quadrant.

The range of $t$ is $\left(2,3\right]$ that produces the graph in the fourth quadrant.

Given data:

The parametric equations are $x=2-\left|t\right|\text{ and }y=t-0.5\text{ where}-3\le t\le 3$ .

Formula used:

The graph of the ordered pairs $\left(x,y\right)$ where $x=f\left(t\right)\text{ and }y=g\left(t\right)$ are functions defined on an interval $l$ of $t$ values, is a parametric curve.

Calculation:

From graph, it can be seen that the curve is a line segment in the fourth quadrant.

Let us suppose the extreme points of line segment are $\left(-x_1,y_1\right)$ and $\left(0,y_{{}_2}\right)$ .

Calculate $t$ for which for $\left(-x_1,y_1\right)$ . The range of $t$ is $-3\le t\le 3$ .

Compute $x$ and $y$ for $t=-3$ .

  $\begin{array}{l}〈x,y〉=〈2-\left|-3\right|,-3-0.5〉\\ 〈x,y〉=〈-1,-3.5〉\end{array}$

Compute $x$ and $y$ for $t=-2$ .

  $\begin{array}{l}〈x,y〉=〈2-\left|-2\right|,-2-0.5〉\\ 〈x,y〉=〈0,-2.5〉\end{array}$

Compute $x$ and $y$ for $t=-2$ .

  $\begin{array}{l}〈x,y〉=〈2-\left|-2\right|,-2-0.5〉\\ 〈x,y〉=〈0,-2.5〉\end{array}$

Compute $x$ and $y$ for $t=-1$ .

  $\begin{array}{l}〈x,y〉=〈2-\left|-1\right|,-1-0.5〉\\ 〈x,y〉=〈1,-1.5〉\end{array}$

Compute $x$ and $y$ for $t=0$ .

  $\begin{array}{l}〈x,y〉=〈2-\left|0\right|,0-0.5〉\\ 〈x,y〉=〈2,-0.5〉\end{array}$

Compute $x$ and $y$ for $t=1$ .

  $\begin{array}{l}〈x,y〉=〈2-\left|1\right|,1-0.5〉\\ 〈x,y〉=〈1,0.5〉\end{array}$

Compute $x$ and $y$ for $t=2$ .

  $\begin{array}{l}〈x,y〉=〈2-\left|2\right|,2-0.5〉\\ 〈x,y〉=〈0,1.5〉\end{array}$

Compute $x$ and $y$ for $t=3$ .

  $\begin{array}{l}〈x,y〉=〈2-\left|3\right|,3-0.5〉\\ 〈x,y〉=〈-1,2.5〉\end{array}$

It is calculated that for $t=3$ , the value of $x$ and $y$ lies in fourth quadrant.

For $\left(0,y_1\right)$ , calculate $t$ for which $x=0$ .

  $\begin{array}{l}x=2-\left|t\right|\\ 0=2-\left|t\right|\\ \left|t\right|=2\\ t=\pm 2\end{array}$

At $t=2,〈x,y〉=〈2-\left|2\right|,2-0.5〉=〈0,1.5〉$

At $t=-2,〈x,y〉=〈2-\left|-2\right|,-2-0.5〉=〈0,-2.5〉$

The coordinate at $t=-2,$ lies on the negative y-coordinate axis that’s why this is invalid.

Therefore, the range of $t$ is $\left(2,3\right]$ that produces the graph in the fourth quadrant.