Chapter 11.1, Problem 17E

(a)

To determine

To find: The sketch for the curve over the given $t$ interval indicating the direction in which it is traced.

Answer to Problem 17E

The required sketch is shown in Figure 1.

Explanation of Solution

Given Data:

The given equations are,

  $\begin{array}{l}x=t+1\\ y=t^2+t\\ -2\le t\le 2\end{array}$

Calculation:

Consider the first equation as,

  $\begin{array}{l}x=t+1\\ t=x-1\end{array}$

Substitute the values in the second equation as,

  $\begin{array}{c}y={\left(x-1\right)}^2+x-1\\ =x^2-2x+1+x-1\\ =x^2-x\\ ={\left(x-\frac{1}{2}\right)}^2-\frac{1}{4}\left(-1\le x\le 3\right)\end{array}$

The sketch for the above function is shown in Figure 1

  

Figure 1

(b)

To determine

To find: The identification for the requested points.

Answer to Problem 17E

The lowest point on the graph is $\left(\frac{1}{2},-\frac{1}{4}\right)$ .

Explanation of Solution

Consider the graph is shown in Figure 1 the graph shows that the lowest point is $\left(\frac{1}{2},-\frac{1}{4}\right)$ .

(c)

To determine

To find: The justification that you have found the requested point by analysing an appropriate derivative.

Answer to Problem 17E

The lowest point on the graph is $\left(\frac{1}{2},-\frac{1}{4}\right)$ .

Explanation of Solution

To obtain the lowest point on the graph differentiate the given equation as,

  $\begin{array}{l}y=x^2-x\\ \frac{dy}{dx}=2x-1\\ 0=2x-1\\ x=\frac{1}{2}\end{array}$

Then,

  $\begin{array}{c}y={\left(\frac{1}{2}\right)}^2-\left(\frac{1}{2}\right)\\ =\frac{1}{2}\end{array}$

Thus, lowest point on the graph is $\left(\frac{1}{2},-\frac{1}{4}\right)$ .