Chapter 11.1, Problem 22E

(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 22E

The required sketch is shown in Figure 1

Explanation of Solution

Given Data:

The given equations are,

  $\begin{array}{l}x=\ln \left(5t\right)\\ y=\ln \left(4t^2\right)\\ 0\le t\le 10\end{array}$

Calculation:

Consider the given equation as,

  $\begin{array}{l}y-2x=\left(\ln 4-2\ln 5\right)\\ y-2x=2\ln 2-2\ln 5\\ y=2x-2\ln \left(\frac{5}{2}\right)\end{array}$

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 22E

The rightmost point on the graph is $\left(\ln 50,\ln 400\right)$ .

Explanation of Solution

Consider the graph is shown in Figure 1 the graph shows that the rightmost point on the graph is $\left(\ln 50,\ln 400\right)$ .

(c)

To determine

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

Answer to Problem 22E

The rightmost point on the graph is $\left(\ln 50,\ln 400\right)$ .

Explanation of Solution

Since, the equation is for the straight line so we do not use derivatives this equation of the line is continuous increasing function so the right most point is at $10$ .

  $x=\ln 50$

Then, the value of $y$ is,

  $\begin{array}{c}y=\ln \left(4\times {10}^2\right)\\ =\ln 400\end{array}$

Thus, rightmost point on the graph is $\left(\ln 50,\ln 400\right)$ .