Chapter 6, Problem 69EP

$\text{(a)}$

To determine

To find: The expression for $v$ in terms of $t$ .

Answer to Problem 69EP

The answer is $v=-30e^{-2t}-17$

Explanation of Solution

Given information:

The differential equation is $dv/dt=-2(v+17)$ and

the initial condition is $v(0)=-47$ .

Calculation:

  $\begin{array}{c}\frac{dv}{dt}=-2(v+17)\\ \to \frac{dv}{v+17}=-2dt\\ \to \ln |v+17|=-2t+C\\ \to v=C{e}^{-2t}-17\\ \to -47=C-17\\ \to C--30\\ \to v=-30e^{-2t}-17\end{array}$

Therefore, the required expression for $v$ in terms of $t$ is $v=-30e^{-2t}-17$ .

$\text{(b)}$

To determine

To find: The terminal velocity of the skydiver to the nearest foot per second.

Answer to Problem 69EP

The answer is $-17\text{ feet}$ per second.

Explanation of Solution

Given information:

The differential equation is $dv/dt=-2(v+17)$ and

the initial condition is $v(0)=-47$ .

Calculation:

  $\begin{array}{c}\underset{t\to \infty }{\lim }v(t)=\underset{t\to \infty }{\lim }\left(-30e^{-2t}-17\right)\\ =\underset{t\to \infty }{\lim }\left(-30e^{-2t}\right)-17\\ =-30\underset{t\to \infty }{\lim }\left(e^{-2t}\right)-17\\ =-30\cdot 0-17\\ =-17\text{ feet per second }\end{array}$

Therefore, the required terminal velocity of the skydiver to the nearest $-17\text{ feet}$ per second.

$(c)$

To determine

To find: The time $t$ does she reach this speed.

Answer to Problem 69EP

The answer is: About $1.151\text{ seconds }$

Explanation of Solution

Given information:

The differential equation is $dv/dt=-2(v+17)$ and

the initial condition is $v(0)=-47$ .

Calculation:

  $\begin{array}{c}v(t)=-30e^{-2t}-17\\ \to -20=-30e^{-2t}-17\\ \to -3=-30e^{-2t}\\ \to \ln \frac{1}{10}=\ln e^{-2t}\\ \to \ln \frac{1}{10}=-2t\\ \to t=-\frac{\ln 1/10}{2}\\ \to t\approx 1.151\text{ seconds }\end{array}$

Therefore, the required about $1.151\text{ seconds }$ does she reach this speed.