Chapter 7, Problem 69RE

a. $$

To determine

Tofind the value of $v$ in terms of $t$ using separation of variables.

Answer to Problem 69RE

The value of $v$ is $v=-30e^{-2t}-17$ .

Explanation of Solution

Given information:

The given differential equation is $\frac{dv}{dt}=-2\left(v+17\right)$ with initial condition $v\left(0\right)=-47$ .

The given differential equation is

  $\frac{dv}{dt}=-2\left(v+17\right)$ where $v\left(t\right)$ -velocity in feet per second of a skydiver at time $t$ seconds.

Consider the given equation,

  $\begin{array}{l}\frac{dv}{dt}=-2\left(v+17\right)\\ \frac{dv}{v+17}=-2dt\end{array}$

Integrating on both sides,

  $\begin{array}{l}\ln \left|v+17\right|=-2t+c\\ v+17=c{e}^{-2t}\\ v=c{e}^{-2t}-17\end{array}$

Substitute the initial value $v\left(0\right)=-47$ ,

  $\begin{array}{l}v=c{e}^{-2t}-17\\ -47=c-17\\ \Rightarrow c=-30\end{array}$

Substituting back we get,

  $\Rightarrow v=-30e^{-2t}-17$

Therefore, the value of $v$ is $v=-30e^{-2t}-17$ .

b. $$

To determine

To find theterminal value of the skydiver to the nearest foot per second if the terminal velocity is defined as $\underset{t\to \infty }{\lim }v\left(t\right)$ .

Answer to Problem 69RE

The terminal value is $-17ft/\sec$ .

Explanation of Solution

Given information:

The given differential equation is $\frac{dv}{dt}=-2\left(v+17\right)$ with initial condition $v\left(0\right)=-47$ .

From part (a) we know that $v=-30e^{-2t}-17$ ,

Appling limit,

  $\begin{array}{l}\underset{t\to \infty }{\lim }v\left(t\right)=-30e^{-2\times \infty }-17\\ \underset{t\to \infty }{\lim }v\left(t\right)=-17\end{array}$

Therefore, the terminal value is $-17ft/\sec$ .

c. $$

To determine

To find time $t$ when the skydiver reaches the land safely with the speed of $20ft/\sec$ .

Answer to Problem 69RE

At $1.15\sec$ after opening her parachute the skydiver reaches the land safely.

Explanation of Solution

Given information:

The given differential equation is $\frac{dv}{dt}=-2\left(v+17\right)$ with initial condition $v\left(0\right)=-47$ .

From part (a)we know that $v=-30e^{-2t}-17$ and $v=20$ ,

  $\begin{array}{l}v=-30e^{-2t}-17\\ -20=-30e^{-2t}-17\\ -3=-30e^{-2t}\\ 1=10e^{-2t}\\ e^{-2t}=\frac{1}{10}\\ \ln \left(e^{-2t}\right)=\ln \left(\frac{1}{10}\right)\\ -2t\ln \left(e\right)=\ln \left(\frac{1}{10}\right)\\ -2t=\ln \left(\frac{1}{10}\right)\\ t=\frac{\ln \left(10\right)}{2}\\ \Rightarrow t\approx 1.15\end{array}$

Therefore, at $1.15\sec$ after opening her parachute the skydiver reaches the land safely.