Chapter 13.4, Problem 36E

a.

To determine

To find : The terminal velocity of the raindrop by evaluating $\underset{t\to \infty }{\lim }v\left(t\right)$.

Answer to Problem 36E

The terminal velocity of the raindrop is 1.2.

Explanation of Solution

Given information:

The downward velocity of a falling raindrop at time t is modeled by the function

  $v\left(t\right)=1.2\left(1-e^{-8.2t}\right)$.

Calculation :

Find $\underset{t\to \infty }{\lim }v\left(t\right)$.

  $\begin{array}{l}\underset{t\to \infty }{\lim }1.2\left(1-e^{-8.2t}\right)=\underset{t\to \infty }{\lim }1.2\left(1-\frac{1}{e^{-8.2t}}\right)\\ \underset{t\to \infty }{\lim }1.2\left(1-e^{-8.2t}\right)=\underset{t\to \infty }{\lim }\left(1.2-\frac{1.2}{e^{8.2t}}\right)\\ \underset{t\to \infty }{\lim }1.2\left(1-e^{-8.2t}\right)=\underset{t\to \infty }{\lim }1.2+\underset{t\to \infty }{\lim }\frac{1.2}{e^{8.2t}}\\ \underset{t\to \infty }{\lim }1.2\left(1-e^{-8.2t}\right)=1.2+\frac{1.2}{e^{\infty }}\\ \underset{t\to \infty }{\lim }1.2\left(1-e^{-8.2t}\right)=1.2+\frac{1.2}{\infty }\text{ (Since, }{e}^{\infty }=0\text{)}\\ \underset{t\to \infty }{\lim }1.2\left(1-e^{-8.2t}\right)=1.2+0\\ \underset{t\to \infty }{\lim }1.2\left(1-e^{-8.2t}\right)=1.2\end{array}$

Hence,

The terminal velocity of the raindrop is 1.2.

b.

To determine

To estimate : Graph $v\left(t\right)$ , and use the graph to estimate how long it takes for the velocity of the raindrop to reach 99% of its terminal velocity.

Answer to Problem 36E

The time at which the raindrop reach 99% of its terminal velocity is approximately 0.562.

Explanation of Solution

Given information:

The downward velocity of a falling raindrop at time t is modeled by the function

  $v\left(t\right)=1.2\left(1-e^{-8.2t}\right)$ and the terminal velocity is 1.2.

Calculation :

From part (a) terminal velocity is 1.2. Therefore,

  $\begin{array}{l}99\%\text{ of the termianal velocity}=1.2\times 0.99\\ 99\%\text{ of the termianal velocity}=1.188\end{array}$

The graph of $v\left(t\right)=1.2\left(1-e^{-8.2t}\right)$ is drawn below:

  

From the above velocity graph when $y=1.188$ , then $x\approx 0.562$

Hence,

The time at which the raindrop reach 99% of its terminal velocity is approximately 0.562.