Chapter 3.4, Problem 46E

To determine

To calculate:the growth rate of bacteria for three values of time.

Answer to Problem 46E

At $t=0$ population of bacteria’s is increasing at the rate of $\underset{\_}{10000\text{bacterias}\text{per}\text{hour}}$ , at $t=5$ , there is no increase or decrease of population and at $t=10$ population of bacteria’s is decreasing at the rate of $\underset{\_}{10000\text{bacterias}\text{per}\text{hour}}$ .

Explanation of Solution

Given information:

1. Function representing size of the population in terms of time $t$ in his as follows

$b\left(t\right)={10}^6+{10}^{4}t-{10}^3{t}^2$

2. Need to calculate growth rate at $t=0$ , $t=5$ and $t=10$

Formula used:

Growth rate of population at any time $t$ that is $r\left(t\right)$ having position of size of population as $b\left(t\right)$ is as follows.

  $r\left(t\right)=\frac{d}{dt}b\left(t\right)$

Calculation:

Function of size of population is $b\left(t\right)={10}^6+{10}^{4}t-{10}^3{t}^2$

Evaluate Growth rate function $r\left(t\right)$ by differentiating $b\left(t\right)$ with respect to $t$ .

  $r\left(t\right)=\frac{d}{dt}b\left(t\right)$

  $=\frac{d}{dt}\left({10}^6+{10}^{4}t-{10}^3{t}^2\right)$

  $=\frac{d}{dt}{10}^6+{10}^4\frac{d}{dt}t-{10}^3\frac{d}{dt}{t}^2$

  $=0+{10}^4-{10}^3\times 2t$

  $r\left(t\right)=10000-2000t$

Calculate Growth rate at $t=0$ that is $r\left(0\right)$

  $r\left(0\right)=10000-2000\times 0$

  $r\left(0\right)=10000$

Calculate Growth rate at $t=5$ that is $r\left(5\right)$

  $r\left(5\right)=10000-2000\times 5$

  $r\left(5\right)=0$

Calculate Growth rate at $t=10$ that is $r\left(10\right)$

  $r\left(10\right)=10000-2000\times 10$

  $r\left(10\right)=-10000$

Hence at $t=0$ population of bacteria’s is increasing at the rate of $\underset{\_}{10000\text{bacterias}\text{per}\text{hour}}$ , at $t=5$ , there is no increase or decrease of population and at $t=10$ population of bacteria’s is decreasing at the rate of $\underset{\_}{10000\text{bacterias}\text{per}\text{hour}}$ .