Chapter 5.8, Problem 2AYU

Growth of Bacteria The number $N$ of bacteria present in a culture at time $t$ (in hours) obeys the model $N(t)={\text{1000e}}^{0.01t}$.

(a) Determine the number of bacteria at $t=0$ hours.

(b) What is the growth rate of the bacteria?

(c) Graph the function using a graphing utility.

(d) What is the population after 4 hours?

(e) When will the number of bacteria reach 1700?

(f) When will the number of bacteria double?

To determine

To find:

a. Determine the number of bacteria at $t=0$ hours.

Answer to Problem 2AYU

Solution:

a. 1000

Explanation of Solution

Given:

$N\left(t\right)=1000e^{0.01t}$

Calculation:

Let us consider 1000 bacteria initially and the number of bacteria after $t$ hours will be

$N\left(t\right)=1000e^{0.01t}$

a. Let us consider the initial state. Then

$N\left(t\right)=1000e^{0.01t}$   $\left(t=0\right)$

$N\left(t\right)=1000e^{0.01\left(0\right)}$

$N\left(t\right)=1000$

To determine

To find:

b. What is the growth rate of the bacteria?

Answer to Problem 2AYU

Solution:

b. $1\%$

Explanation of Solution

Given:

$N\left(t\right)=1000e^{0.01t}$

Calculation:

Let us consider 1000 bacteria initially and the number of bacteria after $t$ hours will be

$N\left(t\right)=1000e^{0.01t}$

b. Let us compare the above model with $N\left(t\right)=1000e^{kt }$ where $k$ defines the growth rate of the model. Then $k=0.01 =1\%$.

To determine

To find:

c. Graph the function using a graphing utility.

Answer to Problem 2AYU

Solution:

c.

Explanation of Solution

Given:

$N\left(t\right)=1000e^{0.01t}$

Calculation:

Let us consider 1000 bacteria initially and the number of bacteria after $t$ hours will be

$N\left(t\right)=1000e^{0.01t}$

c. Graph:

To determine

To find:

d. What is the population after 4 hours?

Answer to Problem 2AYU

Solution:

d. 1041

Explanation of Solution

Given:

$N\left(t\right)=1000e^{0.01t}$

Calculation:

Let us consider 1000 bacteria initially and the number of bacteria after $t$ hours will be

$N\left(t\right)=1000e^{0.01t}$

d. Let us consider $t=4$ hours. Then

$N\left(t\right)=1000e^{0.01t}$

$N\left(t\right)=1000e^{0.01\left(4\right)}$

$N\left(t\right)=1000e^{0.01t}$

$=1000\left(1.0408\right)$

$=1041$

To determine

To find:

e. When will the number of bacteria reach 1700?

Answer to Problem 2AYU

Solution:

e. $53.06$

Explanation of Solution

Given:

$N\left(t\right)=1000e^{0.01t}$

Calculation:

Let us consider 1000 bacteria initially and the number of bacteria after $t$ hours will be

$N\left(t\right)=1000e^{0.01t}$

e. Let us consider $N\left(t\right)=1700$. Then

$N\left(t\right)=1000e^{0.01t}$

$1700=1000e^{0.01t}$

$1.7= e^{0.01t}$

Taking log on both sides we get,

$\text{ln}\left(1.7\right)=0.01 t$

$t=\frac{0.5306}{0.01}$

$t=53.06$ hours

To determine

To find:

f. When will the number of bacteria double?

Answer to Problem 2AYU

Solution:

f. $69.31$

Explanation of Solution

Given:

$N\left(t\right)=1000e^{0.01t}$

Calculation:

Let us consider 1000 bacteria initially and the number of bacteria after $t$ hours will be

$N\left(t\right)=1000e^{0.01t}$

f. Let us consider $N\left(t\right)=2000$.

$N\left(t\right)=1000e^{0.01t}$

$2000=1000e^{0.01t}$

$2= e^{0.01t}$

Taking log on both sides we get,

$\text{ln}2= \text{ln}{e}^{0.01t}$

$t= \frac{0.6931}{0.01}=69.31$ hours