Chapter 5.8, Problem 1AYU

Growth of an Insect Population The size $P$ of a certain insect population at time $t$ (in days) obeys the model $P\left(t\right)=500e^{0.02t}$.

(a) Determine the number of insects at $t=0$ days.

(b) What is the growth rate of the insect population?

(c) Graph the function using a graphing utility.

(d) What is the population after 10 days?

(e) When will the insect population reach 800?

(f) When will the insect population double?

To determine

To find:

a. Determine the number of insects at $t=0$ days.

Answer to Problem 1AYU

Solution:

a. 500

Explanation of Solution

Given:

$P\left(t\right)=500e^{0.02t}$

Calculation:

a. $P\left(0\right)=500e^{0.02\left(0\right)}=500e^0=500$

To determine

To find:

b. What is the growth rate of the insect population?

Answer to Problem 1AYU

Solution:

b. $2\%$

Explanation of Solution

Given:

$P\left(t\right)=500e^{0.02t}$

Calculation:

b. Growth rate is $0.02$ or $2\%$.

To determine

To find:

c. Graph the function using graphing utility.

Answer to Problem 1AYU

Solution:

c.

Explanation of Solution

Given:

$P\left(t\right)=500e^{0.02t}$

Calculation:

c. Graph:

To determine

To find:

d. What is the population after 10 days?

Answer to Problem 1AYU

Solution:

d. 611

Explanation of Solution

Given:

$P\left(t\right)=500e^{0.02t}$

Calculation:

d. After 10 days $= 500e^{0.2} =500\left(1.221\right) = 610.5 \simeq 611\text{days}$.

To determine

To find:

e. When will the insect population reach 800?

Answer to Problem 1AYU

Solution:

e. $23.5$

Explanation of Solution

Given:

$P\left(t\right)=500e^{0.02t}$

Calculation:

e. $800=500e^{0.02t}$

$1.6=e^{0.02t}$

Taking log on both sides.

$\text{ln}1.6 =\text{ln}{e}^{0.02t}$

$0.470=0.02t$

$t=\frac{0.470}{0.02}=23.5$

To determine

To find:

f. When will the insect population double?

Answer to Problem 1AYU

Solution:

f. $34.7$

Explanation of Solution

Given:

$P\left(t\right)=500e^{0.02t}$

Calculation:

f. $1000=500e^{0.02t}$

$2= e^{0.02t}$

Taking log on both sides.

$\ln 2 =\text{ln}{e}^{0.02t}$

$0.6931=0.02t$

$t=\frac{0.6931}{0.02}=34.7$