Chapter 5: Q 5 (page 335)

Growth of a Colony of Mosquitoes
The population of a colony of mosquitoes obeys the law of uninhibited growth.
(a) If N is the population of the colony and t is the time in days, express N as a function of t.
(b) If there are 1000 mosquitoes initially and there are 1800 after 1 day, what is the size of the colony after 3 days?
(c) How long is it until there are 10,000 mosquitoes?

Short Answer

Expert verified

(a) N as a function of t can be expressed as $N (t) = N_{0} e^{k t}$

(b) Size of the colony after 3 days will be 5832

Step by step solution

Step 1. Given information

The population of a colony of mosquitoes obeys the law of uninhibited growth.

Part (a) of step 1. N as a function of t.

If N is the population of the colony and t is the time in days, then N as a function of t can be expressed as $N (t) = N_{0} e^{k t}$

Part (b) of Step 1. Size of the colony after 3 days

Substitute $N (t) = 1800, N_{0} = 1000, t = 1$ , in the function

role="math" localid="1647280123581" $N (t) = N_{0} e^{k t} 1800 = 1000 e^{k \times 1} 1.8 = e^{k} \ln (1.8) = \ln (e^{k}) \ln (1.8) = k k = 0.58779$

Now substitute $N_{0} = 1000, k = 0.58779, t = 3$ in the function

$N (t) = N_{0} e^{k t} N (t) = 1000 e^{0.58779 \times 3} N (t) = 1000 \times 5.832 N (t) = 5832$

The population of the colony after 3 days will be 5832.

Part (c) of Step 1. Time required for the population to be 10,000

Substitute $N (t) = 10000, N_{0} = 1000$ in the function

role="math" localid="1647280306723" $N (t) = N_{0} e^{0.58779 t} 10000 = 1000 e^{0.58779 t} 10 = e^{0.58779 t} \ln (10) = \ln (e^{0.58779 t}) \ln (10) = 0.58779 t t = \frac{\ln (10)}{0.58779} t = 3.9$

The time required for the population to be 10,000 is 3.9 days

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Short Answer

Step by step solution

Step 1. Given information

Part (a) of step 1. N as a function of t.

Part (b) of Step 1. Size of the colony after 3 days

Part (c) of Step 1. Time required for the population to be 10,000

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