The population of a certain species of insect is given by a differentiable function P, where P(t) is the number of insects in the population, in millions, at time t, where t is measured in days. When the environmental conditions are right, the population increases with respect to time at a rate that is directly proportional to the population. Starting August 15, the conditions were favorable and the population began increasing. On August 20, five days later, there were an estimated 10 million insects and the population was increasing at a rate of 2 million insects per day. Which of the following is a differential equation that models this situation?

a. P=2(t−5)+10
b. dP/dt=2/5t
c. dP/dt=1/5P
d. dP/dt=5P

Question

Answered

Answer :

OLADAYOADEMOSUGO OLADAYOADEMOSUGO · Answer 1

Answer:

c. dP/dt = (1/5)P

Step-by-step explanation:

Given that the rate of change of population with respect to time dP/dt is directly proportional to the population, P, we have

dP/dt ∝ P

dP/dt = kP

Given that dP/dt = 2 million insects per day and P = 10 million insects after 5 days, So,

2 = k × 10

k = 2/10

k = 1/5

So, dP/dt = kP

dP/dt = (1/5)P

Answer Link

JAINVEENAMRATAGO JAINVEENAMRATAGO · Answer 2

The option C is correct

The rate of change of a function with respect to the given variable.

The rate of change of population with respect to time is directly proportional to the population P. We have

[tex]\dfrac{\mathrm{d} P}{\mathrm{d} t} \propto P\\\\\dfrac{\mathrm{d} P}{\mathrm{d} t} = kP[/tex]

[tex]\dfrac{\mathrm{d} P}{\mathrm{d} t}[/tex] is 2 million insects per day and P = 10 million insects after 5 days. So

[tex]\begin{aligned} \dfrac{\mathrm{d} P}{\mathrm{d} t} &= kP\\2 &= 10k\\k &= \dfrac{1}{5} \\\end{aligned}[/tex]

Then

[tex]\dfrac{\mathrm{d} P}{\mathrm{d} t} = kP\\\\ \dfrac{\mathrm{d} P}{\mathrm{d} t} = \dfrac{1}{5} P[/tex]

Thus, option C is correct.

More about the Differentiation link is given below.

https://brainly.com/question/24062595