The size P of a certain insect population at time t (in days) obeys the function P(t)=900e^0.02t A. Determine the number of insects at t=0 days. B. What is the growth rate of the insect population? C. What is the population after 10 days? D. When will the insect population reach 1350? E. When will the insect population double?

A. To determine the number of insects at t = 0 days, we substitute t = 0 into P(t) = 900e^{0.02t} :

P(0) = 900e^{0.02 \cdot 0} = 900e^0 = 900 \times 1 = 900

So, the number of insects at t = 0 days is 900 .

B. The growth rate in the function P(t) = 900e^{0.02t} is given by the exponent's coefficient, 0.02 . This corresponds to a growth rate of 2\% .

C. To determine the population after 10 days, we substitute t = 10 into P(t) = 900e^{0.02t} :

P(10) = 900e^{0.02 \cdot 10} = 900e^{0.2}

Using a calculator to evaluate e^{0.2} \approx 1.2214 :

P(10)=900\times1.2214\approx1099.26

So, the population after 10 days is approximately 1099.26 insects.

D. To find when the insect population will reach 1350 , we set P(t) = 1350 and solve for t :

1350 = 900e^{0.02t}

Dividing both sides by 900 gives:

1.5 = e^{0.02t}

Taking the natural logarithm of both sides:

\ln(1.5) = 0.02t

So:

t = \frac{\ln(1.5)}{0.02} \approx \frac{0.4055}{0.02} \approx 20.27

Therefore, the insect population will reach 1350 in approximately 20.27 days.

E. To find when the insect population will double, we set P(t) = 2 \times 900 = 1800 and solve for t :

1800 = 900e^{0.02t}

Dividing both sides by 900 gives:

2 = e^{0.02t}

Taking the natural logarithm of both sides:

\ln(2) = 0.02t

So:

t = \frac{\ln(2)}{0.02} \approx \frac{0.6931}{0.02} \approx 34.66

Therefore, the insect population will double in approximately 34.66 days.