Question

Sterile fruit flies are used in an experiment where the proportion that survives at least t days is given by e^−0.15t. If the experiment begins with 400 fruit flies, and flies are added at the rate of 6 per hour, how many flies are present 15 days after the start of the experiment? (Round your answer to the nearest integer.)

Answer 1

Answer:

901 flies

Step-by-step explanation:

Let the total population of flies is P,

While $P_s$ and $P_i$ represent the number of flies that survived and that are initially respectively.

Here,

$\frac{P_s}{P_i}=e^{-0.15t}$

$\implies P_s = P_i e^{-0.15t}$

Differentiating w. r. t. t ( time ),

$\frac{dP_s}{dt}=-0.15 P_i e^{-0.15t}=-0.15 P_s-----(1)$

Now, if 6 flies are added per hour,

Then the number of flies added per day = 24 × 6 = 144 ( ∵ 1 day = 24 hours),

So,

$\frac{dP}{dt}=144 + \frac{dP_s}{dt}$

$\frac{dP}{dt}=144 -0.15 P_s$

$\implies \frac{dP}{dt}=144 -0.15 P$

When P = $P_s$ = $P_i$

$\implies \frac{dP}{144 -0.15 P}=dt$

∵ if t = 0, P = 400,

$\int_{400}^{P}\frac{dP}{144 -0.15 P}=\int_{t=0}^{15}dt$

$[\frac{\ln(144-0.15P)}{-0.15}}]_{400}^{P} = [t]_{0}^{15}$

$\frac{\ln (144-0.15P)- \ln(144-60)}{-0.15}}=15$

$\ln (144-0.15P) - \ln(84) = -2.25$

$-\ln (144-0.15P) + \ln(84) = 2.25$

$\ln (\frac{84}{144-0.15P})= 2.25$

$\frac{84}{144-0.15P} = e^{2.25}$

$\frac{84}{e^{2.25}} =144-0.15P$

$8.853 = 144 - 0.15P$

$0.15P = 144 - 8.853$

$\implies P = \frac{135.147}{0.15}=\frac{135.147}{0.15}\approx 901$

i.e. 901 flies would present.