$\text { Suppose a population of bacteria is changing at a rate of } \frac { d p } { d t } = \frac { 4,000 } { 1 + 0.5 t } \text {, where } t \text { is }$ the time in days. The initial population (when $t = 0$ ) is 1,500 . Identify an equation that gives the population at any time $t$ and use it to find the population when $t = 5$ days. Round your answer to the nearest integer.
A) 12,783
B) 11,525
C) 13,533
D) 10,772
E) 11,522

Question

Quizplus · Accepted Answer

To find the population at any time $t$, we integrate the rate of change $\frac{dp}{dt} = \frac{4000}{1+0.5t}$ with respect to $t$. The integral of $\frac{4000}{1+0.5t}$ with respect to $t$ is $8000\ln(1+0.5t) + C$, where $C$ is the constant of integration. To find $C$, we use the initial condition that when $t=0$, $p=1500$. Substituting these values gives $1500 = 8000\ln(1+0(0.5)) + C$, which simplifies to $1500 = 8000\ln(1) + C$, and since $\ln(1) = 0$, we find $C = 1500$. Thus, the equation for the population at any time $t$ is $p(t) = 8000\ln(1+0.5t) + 1500$.To find the population when $t = 5$ days, we substitute $5$ for $t$ in the equation: $p(5) = 8000\ln(1+0.5(5)) + 1500 = 8000\ln(3.5) + 1500$. Calculating this gives a value close to 11,522, rounding to the nearest integer.

The Time in Days t=0t = 0t=0 ) Is 1,500 ttt

The Time in Days $t = 0$ ) Is 1,500 $t$