Question

the number of bacteria in a culture is growing at a rate of 1500e^(3t/4) per unit of time t. at t=0, the number of bacteria present was 2000. find the number present at t=4.

Answer 1

\[P=\int\limits_{}^{}1500e^{\frac{3}{4}t}dt=1500\frac{4}{3}e^{\frac{3}{4}t}+C\]

at t=0 P=2000
2000=2000e^0+C
C=0

P(t)=2000e^{3t/4}
P(4)=2000e^3
P(4)=40171.1