Question

The number of bacteria in a culture is growing at a rate of At 3000e^2t/5 per unit of time t. At t=0,   the number of bacter ia present was 7500. Find the number present at t=5

           (A) 1200e^2 (B)3000^2 (C) 7500e^2 (D) 7500e^2 (E) 15000/7 e^7

Answer 1

Given ----> The number of bacteria in a culture is growing at a rate of At 3000e^2t/5 per unit of time t.

dB / dt = 3000 e^(2/5 t)

=> dB = 3000 e^(2/5 t) dt

=> B = ∫(3000 e^(2/5 t) dt)

=> B = (5/2) 3000 e^(2/5 t) + C

=> B = 7500 e^(2/5 t) + C

So the given equation is -----> B = 7500 e^(2/5 t) + C

At t = 0,   the number of bacter ia present was 7500

So,

7500 = 7500 + C

C = 0

At t = 5 The number of bacteria present

B = 7500 e^(2/5 *5) = 7500 e^2