Question

I need some help with this Calculus problem.

A bacteria population starts with 500 bacteria and grows at a rate of r(t) = 548e^6.5t bacteria per hour.

A. Determine the function P(t) which gives the population at time t.

C. How long does it take the initial population to triple in size?

Answer 1

dP/dt = 548 e^(6.5 t) ????
if I understand you that means
P = (548/6.5) e^(6.5 t) + constant
but we know when t = 0, P = 500
so
500 = 84.3 + c
c = 415.7
so
P = 84.3 e^(6.5 t) + 415.7

when is it 1500?

1500 - 415.7 = 84.3 e^(6.5 t)
12.86 = e^6.5 t
ln 12.86 = 6.5 t

Answer 2

Since r(t) is the rate,
r(t) = 548e^6.5t is the derivative of P(t) , then
P(t) = 548/6.5 e^6.5t + c , where c is a constant
when t = 0, P(0) = 500
500 = 548/6.5 (e^0) + c , but e^0 = 1 , so ....
c = appr 415.69

P(t) = (548/6.5) e^6.5t + 415.69

Then when is P(t) = 1500 ?

1500 = (548/6.5) e^6.5t + 415.69
e^6.5t = 12.86131...
6.5t = ln 12.86131...
t = .393 hrs or appr 23.6 minutes

Answer 3

Awesome, thank you! There is one more part to this question I need help with. I tried figuring it out but the number of bacteria I calculated didn't make any sense. It says to determine the number of bacteria after one hour.

Answer 4

P = 84.3 e^(6.5 t) + 415.7
if t = 1
P(1) = 84.3 e^6.5 + 415.7
= 56,487