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?
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
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
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.
P = 84.3 e^(6.5 t) + 415.7
if t = 1
P(1) = 84.3 e^6.5 + 415.7
= 56,487
A. To determine the function P(t) which gives the population at time t, you first need to know that the rate of growth of the bacteria population is given by the differential equation:
dP/dt = r(t)
Integrating both sides of the equation will give you the population function:
∫dP = ∫r(t) dt
Assuming the initial population P(0) is 500, you can integrate the expression for r(t) to find P(t). In this case, r(t) = 548e^(6.5t). Thus, the integration becomes:
∫dP = ∫548e^(6.5t) dt
The integral of e^(6.5t) is:
(1/6.5)e^(6.5t) + C
where C is the constant of integration. Evaluating the integral with the given initial condition P(0) = 500:
500 = (1/6.5)e^0 + C
C = 500 - (1/6.5)
Therefore, the population function P(t) is:
P(t) = (1/6.5)e^(6.5t) + (500 - 1/6.5)
Simplifying the expression will give you the final form of the function P(t).
B. To determine how long it takes for the initial population to triple in size, you need to find the time t for which the population P(t) is three times the initial population P(0).
Set up the equation:
P(t) = 3P(0)
Substitute P(t) with its expression (found in part A):
(1/6.5)e^(6.5t) + (500 - 1/6.5) = 3(500)
Solve for t by isolating e^(6.5t):
(1/6.5)e^(6.5t) = 1500 - (1/6.5)
Now, take the natural logarithm (ln) of both sides:
ln[(1/6.5)e^(6.5t)] = ln[1500 - (1/6.5)]
Using logarithmic properties:
ln(1/6.5) + ln(e^(6.5t)) = ln(1500 - 1/6.5)
Simplifying:
ln(1/6.5) + 6.5t = ln(1500 - 1/6.5)
Now, isolate t:
6.5t = ln(1500 - 1/6.5) - ln(1/6.5)
Finally, solve for t:
t = [ln(1500 - 1/6.5) - ln(1/6.5)]/6.5
Evaluating this expression will give you the time it takes for the initial population to triple in size.