I just need help with one question.
A bacteria population starts with 500 bacteria and grows at a rate of r(t) = 548e^(6.5t) bacteria per hour.
Determine the number of bacteria after 1 hour.
Thank you!
Please check where I did this below
Calculus - Damon today at 3:57pm
P = 84.3 e^(6.5 t) + 415.7
if t = 1
P(1) = 84.3 e^6.5 + 415.7
= 56,487
To determine the number of bacteria after 1 hour, we need to find the value of the population function at t = 1.
The population function is given by:
P(t) = P(0) + ∫[0 to t] r(u) du
where P(t) is the population at time t, P(0) is the initial population, r(u) is the rate function, and ∫[0 to t] r(u) du represents the integral of the rate function from 0 to t.
In this case, the initial population, P(0), is given as 500. The rate function, r(t), is given as 548e^(6.5t).
Substituting these values into the population function, we have:
P(t) = 500 + ∫[0 to t] 548e^(6.5u) du
To find the value of the population after 1 hour, we need to evaluate the definite integral from 0 to 1:
P(1) = 500 + ∫[0 to 1] 548e^(6.5u) du
To evaluate this integral, we can use the power rule of integration:
∫e^(ax) dx = (1/a)e^(ax) + C,
where a is a constant and C represents the constant of integration.
In this case, the integral becomes:
P(1) = 500 + (548/6.5)e^(6.5u) | [0 to 1]
To evaluate the integral, we plug in the upper limit (1) and subtract the value at the lower limit (0):
P(1) = 500 + (548/6.5)e^(6.5 * 1) - (548/6.5)e^(6.5 * 0)
P(1) = 500 + (548/6.5)e^(6.5) - (548/6.5)e^0
Since e^0 = 1, the expression simplifies to:
P(1) = 500 + (548/6.5)e^(6.5) - (548/6.5)
Now we can calculate the value of P(1).
P(1) = 500 + (548/6.5)(e^(6.5) - 1)
Using a calculator, we find:
P(1) ≈ 500 + (548/6.5)(1388.73425 - 1)
P(1) ≈ 500 + (548/6.5)(1387.73425)
P(1) ≈ 500 + (848/6.5)(1387.73425)
P(1) ≈ 500 + 848(213.36065)
P(1) ≈ 500 + 180694.7304
P(1) ≈ 181194.7304
Therefore, the number of bacteria after 1 hour is approximately 181,194.