Question

(15 points) Problem (7) A culture of living cells in a lab has a population of 100 cells when nutrients are added at time t = 0. Suppose the population P(t) (in cells) increases at a rate given by P' (t) = 90e-0.16 (in cells per hour) Find the total number of cells after 1 hour.

Answer 1

we have

$P'(t)=90e^{-0.1t}$

$\int P'(t)=\int 90e^{-0.1t}dt$

$P(t)=\frac{90e^{-0.1t}}{-0.1}+C$

$P(t)=-900e^{-0.1t}+C$ ......................1)

put P(t) = 100, t =0,

$100=-900e^{-0.1(0)}+C$

$100=-900(1)+C$

$100+900=C$

$C=1000$

put C = 1000 in equation 1),

$P(t)=-900e^{-0.1t}+1000$

total number of cell after 1 hour(t = 1),

$P(1)=-900e^{-0.1(1)}+1000$

$P(1)=-\frac{900}{e^{0.1}}+1000$

total number of cell after 1 hour is 185.64632 cells.