Question

At any time t ≥ 0, in hours, the rate of growth of a population of bacteria is given by dy/dt = 0.5y. Initially, there are 200 bacteria.
(a) Solve for y, the number of bacteria present, at any time t ≥ 0.

(b) Write and evaluate an expression to find the average number of bacteria in the population for 0 ≤ t ≤ 10.

(c) Write an expression that gives the average rate of bacteria growth over the first 10 hours of growth. Indicate units of measure.

Answer 1

Answer:

(a) The equation for the number of bacteria at time t ≥ 0 is $y = 200\cdot e^{0.5 \cdot t }$

(b) The average population in the bacteria for 0 ≤ t ≤ 10 is 29,482 bacteria

(c)  The growth rate of the bacteria in the first 10 hours is 14,841 Bacteria/hour

Step-by-step explanation:

(a) The given rate of growth of the bacteria population is;

$\dfrac{dy}{dt} = 0.5 \cdot y$

The initial amount of bacteria = 200

At time, t ≥ 0, we have;

$\dfrac{dy}{y} = 0.5 \cdot dt$

$\int\limits {\dfrac{dy}{y} } = \int\limits {0.5} \, dt$

ln(y) = 0.5·t + C

$y = e^{0.5 \cdot t + C}$

$y = C_1 \cdot e^{0.5 \cdot t }$

When, t = 0, y = 200, we have;

$200 = C_1 \cdot e^{0.5 \times 0 } = C_1$

C₁ = 200

The equation for the number of bacteria at time t ≥ 0 is therefore given as follows;

$y = 200\cdot e^{0.5 \cdot t }$

(b) The expression for the average number of bacteria in the population for 0 ≤ t ≤ 10 is given as follows;

$y = \int\limits^t_0 {0.5 \cdot y} \, dt$

$y = \int\limits^{10}_0 {0.5 \cdot 200\cdot e^{0.5 \cdot t }} \, dt = \left[200\cdot e^{0.5 \cdot t }\right]^{10}_0 \approx 2,9682.6 - 200 = 29,482.6$

The average population in the bacteria for 0 ≤ t ≤ 10 is therefore, y = 29,482

(c)  The average rate of growth after 10 hours of growth is given as follows;

$\dfrac{dy}{dt} = 0.5 \cdot y$

$\dfrac{dy}{dt} = 0.5 \cdot 200\cdot e^{0.5 \cdot 10 } = 14,841.32 \ Bacteria/hour$

The growth rate of the bacteria in the first 10 hours is 14,841 Bacteria/hour