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

(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

Explanation:

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

`(dy)/(dt) = 0.5 \cdot y`

The initial amount of bacteria = 200

At time, t ≥ 0, we have;

`(dy)/(y) = 0.5 \cdot dt`

`\int\limits {(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 * 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;

`(dy)/(dt) = 0.5 \cdot y`

`(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

Answer 2

a. The equation for the number of bacteria at any time t is: y = 200e^0.5t

b. The average number of bacteria in the population for 0 ≤ t ≤ 10 is approximately 1653.84

c. At t = 10, the average rate of bacteria growth is approximately 110.52 bacteria per hour.

(a) The rate of growth of the population of bacteria is given by the equation dy/dt = 0.5y.

We can solve this differential equation to find y, the number of bacteria present at any time t.

To do this, we separate the variables and integrate:

dy/y = 0.5dt

ln|y| = 0.5t + C

By exponentiating both sides, we get:

|y| = e^(0.5t + C) = Ce^0.5t

Since initially there are 200 bacteria, we can substitute y = 200 and solve for C:

|200| = C * e^0

C = 200

Therefore, the equation for the number of bacteria at any time t is:

y = 200e^0.5t

(b) To find the average number of bacteria for 0 ≤ t ≤ 10, we can integrate the equation y = 200e^0.5t over this interval and divide by the length of the interval:

Average number of bacteria = (1/10) * ∫010200e^0.5tdt

= (1/10) * [400e^0.5t] from 0 to 10

= (1/10) * (400e^5 - 400)

≈ 1653.84

(c) The average rate of bacteria growth over the first 10 hours can be calculated by finding the derivative of the average number of bacteria with respect to time:

Average rate of bacteria growth = d/dt [(1/10) * (400e^0.5t)]

= (1/10) * (200e^0.5t)

At t = 10, the average rate of bacteria growth is approximately 110.52 bacteria per hour.