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:

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

(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;

[tex]\dfrac{dy}{dt} = 0.5 \cdot y[/tex]

The initial amount of bacteria = 200

At time, t ≥ 0, we have;

[tex]\dfrac{dy}{y} = 0.5 \cdot dt[/tex]

[tex]\int\limits {\dfrac{dy}{y} } = \int\limits {0.5} \, dt[/tex]

ln(y) = 0.5·t + C

[tex]y = e^{0.5 \cdot t + C}[/tex]

[tex]y = C_1 \cdot e^{0.5 \cdot t }[/tex]

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

[tex]200 = C_1 \cdot e^{0.5 \times 0 } = C_1[/tex]

C₁ = 200

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

[tex]y = 200\cdot e^{0.5 \cdot t }[/tex]

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

[tex]y = \int\limits^t_0 {0.5 \cdot y} \, dt[/tex]

[tex]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[/tex]

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;

[tex]\dfrac{dy}{dt} = 0.5 \cdot y[/tex]

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

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

Population functions can either represent growth or decays.

The given parameter is:

[tex]\frac{dy}{dt}= 0.5y[/tex]

(a) Solve for y

We have:

[tex]\frac{dy}{dt}= 0.5y[/tex]

Integrate both sides

[tex]\int \frac{dy}{dt}= \int 0.5y[/tex]

Divide both sides by y, and multiply both sides by dt

[tex]\int \frac{dy}{y}= \int 0.5\ dt[/tex]

Integrate both sides of the equation

[tex]\ln(y) = 0.5t +c[/tex]

Take the exponent of both sides

[tex]y = e^{0.5t +c}[/tex]

This gives

[tex]y = ce^{0.5t}[/tex]

When t = 0, and y = 200.

So, we have:

[tex]200 = ce^{0.5 \times 0}[/tex]

[tex]200 = ce^{0}[/tex]

Evaluate the exponent

[tex]200 = c[/tex]

Rewrite as:

[tex]c = 200[/tex]

Substitute 200 for c in [tex]y = ce^{0.5t}[/tex]

[tex]y = 200e^{0.5t}[/tex]

Hence, the equation for y is [tex]y = 200e^{0.5t}[/tex]

(b) The average number of bacterial for 0 ≤ t ≤ 10.

The average value of the function is calculated as:

[tex]y' = y(10) - y(0)[/tex]

Calculate y(0) and y(10)

[tex]y = 200e^{0.5t}[/tex]

[tex]y(0) = 200e^{0.5(0)}= 200[/tex]

[tex]y(10) = 200e^{0.5\times 10} =29682.6318205[/tex]

So, we have

[tex]y' = y(10) - y(0)[/tex]

[tex]y' = 29682.6318205 - 200[/tex]

[tex]y' = 29482.6318205}[/tex]

[tex]y' = 29483[/tex]

Hence, The average number of bacterial for 0 ≤ t ≤ 10 is 29483

(c) The average rate over the first 10 hours of growth

We have:

[tex]\frac{dy}{dt}= 0.5y[/tex]

Substitute [tex]y = 200e^{0.5t}[/tex]

[tex]\frac{dy}{dt}= 0.5 \times 200e^{0.5t}[/tex]

[tex]\frac{dy}{dt}= 100e^{0.5t}[/tex]

When t = 10, we have:

[tex]\frac{dy}{dt}= 100e^{0.5 \times 10}[/tex]

[tex]\frac{dy}{dt}= 100e^{5}[/tex]

[tex]\frac{dy}{dt}= 14841[/tex]

Hence, the average rate over the first 10 hours of growth is 14841

Read more about population growth functions at:

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