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In 2012, the world population was 7.02 billion. The birth rate had stabilized to 134 million per year and is projected to remain constant. The death rate is projected to increase from 56 million per year in 2012 to 80 million per year in 2040 and to continue increasing at the same rate.
(a) Assuming the death rate increases linearly, write a differential equation for P(t), the world population in billions t years from 2012.
(b) Solve the differential equation.
(c) Find the population predicted for 2050.

Sagot :

MrRoyal

Answer:

[tex](a)\ \frac{dP}{dt} = 0.078- 0.000857t[/tex]

[tex](b)\ P = 0.078t- 0.0004285t^2 + 7.02[/tex]

[tex](c)\ 9.37\ billion[/tex]

Step-by-step explanation:

Given

In 2012

[tex]P(t) = 7.02\ billion[/tex]

[tex]R_1 = 134\ million/year[/tex] --- Birth rate

[tex]R_2 = 56\ to\ 80[/tex] -- Death rate from '12 to '40

Solving (a): Equation for population from 2012

In 2012, we have the addition in population to be:

[tex]\triangle P= Birth - Death[/tex]

[tex]\triangle P= 134 - 56[/tex]

[tex]\triangle P = 78[/tex] million

From 2012 to 2040, we have the death rate per year to be:

[tex]Rate/yr = \frac{80 - 56}{2040 - 2012}[/tex]

[tex]Rate/yr = \frac{24}{28}[/tex]

[tex]Rate/yr = \frac{6}{7}[/tex] million/year

So, the differential equation is:

[tex]\frac{dP}{dt} = \triangle P - Rate/yr * t[/tex] where t represents time

[tex]\frac{dP}{dt} = 78 - \frac{6}{7} * t[/tex]

[tex]\frac{dP}{dt} = 78 - \frac{6t}{7}[/tex]

Express in millions

[tex]\frac{dP}{dt} = \frac{78}{1000} - \frac{6t}{7000}[/tex]

[tex]\frac{dP}{dt} = 0.078- 0.000857t[/tex]

Solving (b): The solution to the equation in (a)

[tex]\frac{dP}{dt} = 0.078- 0.000857t[/tex]

Make dP the subject

[tex]dP = (0.078- 0.000857t)dt[/tex]

Integrate both sides

[tex]\int dP = \int (0.078- 0.000857t)dt[/tex]

[tex]P = \int (0.078- 0.000857t)dt[/tex]

This gives:

[tex]P = 0.078t- \frac{0.000857t^2}{2} + c[/tex]

[tex]P = 0.078t- 0.0004285t^2 + c[/tex]

Solve for c.

When t = 0; P = 7.02

So, we have:

[tex]7.02 = 0.078*0- 0.0004285*0^2 + c[/tex]

[tex]7.02 =0-0 + c[/tex]

[tex]7.02 =c[/tex]

[tex]c = 7.02[/tex]

So:

[tex]P = 0.078t- 0.0004285t^2 + c[/tex]

[tex]P = 0.078t- 0.0004285t^2 + 7.02[/tex]

Solving (c): The population in 2050

We have:

[tex]P = 0.078t- 0.0004285t^2 + 7.02[/tex]

In 2050, the value of t is:

[tex]t = 2050 - 2012[/tex]

[tex]t = 38[/tex]

So, the expression becomes

[tex]P = 0.078*38- 0.0004285*38^2 + 7.02[/tex]

[tex]P \approx 9.37\ billion[/tex]

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