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]
