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In 2010, the world's population reached 6.91 billion and was increasing at a rate of 1.1% per year. Assume that this growth rate remains constant. (In fact, the growth rate has decreased since 1987.)

(a)

Write a formula for the world population (in billions) as a function of the number of years since 2010.

(b)

Estimate the population of the world in the year 2020.

Question

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In 2010, the world's population reached 6.91 billion and was increasing at a rate of 1.1% per year. Assume that this growth rate remains constant. (In fact, the growth rate has decreased since 1987.)


(a)  


Write a formula for the world population (in billions) as a function of the number of years since 2010.


(b)  


Estimate the population of the world in the year 2020.

anonymous · Answer

(a) Okay so the question straight away is a reminder of compound interest: for example imagine a compound interest of 2% on $90. After the first year (x=1) there will be $$90+90\times \frac{ 2 }{ 100 }$$ ie (if you factorize)$$90\times1.02$$
After the 2nd year (x=2) you will have the same interest rate, but on what you had after the first year:
$$(90\times1.02)\times1.02$$ which is $$90\times1.02^2$$
So we understand that there an underlying formula here$$PV\times(1+r)^n=FV$$:where PV is the present value, r the interest rate, n the number of years for which the interest will be applied (number of periods) and FV our final value.

With this in mind, we can easily answer the question
$$f(x)=6.91\times10^9\times1.011^x$$where x is the number of years since 2010

(b) Simply input the numbers into the formula, and we get:
$$f(20)=8600053623.63$$
Now you can't have .63 of a human, so the rational thing to do is to round it. Your final answer is then 8600053624.