The world population at the beginning of 1990 was 5.3 billion. Assume that the population continues to grow at the rate of approximately 2%/year and find the function Q(t) that expresses the world population (in billions) as a function of time t (in years), with t = 0 corresponding to the beginning of 1990.(a) If the world population continues to grow at approximately 2%/year, find the length of time t0 required for the population to triple in size.
t0 = 1 yr
(b) Using the time t0 found in part (a), what would be the world population if the growth rate were reduced to 1.1%/yr?

First of all, note that when you increase certain amount by x percent, you only have to multiply that amount for a decimal number following this rule:

[tex]New=Original(1+\frac{x}{100})[/tex]

For example, if you increase 5.3 billion by 20%, then:

[tex]New=5.3billion(1+\frac{20}{100})\\ New=5.3billion(1.2)[/tex]

[tex]New=6.36 billion[/tex]

In the problem you need to increase the population by 2%/year, then after one year you'll have:

[tex]Q=5.3(1+\frac{2}{100})billions\\Q=5.3(1.02) billions\\Q=5.406 billions[/tex]

Note that this last quantity will increase 2% in the second year, then:

[tex]Q(2)=5.3(1.02)(1.02) billions\\Q(2)=5.3(1.02)^{2} billions[/tex]

In the third year the population will be:

[tex]Q(3)=5.3 (1.02)(1.02)(1.02)\\Q(3)=5.3(1.02)^{3} billions[/tex]

Then, the function Q(t) that expresses the world population (in billions) is given by:

[tex]Q(t)=5.3 (1.02)^{t}[/tex]

where t=0 corresponds to the beginning of 1990 (5.3 billions).

a)The time necessary for the population to triple in size is given by:

[tex]3(5.3)=5.3(1.02)^{t}\\ 3=1.02^{t}[/tex]

To solve for t, you need to apply the natural logarithm or the common logarithm in both sides of the equation:

[tex]ln(3)=ln[(1.02)^{t}]\\ln(3)=t(ln(1.02))\\t=\frac{ln(3)}{ln(1.02)}\\ t=55.48 years[/tex]

Then, the time required to the population to triple in size is 55.48 years.

b)If the growth rate were reduced to 1.1%/year, the function would be:

[tex]Q(t)=5.3(1+\frac{1.1}{100} )^{t}\\Q(t)=5.3(1.011)^{t}[/tex]

The world population at the time obtained in a) would be:

[tex]Q(55.48years)=5.3(1.011)^{55.48}\\ Q(55.48years)=9.725 billions[/tex]

eudora eudora · Answer 2 · 2019-11-12T09:40:20+01:00

Answer:

a). 55.47 years

b). population = 9.72 billion

Step-by-step explanation:

Since population growth is an exponential phenomenon, population after t years will be represented by,

[tex]P_{t}=P_{0}(1+0.02)^{t}[/tex]

[Since rate of population growth is 2%]

[tex]P_{t}=P_{0}(1.02)^{t}[/tex]

Where [tex]P_{t}[/tex] = Population after t years

[tex]P_{0}[/tex] = Population at t = 0 years

A). Now we have to find the time by which the population will triple in size.

Therefore, for [tex]P_{t}=3P_{0}[/tex]

[tex]3P_{0}=P_{0}(1.02)^{t}[/tex]

[tex]3=(1.02)^{t}[/tex]

By taking log on both the sides of the equation.

[tex]log3=log(1.02)^{t}[/tex]

0.4771212 = tlog(1.02)

0.4771212 = t(0.0086)

t = [tex]\frac{0.4771212}{0.0086}[/tex]

= 55.47 years

B). If growth rate was reduced to 1.1% per year then we have to find the world population after t = 55.47 years and [tex]P_{0}[/tex] = 5.3 billion

[tex]P_{t}=5.3(1+0.011)^{55.47}[/tex]

[tex]P_{55.47}=5.3(1.011)^{55.47}[/tex]

= 5.3×1.8346

= 9.72 billion