Question

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?

Answer 1

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,

$P_{t}=P_{0}(1+0.02)^{t}$

[Since rate of population growth is 2%]

$P_{t}=P_{0}(1.02)^{t}$

Where $P_{t}$ = Population after t years

$P_{0}$ = Population at t = 0 years

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

Therefore, for $P_{t}=3P_{0}$

$3P_{0}=P_{0}(1.02)^{t}$

$3=(1.02)^{t}$

By taking log on both the sides of the equation.

$log3=log(1.02)^{t}$

0.4771212 = tlog(1.02)

0.4771212 = t(0.0086)

t = $\frac{0.4771212}{0.0086}$

 = 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 $P_{0}$ = 5.3 billion

$P_{t}=5.3(1+0.011)^{55.47}$

$P_{55.47}=5.3(1.011)^{55.47}$              

              = 5.3×1.8346

              = 9.72 billion

Answer 2

Answer:

a)55.48 years

b)9.725 billions

Step-by-step explanation:

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:

$New=Original(1+\frac{x}{100})$

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

$New=5.3billion(1+\frac{20}{100})\\ New=5.3billion(1.2)$

$New=6.36 billion$

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

$Q=5.3(1+\frac{2}{100})billions\\Q=5.3(1.02) billions\\Q=5.406 billions$

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

$Q(2)=5.3(1.02)(1.02) billions\\Q(2)=5.3(1.02)^{2} billions$

In the third year the population will be:

$Q(3)=5.3 (1.02)(1.02)(1.02)\\Q(3)=5.3(1.02)^{3} billions$

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

$Q(t)=5.3 (1.02)^{t}$

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:

$3(5.3)=5.3(1.02)^{t}\\ 3=1.02^{t}$

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

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

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:

$Q(t)=5.3(1+\frac{1.1}{100} )^{t}\\Q(t)=5.3(1.011)^{t}$

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

$Q(55.48years)=5.3(1.011)^{55.48}\\ Q(55.48years)=9.725 billions$