the population of brazil increased form 122 million to 1980 to 158 million in 1992. At what continuous rate was the population growing during this period?

Question

b) assuming that brazils population continues to grow at this rate, when will it reach 250 million?

Answer 1

This is an exponential growth problem. Here's how to set it up. Remember that the formula for exponential growth is y = a(1 + r)^x, where y is the population, a is the starting population, r is the growth rate we want to find, and x is the number of years.

158,000,000 = 122,000,000 * (1 + r)¹²

1.29508197 = (1 + r)¹²

Now we want to undo the exponent on the right side, that is, make the exponent 1. To do so, multiply both sides by the reciprocal of the exponent, which is 1/12.

1.29508197^1/12 = 1 + r

Using my TI-84 graphing calculator:

1.021781664 = 1 + r

Therefore the continuous growth rate is 0.021781664 per year.

Now that we know what r is, we can find out when it will reach 250 million:

250,000,000 = 122,000,000 * 1.021781664^x

2.04918 = 1.021781664^x

To solve for the exponent, take the natural logarithm of both sides:

ln(2.04918) = x * ln(1.021781664)

33.295 = x

So Brazil's population will reach 250 million 33.295 years after 1980, or in the year 2013.

Answer 2

a) P = P₀(1 + r)^t, where P = 158 mln is population in 1992, P₀ = 122 mln is population in 1980, t = 1992 - 1980 = 12 years, r is rate of population growth. We have equation for r: 158 = 122(1 + r)^t, r = (158/122)^1/12- 1 = 0.02178 = 2.178%;

b) P = 250 mln we have equation for t: 250 = 122(1 + r)^t, 250/122 = (1 + r)^t, ln(250/122) = t ln(1 + r).

So, t = ln(250/122)/ln(1 + 0.02178) = 33.3 years.

So in 1980 + 34 = 2014 year population of brazil was 250 mln

Answer 3

You need to use the continuous growth formula f(t)= (a)e^kt where

a = initial amount k= growth constant and t = time

Initial amount is 122 time is 12 years from 1980 to 1992 and f(t) is 158 so

158 = 122e^12k

158/122 = e^12 k Take the natural log of both sides which gives you

ln(158/122) = ln e^12k

.25857= 12k

k=.021547

To solve for t when f(t)=250 we would have 250= 122e^(.021547t)

250/122 = e ^.021547t

ln(250/122)= .021547t

t= 33.3 years