Question

The population of a community is known to increase at a rate proportional to the number of people present at time t. The initial population P0 has doubled in 5 years. Suppose it is known that the population is 9,000 after 3 years. What was the initial population P0? (Round your answer to one decimal place.) P0 = What will be the population in 10 years? (Round your answer to the nearest person.) persons How fast is the population growing at t = 10? (Round your answer to the nearest person.) persons/year

Answer 1

Answer:

Part 1) $p_0=5,937.8\ people$

Part 2) $23,752\ people$

Part 3) $1,781\ persons/year$

Step-by-step explanation:

In this problem we have a exponential growth function of the form

$p(t)=p_0(1+r)^t$

where

p_0 is the initial population

t is the number of years

r is the rate of change

we have that

The initial population p_0 has doubled in 5 years

so

$p(5)=2p_0$

substitute in the equation above

$2p_0=p_0(1+r)^5$

solve for r

$2=(1+r)^5$

elevated both sides to 1/5

$r=2^{\frac{1}{5}} -1$

$r=0.1487$

substitute

$p(t)=p_0(1+0.1487)^t$

$p(t)=p_0(1.1487)^t$

Find the value of p_0

Remember that

the population is 9,000 after 3 years

so

substitute

$9,000=p_0(1.1487)^3$

$p_0=5,937.8\ people$

so    

$p(t)=5,937.8(1.1487)^t$

What will be the population in 10 years?

For t=10 years

substitute

$p(t)=5,937.8(1.1487)^{10}= 23,752\ people$

How fast is the population growing at t = 10?

we know that

For t=0 ----> P=5,937.8 people

For t=10 ---> P=23,752 people

so

$(23,752-5,937.8)/(10-0)=1,781\ persons/year$