he 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. (Round your answers to two decimal places.)
(a) If the world population continues to grow at approximately 2%/year, find the length of time t0 required for the population to double in size.
t0 = yr
(b) Using the time t0 found in part (a), what would be the world population if the growth rate were reduced to 1.6%/yr?
billion people
(a) Oh dear, the world population doubling? That's a lot of cake to bake! Let's calculate the time it takes, using the growth rate of 2% per year.
To find the time required for the population to double, we can use the formula:
t0 = (ln(2) / ln(1 + r))
Where r is the growth rate in decimal form. Plugging in r = 0.02, we get:
t0 = (ln(2) / ln(1 + 0.02))
Let me crunch the numbers for you... *whirring sounds* ...
t0 ≈ 34.66 years
So, it would take approximately 34.66 years for the world population to double if it continues to grow at a rate of 2% per year.
(b) Now, let's see what would happen if we reduce the growth rate to 1.6% per year. We'll use t0 from part (a) to find the new population size.
Using the formula Q(t) = Q0 * (1 + r)^t, where Q0 is the initial population size and t is the time in years, we have:
Q(t) = 5.3 * (1 + 0.016)^34.66
Oh, the suspense is killing me... *calculating* ...
Q(t) ≈ 7.97 billion people
So, if the growth rate were reduced to 1.6% per year, the world population would be approximately 7.97 billion people after approximately 34.66 years. Mind-boggling, isn't it?
To find the function Q(t) that expresses the world population as a function of time, we can use the formula for exponential growth:
Q(t) = Q₀ * (1 + r)^t
Where Q(t) represents the world population at time t, Q₀ is the initial population, r is the growth rate (as a decimal), and t is the time in years.
In this case, the initial population Q₀ is given as 5.3 billion, and the growth rate r is 2% per year (or 0.02 as a decimal).
(a) To find the length of time t₀ required for the population to double in size, we need to determine when Q(t₀) = 2 * Q₀.
2 * Q₀ = Q(t₀)
2 * 5.3 billion = 5.3 billion * (1 + 0.02)^t₀
Dividing both sides of the equation by 5.3 billion, we get:
2 = (1.02)^t₀
To solve for t₀, we can take the natural logarithm (ln) of both sides:
ln(2) = ln((1.02)^t₀)
ln(2) = t₀ * ln(1.02)
Now we can solve for t₀:
t₀ = ln(2) / ln(1.02)
Using a calculator, we find t₀ ≈ 35.00 years (rounded to two decimal places).
(b) Using the value of t₀ found in part (a), we can determine the world population if the growth rate were reduced to 1.6% per year (or 0.016 as a decimal).
Q(t) = Q₀ * (1 + r)^t
Q(t₀) = 5.3 billion * (1 + 0.016)^35
Calculating this expression using a calculator, we find that the world population would be approximately 7.64 billion people.
part a is incorrect, the proper answer is:
Q(t)=Qoe^kt
Q(t)= 5.3e^0.02t
2=e^0.02t
ln2=0.02t
t=34.6yr
Q(t) = 5.3 + 0.02*5.3t,
Q(t) = 5.3 + 0.106t.
a. Q(t) = 5.3 + 0.106t = 10.6 Billion,
5.3 + 0.106t = 10.6,
0.106t = 10.6 - 5.3 = 5.3,
t = 50 years.
b. Q(t) = 5.3 + 0.016t*5.3,
Q(t) = 5.3 + 0.0848t.
Q(t) = 5.3 + 0.0848*50 = 9.54 Billion.