MATH 131 Quiz #2
June 3, 2014
Print Last Name: SOLUTION
First Name: SOLUTION
Maximum Points: 10
Grade: 10
Problem 1: The Power of Compounding Interest
Recall from class that if you invest a principal value of M (0) in an account that
earns continuously compounding interest at a rate r, then the amount of money in
your account after t years will be
M (t) = M (0)ert .
Fry has $0.01 in his bank account, but is accidentally transported 1,000 years into
the future. He returns to his bank (which luckily still exists) to see how much money
he now has. His account earns continuously compounding interest at a rate of 5%.
a) How much money is in his account? (Write the formula also; you may use a
calculator to evaluate the answer)
SOLUTION: Here the principal amount is $0.01, and r = .05, so our formula is
M (t) = .01e.05t .
Therefore, after 1,000 years, he will have
M (1, 000) = .01e.05·1000 = .01e50 ≈ $5.18471 × 1019 .
b) Graph the function from part (a), and estimate how many years it will take
Fry’s account to reach $1,000,000. (Sketch the graph here as well)
SOLUTION:
6
5
M(t)
x 10
4.5
Money in Fry’s Account, M (in $)
4
3.5
3
2.5
2
1.5
1
0.5
0
300
310
320
330
340
350
time, t (in years)
360
370
380
390
400
From my graph above, it looks like Fry’s account will reach $1,000,000 around 369
years, give or take a couple.
c) To illustrate the power of compounding interest, consider the formula for simple
interest, in which you only earn interest on the principal value, P (0). The formula is
P (t) = P (0)(1 + rt).
If Fry’s account earned 5% in simple interest, how much money would be in his account
after 1,000 years? Which interest method is better?
SOLUTION:
P (1000) = .01(1 + .05 · 1000) = .01(1 + 50) = .01(51) = $0.51
So using simple interest, Fry only has 51 cents in his account rather than the enormous
amount from compounding interest!
Problem 2: Exponential vs. Power Growth
Assume that a planet has a population of 1,000 people, and the total population
doubles every 10 years.
a) Come up with a formula for the population, P , as a function of time, t (in
years).
SOLUTION: Start by noticing that P (0) = 1000, and then P (10) = 2P (0), and
P (20) = 2P (10) = 22 P (0). So the pattern we notice is that
P (10t) = 2t P (0) = 2t · 1000.
To get P (t), just divide by 10:
t
P (t) = 2 10 · 1000.
b) What is P (200)? That is, what will the population of the planet be 200 years
from now.
SOLUTION:
By our formula for part (a),
200
P (200) = 2 10 · 1000 = 22 0 · 1000 = 1, 048, 576, 000.
This means that after 200 years, the population of the planet has reached over 1
billion people.
c) Assume instead that the population grows like t2 , and our population model is
P (t) = 1000 + t2 .
What would the population be in this case after 200 years? What do you note about
the answers for parts (b) and (c)?
SOLUTION:
Using this model,
P (200) = 1000 + 2002 = 1, 000 + 40, 000 = 41, 000.
From parts (b) and (c), it should be clear that population growing like 2t is much
faster growth than that of t2 .