MATH 131:100 Section 1.5 Examples
3 June, 2014
Compound Interest Problem
If you put P dollars (for principal) into an account that earns you continuously compounding interest at a rate of 4%, then after t years, the amount of money you will have
is
M (t) = P e.04t .
If you invest P =$1,000,000, then after 40 years, how much money will you have?
SOLUTION:
Just plugging in 40 for t into the formula gives
M (40) = 1000000e.04(40) ≈ $4, 953, 032.
You made a lot of money over those 40 years.
Population/Bacteria Growth Problem
Suppose you start with a population of 1,000 bacteria, and that every hour, the population doubles.
a) Find a formula for population, P , in terms of time, t (in hours).
SOLUTION:
At the beginning (t = 0 hours), the population is 1,000. So
P (0) = 1000.
After 1 hour, the population doubles, so
P (1) = 2P (0) = 2000.
Then, after the second hour, the population doubles again, so using the previous formula
P (2) = 2P (1) = 2 (2P (0)) = 22 P (0).
Then after the third hour,
P (3) = 2P (2) = 2 22 P (0) = 23 P (0).
Following the pattern, we see that after t years, the population is given by
P (t) = 2t P (0) = 2t · 1000.
b) What is the population after 10 hours?
SOLUTION:
Plugging into our equation from part (a),
P (10) = 210 · 1000 = (1024)(1000) = 1, 024, 000.
Radioactive Decay/Half-Life Problem
The Half-Life of Strontium-90 is 25 years. This means that half of any given quantity of
Strontium-90 will disintegrate in 25 years.
a) If you start with 24mg of Strontium-90, find a formula for the mass, m, that remains
after t years.
SOLUTION:
Again, start with the fact that at 0 years, there are 24mg. So
m(0) = 24.
Then we know that at 25 years, there will be half the amount left:
m(25) =
1
m(0).
2
Then after another 25 years, there will be half of that amount left:
m(50) =
1
11
m(25) =
m(0) =
2
22
2
1
m(0).
2
Then after 25 more years, there will be half of this amount:
m(75) =
1
1
m(50) =
2
2
2
3
1
1
m(0) =
m(0).
2
2
Now we write our equation in terms of these 25 year increments as follows.
m(25t) =
t
t
1
1
m(0) =
· 24.
2
2
In order to find the equation for t, we divide t by 25 in order to cancel that factor on the
left hand side.
25t
1
· 24.
m(t) =
2
And since
1
2
= 2−1 , we can also write this as
t
m(t) = 2− 25 · 24.
b) Find the mass remaining after 40 years.
SOLUTION:
Plugging in,
40
m(40) = 2− 25 · 24 ≈ 7.9 mg.
2