Page 315 - Calculus Demystified

Solutions to Exercises
302
Taking March 10 to be about t = 2, we find that the amount of radioactive
material present on March 10 is
2
3 9
R(2) = 5 · = .
5 5
7. Let the amount of bacteria present at time t be
B(t) = P · e Kt .

Let t = 0 be 10:00 a.m. We know that B(0) = 10000 and B(3) = 15000.
Thus

10000 = B(0) = P · e K·0
so P = 10000. Also
15000 = B(3) = 10000 · e K·3
hence
1
K = · ln(3/2).
3
As a result,
B(t) = 10000 · e t·[1/3] ln(3/2)

or
t/3
3
B(t) = 10000 · .
2
We find that, at 2:00 p.m., the number of bacteria is
4/3
3
B(4) = 10000 · .
2
8. If M(t) is the amount of money in the account at time t then we know that
M(t) = 1000 · e 6t/100 .
Here t = 0 corresponds to January 1, 2005. Then, on January 1, 2009, the
amount of money present is
M(4) = 1000 · e 6·4/100 ≈ 1271.25.
1
x
9. (a) · 1 · e + x · e x
x 2
1 − (x · e )
1 1
(b) ·
1 + (x/[x + 1]) 2 (x + 1) 2