Page 187 - Calculus Demystified

Transcendental Functions
CHAPTER 6
174
where C and C are constants of integration. We thus obtain

ln |B(t)|= Kt + D,
where we have amalgamated the two constants into a single constant D. Exponen-
tiating both sides gives
|B(t)|= e Kt+D
or
D
B(t) = e · e Kt = P · e Kt . (.)
Notice that we have omitted the absolute value signs since the number of bacteria is
D
always positive.Also we have renamed the constant e with the simpler symbol P .
Equation (.) will be our key to solving exponential growth and decay problems.
We motivated our calculation by discussing bacteria, but in fact the calculation
applies to any function which grows at a rate proportional to the size of the function.
Next we turn to some examples.

6.5.2 BACTERIAL GROWTH
EXAMPLE 6.30
A population of bacteria tendsto double every four hours. If there are 5000
bacteria at 9:00 a.m., then how many will there be at noon?

SOLUTION
To answer this question, let B(t) be the number of bacteria at time t. For con-
venience, let t = 0 correspond to 9:00 a.m. and suppose that time is measured
in hours. Thus noon corresponds to t = 3.
Equation (.) guarantees that

B(t) = P · e Kt
for some undetermined constants P and K. We also know that

5000 = B(0) = P · e K·0 = P.

We see that P = 5000 and B(t) = 5000 · e Kt . We still need to solve for K.
Since the population tends to double in four hours, there will be 10,000
bacteria at time t = 4; hence
10000 = B(4) = 5000 · e K·4 .

We divide by 5000 to obtain

2 = e K·4 .