Differential Calculus Word Problem?

A colony of bacteria doubles in population every 20 minutes starting from an initial population size of y0. Let y(t) denote the population at time t.
1. Express y as an exponential function with base 2
2. Express y as an exponential function with base e
3. What differential equation does y satisfy?
4. At what time has the initial population grown by a factor of 3?

2 Answers
Nov 18, 2016

y(t)=y_0*2^(t/20)
y(t)=y_0*e^((ln2/20*t))
y'(t)=ln2/20*y with the initial condition y(0)=y_0
t_(3y_0)=20*ln3/ln2

Explanation:

The time law describing the growth of such a population has the form
y(t)=y_0*2^(t/20) (1)
with t measured in seconds. Indeed it is easy to see that every 20 seconds the population doubles.

In term of exponential function 2^(t/20) is equivalent to e^((ln2/20*t)) and as a reasult we can rewrite (1) as
y(t)=y_0*e^((ln2/20*t))
If we derive y(t) by the chain rule we get y'(t)=y_0e^((ln2/20*t))*ln2/20 that can be rewritten as the Cauchy problem y'(t)=ln2/20*y for y(0)=y_0

Finally the population's triplication time is given by the equation 3y_0=y_0*e^((ln2/20*t)) that can be solved to find
t_(3y_0)=20*ln3/ln2

Nov 18, 2016
  1. y = y_0 2^(0.05t)
  2. y = y_0 e^(0.05ln2t)
  3. dy/dt = (0.05ln2)y
  4. t ~~ 32 mins

Explanation:

1. Express y as an exponential function with base 2
y(0) = y_0
y(20) = 2y_0 = y_0 2^1
y(40) = 2*2y_0 = y_0 2^2
y(60) = 2*2*2y_0 = y_0 2^3
y(80) = 2*2*2*2y_0 = y_0 2^4
vdots
y(t) = y_0 2^(t/20)
y(t) = y_0 2^(0.05t)

2. Express y as an exponential function with base e
Take natural logarithms
lny = ln{y_0 2^(0.05t)}
lny = lny_0 + 0.05ln2t
y = e^(lny_0 + 0.05ln2t)
y = e^(lny_0)(e^(0.05ln2t))
y = y_0 e^(0.05ln2t)

3. What differential equation does y satisfy?
dy/dt = y_0 e^(0.05ln2t)(0.05ln2)
dy/dt = y(0.05ln2)
dy/dt = (0.05ln2)y

4. At what time has the initial population grown by a factor of 3?
We want
y_0 e^(0.05ln2t) = 3y_0

e^(0.05ln2t) = 3

(0.05ln2)t = ln3

t = (ln3)/(0.05ln2)

t = 31.699 ... ~~ 32 mins