Question

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?

Answer 1

`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`

Answer 2

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