# 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?

Nov 18, 2016

$y \left(t\right) = {y}_{0} \cdot {2}^{\frac{t}{20}}$
$y \left(t\right) = {y}_{0} \cdot {e}^{\left(\ln \frac{2}{20} \cdot t\right)}$
$y ' \left(t\right) = \ln \frac{2}{20} \cdot y$ with the initial condition $y \left(0\right) = {y}_{0}$
${t}_{3 {y}_{0}} = 20 \cdot \ln \frac{3}{\ln} 2$

#### Explanation:

The time law describing the growth of such a population has the form
$y \left(t\right) = {y}_{0} \cdot {2}^{\frac{t}{20}}$ $\left(1\right)$
with $t$ measured in seconds. Indeed it is easy to see that every 20 seconds the population doubles.

In term of exponential function ${2}^{\frac{t}{20}}$ is equivalent to ${e}^{\left(\ln \frac{2}{20} \cdot t\right)}$ and as a reasult we can rewrite $\left(1\right)$ as
$y \left(t\right) = {y}_{0} \cdot {e}^{\left(\ln \frac{2}{20} \cdot t\right)}$
If we derive $y \left(t\right)$ by the chain rule we get $y ' \left(t\right) = {y}_{0} {e}^{\left(\ln \frac{2}{20} \cdot t\right)} \cdot \ln \frac{2}{20}$ that can be rewritten as the Cauchy problem $y ' \left(t\right) = \ln \frac{2}{20} \cdot y$ for $y \left(0\right) = {y}_{0}$

Finally the population's triplication time is given by the equation $3 {y}_{0} = {y}_{0} \cdot {e}^{\left(\ln \frac{2}{20} \cdot t\right)}$ that can be solved to find
${t}_{3 {y}_{0}} = 20 \cdot \ln \frac{3}{\ln} 2$

Nov 18, 2016
1. $y = {y}_{0} {2}^{0.05 t}$
2. $y = {y}_{0} {e}^{0.05 \ln 2 t}$
3. $\frac{\mathrm{dy}}{\mathrm{dt}} = \left(0.05 \ln 2\right) y$
4. $t \approx 32$ mins

#### Explanation:

1. Express y as an exponential function with base 2
$y \left(0\right) = {y}_{0}$
$y \left(20\right) = 2 {y}_{0} = {y}_{0} {2}^{1}$
$y \left(40\right) = 2 \cdot 2 {y}_{0} = {y}_{0} {2}^{2}$
$y \left(60\right) = 2 \cdot 2 \cdot 2 {y}_{0} = {y}_{0} {2}^{3}$
$y \left(80\right) = 2 \cdot 2 \cdot 2 \cdot 2 {y}_{0} = {y}_{0} {2}^{4}$
$\vdots$
$y \left(t\right) = {y}_{0} {2}^{\frac{t}{20}}$
$y \left(t\right) = {y}_{0} {2}^{0.05 t}$

2. Express y as an exponential function with base e
Take natural logarithms
$\ln y = \ln \left\{{y}_{0} {2}^{0.05 t}\right\}$
$\ln y = \ln {y}_{0} + 0.05 \ln 2 t$
$y = {e}^{\ln {y}_{0} + 0.05 \ln 2 t}$
$y = {e}^{\ln {y}_{0}} \left({e}^{0.05 \ln 2 t}\right)$
$y = {y}_{0} {e}^{0.05 \ln 2 t}$

3. What differential equation does y satisfy?
$\frac{\mathrm{dy}}{\mathrm{dt}} = {y}_{0} {e}^{0.05 \ln 2 t} \left(0.05 \ln 2\right)$
$\frac{\mathrm{dy}}{\mathrm{dt}} = y \left(0.05 \ln 2\right)$
$\frac{\mathrm{dy}}{\mathrm{dt}} = \left(0.05 \ln 2\right) y$

4. At what time has the initial population grown by a factor of 3?
We want
${y}_{0} {e}^{0.05 \ln 2 t} = 3 {y}_{0}$

${e}^{0.05 \ln 2 t} = 3$

$\left(0.05 \ln 2\right) t = \ln 3$

$t = \frac{\ln 3}{0.05 \ln 2}$

$t = 31.699 \ldots \approx 32$ mins