Applications of Differential Equations
Differential equations are dynamic, describing instantaneous rates of change in physical, biological, and engineering systems.
Rocket Science
Example
A toy rocket ,consisting of casing and fuel,
is launched at time \(t=0\).
The initial total mass is \(m_0\).
The fuel burns such that the mass of the rocket
satisfies the equation \(
\frac{dm}{dt} = -\frac{1}{4}km
\)
Given that the mass of the casing is \(\tfrac12 m_0\),
for how long does the rocket’s fuel burn?
\[
\begin{aligned}
\frac{dm}{dt} &= -\frac{1}{4}t
\\[10pt]
dm &= -\frac{1}{4}t\,dt
\\[14pt]
\int dm &= \int -\frac{1}{4}t\,dt
\\[14pt]
m &= -\frac{t^{2}}{8} + C
\end{aligned}
\]
\[
\begin{aligned}
\text{When } t &= 0,\; m = m_{0}
\\[12pt]
m_{0} &= -\frac{0^{2}}{8} + C
\\[12pt]
m_{0} &= C
\\[18pt]
m &= -\frac{t^{2}}{8} + m_{0}
\end{aligned}
\]
\[
\begin{aligned}
\text{The rocket has spent its fuel when }
m &= \tfrac12 m_{0}
\\[14pt]
\Rightarrow\quad
m(T) &= \tfrac12 m_{0}
\end{aligned}
\]
\[
m(T) = m_{0} - \frac{t^{2}}{8}
\]
\[
\tfrac12 m_{0} = m_{0} - \frac{t^{2}}{8}
\]
\[
\frac{t^{2}}{8} = \tfrac12 m_{0}
\]
\[
t^{2} = 4m_{0}
\]
\[
t = \sqrt{4m_{0}}
\]
\[
t = 2\sqrt{m_{0}}
\]
Newton’s Law of Cooling
The rate at which an object cools is
proportional to the difference between its
temperature and that of its surroundings
Example
The temperature \(\theta\) °C of a cup of coffee after \(t\) minutes is given by:
\[
\frac{d\theta}{dt} = -k(\theta - 20)
\]
Initially the temperature is \(70^\circ\text{C}\).
After 5 minutes it is \(50^\circ\text{C}\).
When is the temperature \(40^\circ\text{C}\)?
\[
\begin{aligned}
\frac{d\theta}{dt} &= -k(\theta - 20)
\\[12pt]
\frac{d\theta}{\theta - 20} &= -k\,dt
\\[16pt]
\int \frac{d\theta}{\theta - 20} &= \int -k\,dt
\\[16pt]
\ln|\theta - 20| &= -kt + C
\\[16pt]
\ln\left|\frac{\theta - 20}{C}\right| &= -kt
\\[16pt]
\theta - 20 &= C e^{-kt}
\end{aligned}
\]
\[
\begin{aligned}
\text{When } t &= 0,\; \theta = 70
\\[14pt]
50 &= C
\\[18pt]
\theta &= 50\,e^{-kt} + 20
\end{aligned}
\]
\[
\begin{aligned}
\text{When } t &= 5,\; \theta = 50
\\[14pt]
50 &= 50\,e^{-5k} + 20
\\[14pt]
30 &= 50\,e^{-5k}
\\[14pt]
k &= \frac{\ln(3/5)}{-5}
\\[18pt]
\theta &= 50\,e^{-0.1021\,t} + 20
\end{aligned}
\]
For a temperature of40°C
\[
\begin{aligned}
40 &= 50\,e^{-0.1021t} + 20
\\[14pt]
20 &= 50\,e^{-0.1021t}
\\[14pt]
t &= \frac{\ln(2/5)}{-0.1021}
\\[18pt]
t &= 8.97 \text{ minutes after it was made}
\end{aligned}
\]
Growth & Decay
Example
The rate of growth of a population \(P\) is given by \(2P\) per day.
Initially \(P=100\).
How long does it take for the population to double?
\[
\begin{aligned}
\frac{dP}{dt} &= 2P
\\[14pt]
\frac{dP}{2P} &= dt
\\[16pt]
\int \frac{dP}{2P} &= \int dt
\\[16pt]
\frac12 \ln|P| &= t + C
\\[16pt]
\ln\left|\frac{P^{1/2}}{C}\right| &= t
\\[16pt]
P^{1/2} &= C e^{t}
\end{aligned}
\]
\[
\begin{aligned}
\text{When } t &= 0,\; P = 100
\\[14pt]
P^{1/2} &= \sqrt{100}\,e^{t}
\\[14pt]
P^{1/2} &= 10e^{t}
\\[14pt]
P &= 100e^{2t}
\end{aligned}
\]
\[
\begin{aligned}
\text{When } P &= 200
\\[12pt]
200 &= 100e^{2t}
\\[12pt]
2 &= e^{2t}
\\[12pt]
t &= \frac{\ln 2}{2}
\\[12pt]
t &= 0.346 \text{ days}
\\[12pt]
t &= 0.346 \times 24 = 8.3 \text{ hours}
\end{aligned}
\]
Kinematics
Example
The acceleration of a body moving in a straight line
is given as \( a = 3x^{2}\ \text{m/s}^{2} \) , where x is the distance from the
origin in metres.
When \(x=1\), velocity \(v=2 \text{ m/s} \).
Find the velocity when \(x=2 m\).
\[
a = v\,\frac{dv}{dx} = 3x^{2}
\]
\[
v\,dv = 3x^{2}\,dx
\]
\[
\int v\,dv = \int 3x^{2}\,dx
\]
\[
\tfrac12 v^{2} = x^{3} + c
\]
\[
\text{When } v = 2,\ x = 1
\]
\[
2 = 1 + c
\]
\[
c = 1
\]
\[
\tfrac12 v^{2} = x^{3} + 1
\]
\[
\text{When } x = 2
\]
\[
\tfrac12 v^{2} = 2^{3} + 1
\]
\[
v = \sqrt{18}
\]
\[
v = 3\sqrt{2}\,\text{ms}^{-1}
\]
\[
v \approx 4.24\,\text{ms}^{-1}
\]
