Applications of differential equations
Master applications of differential equations for Year 12 Specialist Mathematics in Queensland (QCAA). This sub-topic models real situations with two provided equations — Newton's law of cooling and radioactive decay — then solves and interprets them.
You will learn to separate the variables to reach \(T=T_s+(T_0-T_s)e^{-kt}\) and \(N=N_0 e^{-kt}\), find the constant \(k\) from a data point, and read off half-lives and limiting temperatures — a Unit 4 application that links calculus to biology, physics and everyday cooling.
Theory
Applications of differential equations in Year 12 Specialist Mathematics (QCAA, Queensland) use two provided models: Newton's law of cooling \(\dfrac{dT}{dt}=-k(T-T_s)\) and radioactive decay \(\dfrac{dN}{dt}=-kN\). Separating the variables gives \(T=T_s+(T_0-T_s)e^{-kt}\) and \(N=N_0 e^{-kt}\). This page shows how to solve these models, find \(k\) from data, and interpret half-lives and limiting temperatures.
A differential equation is an equation for an unknown function that involves its rate of change. In this sub-topic the equation is provided and you model a real situation by solving it and interpreting the answer.
Newton's law of cooling says an object's temperature \(T\) changes at a rate proportional to how far it is above (or below) the surrounding temperature \(T_s\): \(\dfrac{dT}{dt}=-k(T-T_s)\), where \(k>0\). Separating the variables and applying \(T=T_0\) at \(t=0\) gives the solution \(T=T_s+(T_0-T_s)e^{-kt}\). The temperature approaches the room temperature \(T_s\), so \(T_s\) is a horizontal asymptote.
Radioactive decay (and unrestricted growth) obeys \(\dfrac{dN}{dt}=-kN\), meaning the amount \(N\) falls at a rate proportional to how much is present. Its solution is \(N=N_0 e^{-kt}\), where \(N_0\) is the initial amount. Replacing \(-k\) with \(+k\) gives exponential growth \(N=N_0 e^{kt}\), the model for a bacterial colony or an investment.
The half-life is the time for a decaying quantity to halve. Setting \(N=\tfrac{1}{2}N_0\) gives \(e^{-kt}=\tfrac{1}{2}\), so the half-life is \(\dfrac{\ln 2}{k}\). The constant \(k\) controls how fast the process runs and is usually found by substituting one measured data point and taking logarithms.
Newton's law of cooling: the provided differential equation and its solution (from separating the variables) are
Radioactive decay and exponential growth:
The half-life of a decaying quantity comes from halving the amount:
Modelling and solving a provided differential equation
- Identify the model and write the solution. Cooling \(\dfrac{dT}{dt}=-k(T-T_s)\) gives \(T=T_s+(T_0-T_s)e^{-kt}\); decay/growth \(\dfrac{dN}{dt}=\pm kN\) gives \(N=N_0 e^{\pm kt}\).
- Apply the initial condition. Put \(t=0\) to find the constant — the amount \(N_0\), or the temperature gap \(T_0-T_s\).
- Find \(k\) from a second data point. Substitute the known \((t,\text{value})\), isolate \(e^{-kt}\), and take logarithms; a half-life or doubling time gives \(k=\dfrac{\ln 2}{t_{1/2}}\).
- Answer and interpret. Substitute a time to predict a value, solve for \(t\) to find when a value is reached, or take \(t\to\infty\) for the long-term (limiting) behaviour.
Separate the variables and integrate both sides:
| \(\dfrac{dN}{dt}\) | \(=\) | \(-kN\) |
| \(\int\dfrac{1}{N}\,dN\) | \(=\) | \(-\int k\,dt\) |
| \(\ln N\) | \(=\) | \(-kt+c\) |
| \(N\) | \(=\) | \(e^{-kt+c}=A e^{-kt}\) |
Apply \(N=N_0\) at \(t=0\) to find the constant \(A\):
| \(N_0\) | \(=\) | \(A e^{0}\) |
| \(A\) | \(=\) | \(N_0\) |
| \(N\) | \(=\) | \(N_0 e^{-kt}\) |
Use the half-life: the amount halves at \(t=6\), so solve for \(k\):
| \(\tfrac{1}{2}N_0\) | \(=\) | \(N_0 e^{-6k}\) |
| \(e^{-6k}\) | \(=\) | \(\dfrac{1}{2}\) |
| \(6k\) | \(=\) | \(\ln 2\) |
| \(k\) | \(=\) | \(\dfrac{\ln 2}{6}\) |
The solution is \(N=N_0 e^{-kt}\) with \(k=\dfrac{\ln 2}{6}\) per year.
Apply the initial condition \(T=85\) at \(t=0\) to find \(A\):
| \(85\) | \(=\) | \(25+A e^{0}\) |
| \(A\) | \(=\) | \(85-25\) |
| \(=\) | \(60\) |
Use \(T=55\) at \(t=10\), isolate the exponential, take logs:
| \(55\) | \(=\) | \(25+60 e^{-10k}\) |
| \(30\) | \(=\) | \(60 e^{-10k}\) |
| \(e^{-10k}\) | \(=\) | \(\dfrac{1}{2}\) |
| \(10k\) | \(=\) | \(\ln 2\) |
| \(k\) | \(=\) | \(\dfrac{\ln 2}{10}\) |
Substitute \(t=20\); note \(e^{-20k}=(e^{-10k})^2=\tfrac{1}{4}\):
| \(T\) | \(=\) | \(25+60 e^{-20k}\) |
| \(=\) | \(25+60\left(\dfrac{1}{4}\right)\) | |
| \(=\) | \(25+15\) | |
| \(=\) | \(40\) |
\(A=60\), \(k=\dfrac{\ln 2}{10}\) per minute, and the coffee is \(40^\circ\mathrm{C}\) after \(20\) minutes.
Doubling every \(3\) hours means \(N=2N_0\) at \(t=3\); solve for \(k\):
| \(2N_0\) | \(=\) | \(N_0 e^{3k}\) |
| \(e^{3k}\) | \(=\) | \(2\) |
| \(3k\) | \(=\) | \(\ln 2\) |
| \(k\) | \(=\) | \(\dfrac{\ln 2}{3}\) |
Substitute \(N_0=2000\) and \(t=9\); \(9\) hours is \(3\) doublings:
| \(N\) | \(=\) | \(2000 e^{9k}\) |
| \(=\) | \(2000\left(e^{3k}\right)^{3}\) | |
| \(=\) | \(2000\times 2^{3}\) | |
| \(=\) | \(16000\) |
\(k=\dfrac{\ln 2}{3}\) per hour, and the population is \(16000\) after \(9\) hours.
Substitute \(T=30\) and isolate the exponential:
| \(30\) | \(=\) | \(25+60 e^{-kt}\) |
| \(5\) | \(=\) | \(60 e^{-kt}\) |
| \(e^{-kt}\) | \(=\) | \(\dfrac{1}{12}\) |
Take natural logs and solve for \(t\) using \(k=\dfrac{\ln 2}{10}\):
| \(-kt\) | \(=\) | \(\ln\dfrac{1}{12}=-\ln 12\) |
| \(t\) | \(=\) | \(\dfrac{\ln 12}{k}\) |
| \(=\) | \(\dfrac{10\ln 12}{\ln 2}\) | |
| \(=\) | \(\dfrac{10(2.48491)}{0.69315}\) | |
| \(=\) | \(35.85\) | |
| \(\approx\) | \(35.8\) |
The coffee reaches \(30^\circ\mathrm{C}\) after about \(35.8\) minutes.
Common pitfalls
Frequently asked questions
What is Newton's law of cooling?
It is the differential equation \(\dfrac{dT}{dt}=-k(T-T_s)\): an object cools (or warms) at a rate proportional to the difference between its temperature \(T\) and the surrounding temperature \(T_s\). Its solution is \(T=T_s+(T_0-T_s)e^{-kt}\).
How do you solve \(\dfrac{dN}{dt}=-kN\)?
Separate the variables to get \(\int\dfrac{1}{N}\,dN=-\int k\,dt\), so \(\ln N=-kt+c\) and \(N=A e^{-kt}\). Applying \(N=N_0\) at \(t=0\) gives \(N=N_0 e^{-kt}\).
What is a half-life and how is it related to \(k\)?
The half-life is the time for a decaying quantity to halve. Setting \(e^{-kt}=\dfrac{1}{2}\) gives the half-life \(\dfrac{\ln 2}{k}\), so \(k=\dfrac{\ln 2}{t_{1/2}}\).
What temperature does a cooling object approach?
As \(t\to\infty\) the term \(e^{-kt}\to 0\), so \(T\to T_s\): the object approaches the surrounding (room) temperature, which is the horizontal asymptote of the cooling curve.
How do you find \(k\) from a data point?
Substitute the measured time and value, isolate the exponential \(e^{-kt}=r\), then take natural logs: \(k=-\dfrac{\ln r}{t}\). One extra data point beyond the initial condition is enough.
What is the difference between the growth and decay differential equations?
Decay is \(\dfrac{dN}{dt}=-kN\) with solution \(N=N_0 e^{-kt}\) (falling to \(0\)); growth is \(\dfrac{dN}{dt}=kN\) with solution \(N=N_0 e^{kt}\) (rising without bound). Only the sign of the exponent changes.