Applications of differential equations
Master applications of differential equations in Year 12 VCE Specialist Mathematics. This sub-topic models real situations with provided equations — Newton’s law of cooling and radioactive decay — then solves and interprets them. It sits in the Calculus area of study of the VCE Mathematics Study Design (VCAA), within the Differential equations topic of Unit 4.
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 — an application that links calculus to biology, physics and everyday cooling.
Theory
Applications of differential equations in Year 12 Specialist Mathematics 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.