Differential equations with related rates
Solve differential equations with related rates in Year 12 Specialist Mathematics for Queensland (QCAA). These multi-step problems use the chain rule to connect two rates — such as a tank's volume rate and its depth rate — and turn that link into a differential equation.
You will learn to set up the rate relation, form the differential equation, solve it by direct integration or separation of variables, and apply the initial condition to find a depth, a time to empty or a long-term amount — a key modelling skill in Unit 4's rates of change and differential equations.
Theory
A differential equation with related rates is a two-step problem in Year 12 Specialist Mathematics (QCAA, Queensland): first the chain rule links two rates, \(\dfrac{dV}{dt}=\dfrac{dV}{dh}\cdot\dfrac{dh}{dt}\), to form a first-order differential equation; then you solve it by direct integration or separation of variables and apply the initial condition to answer the question.
A related-rates problem connects the rates of change of two quantities that share a variable. The chain rule is the link: if the volume \(V\) of water in a tank depends on the depth \(h\), then \(\dfrac{dV}{dt}=\dfrac{dV}{dh}\cdot\dfrac{dh}{dt}\). A given volume rate \(\dfrac{dV}{dt}\) therefore produces a rate for the depth.
A differential equation is an equation involving a derivative. Here the chain-rule link is the differential equation: substituting the geometry (for example \(\dfrac{dV}{dh}=A\)) turns \(\dfrac{dV}{dt}=-k\) into a first-order equation such as \(\dfrac{dh}{dt}=-\dfrac{k}{A}\) or \(\dfrac{dh}{dt}=-\dfrac{\sqrt{h}}{A}\).
First-order equations come in three shapes. \(\dfrac{dh}{dt}=f(t)\) is solved by integrating directly. \(\dfrac{dh}{dt}=g(h)\) and \(\dfrac{dh}{dt}=f(t)g(h)\) are solved by separation of variables: gather every \(h\) on one side and every \(t\) on the other, then integrate both sides.
The initial condition (for example the starting depth \(h(0)\)) fixes the constant of integration, turning the general solution into the particular solution. Common contexts are a draining tank, an evaporating sphere, and a well-mixed brine tank where \(\dfrac{dQ}{dt}=\text{(rate in)}-\text{(rate out)}\).
Link the two rates with the chain rule (the shared variable is the depth \(h\)):
Solve a separable first-order equation \(\dfrac{dh}{dt}=g(h)\) by gathering the variables and integrating:
Useful volumes and areas for the related-rates link:
A well-mixed tank of volume \(V\), inflow concentration \(c\) at rate \(R\text{ in}\) and the same rate out:
How to solve a differential equation with related rates
- Link the rates: write the chain rule \(\dfrac{dV}{dt}=\dfrac{dV}{dh}\cdot\dfrac{dh}{dt}\) and find \(\dfrac{dV}{dh}\) from the geometry (\(A\), \(\pi h^2\), \(4\pi r^2\), \(\dots\)).
- Form the differential equation: substitute the given rate (a leak or drain makes \(\dfrac{dV}{dt}\) negative) and make \(\dfrac{dh}{dt}\) the subject.
- Solve it: integrate directly if the right side depends only on \(t\); otherwise separate the variables and integrate both sides.
- Apply the condition and answer: use the initial value to fix \(c\), then substitute to find the depth, time to empty, or steady amount asked for.
Link the rates with \(\dfrac{dV}{dt}=\dfrac{dV}{dh}\cdot\dfrac{dh}{dt}\); here \(\dfrac{dV}{dh}=A=3\):
| \(V\) | \(=\) | \(3h\) |
| \(\dfrac{dV}{dh}\) | \(=\) | \(3\) |
| \(\dfrac{dV}{dt}\) | \(=\) | \(3\cdot\dfrac{dh}{dt}\) |
| \(-6\) | \(=\) | \(3\cdot\dfrac{dh}{dt}\) |
| \(\dfrac{dh}{dt}\) | \(=\) | \(-2\) |
Integrate the constant-rate equation and apply \(h(0)=10\):
| \(h\) | \(=\) | \(\int -2\,dt\) |
| \(=\) | \(-2t+c\) | |
| \(10\) | \(=\) | \(-2(0)+c \ \Rightarrow\ c=10\) |
| \(h\) | \(=\) | \(10-2t\) |
Set \(h=0\) for an empty tank:
| \(0\) | \(=\) | \(10-2t\) |
| \(2t\) | \(=\) | \(10\) |
| \(t\) | \(=\) | \(5\) |
The tank is empty after \(5\) minutes.
The right side depends on \(h\), so separate the variables and integrate:
| \(\int h^{-1/2}\,dh\) | \(=\) | \(\int -\dfrac{1}{3}\,dt\) |
| \(2\sqrt{h}\) | \(=\) | \(-\dfrac{t}{3}+c\) |
Apply \(h(0)=25\) to fix \(c\):
| \(2\sqrt{25}\) | \(=\) | \(-0+c\) |
| \(c\) | \(=\) | \(10\) |
| \(2\sqrt{h}\) | \(=\) | \(10-\dfrac{t}{3}\) |
Make \(h\) the subject, then set \(h=0\) for empty:
| \(\sqrt{h}\) | \(=\) | \(5-\dfrac{t}{6}\) |
| \(h\) | \(=\) | \(\left(5-\dfrac{t}{6}\right)^{2}\) |
| \(0\) | \(=\) | \(5-\dfrac{t}{6}\) |
| \(t\) | \(=\) | \(30\) |
\(h=\left(5-\dfrac{t}{6}\right)^{2}\); the tank empties after \(30\) minutes.
Differentiate \(V\), then link the rates with \(\dfrac{dV}{dt}=-2\pi\sqrt{h}\):
| \(V\) | \(=\) | \(\dfrac{1}{3}\pi h^{3}\) |
| \(\dfrac{dV}{dh}\) | \(=\) | \(\pi h^{2}\) |
| \(-2\pi\sqrt{h}\) | \(=\) | \(\pi h^{2}\cdot\dfrac{dh}{dt}\) |
| \(\dfrac{dh}{dt}\) | \(=\) | \(-\dfrac{2}{h^{3/2}}\) |
Separate the variables and integrate:
| \(\int h^{3/2}\,dh\) | \(=\) | \(\int -2\,dt\) |
| \(\dfrac{2}{5}h^{5/2}\) | \(=\) | \(-2t+c\) |
Apply \(h=9\) at \(t=0\) (note \(9^{5/2}=3^{5}=243\)), then set \(h=0\):
| \(\dfrac{2}{5}(243)\) | \(=\) | \(c \ \Rightarrow\ c=\dfrac{486}{5}\) |
| \(0\) | \(=\) | \(-2t+\dfrac{486}{5}\) |
| \(2t\) | \(=\) | \(\dfrac{486}{5}\) |
| \(t\) | \(=\) | \(\dfrac{243}{5}=48.6\) |
The tank is empty after \(\dfrac{243}{5}=48.6\) minutes.
Rate in \(=\) concentration \(\times\) flow; rate out \(=\dfrac{Q}{400}\times8\):
| \(\dfrac{dQ}{dt}\) | \(=\) | \(0.5\times8-\dfrac{Q}{400}\times8\) |
| \(=\) | \(4-\dfrac{Q}{50}\) |
Factor the right side, separate the variables and integrate:
| \(\dfrac{dQ}{dt}\) | \(=\) | \(-\dfrac{Q-200}{50}\) |
| \(\int \dfrac{1}{Q-200}\,dQ\) | \(=\) | \(\int -\dfrac{1}{50}\,dt\) |
| \(\ln|Q-200|\) | \(=\) | \(-\dfrac{t}{50}+c\) |
Apply \(Q(0)=0\) to fix the constant:
| \(Q-200\) | \(=\) | \(Ae^{-t/50}\) |
| \(0-200\) | \(=\) | \(A \ \Rightarrow\ A=-200\) |
| \(Q\) | \(=\) | \(200-200e^{-t/50}\) |
Long term \(t\to\infty\), so \(e^{-t/50}\to0\):
| \(Q\) | \(\to\) | \(200\) |
\(Q=200\left(1-e^{-t/50}\right)\); the salt approaches \(200\text{ kg}\).
Common pitfalls
Frequently asked questions
How do related rates lead to a differential equation?
The chain rule \(\dfrac{dV}{dt}=\dfrac{dV}{dh}\cdot\dfrac{dh}{dt}\) links a known rate (say a leak \(\dfrac{dV}{dt}\)) to the rate you want. Substituting the geometry \(\dfrac{dV}{dh}\) and making \(\dfrac{dh}{dt}\) the subject gives a first-order differential equation to solve.
What is separation of variables?
A method for \(\dfrac{dh}{dt}=g(h)\) or \(f(t)g(h)\): move all the \(h\) terms (with \(dh\)) to one side and all the \(t\) terms (with \(dt\)) to the other, then integrate both sides. For example \(\displaystyle\int h^{-1/2}\,dh=\int -\tfrac12\,dt\).
Why is dV/dt negative when a tank drains?
Because the volume is falling. A decreasing quantity has a negative rate of change, so a leak or drain of size \(\sqrt{h}\) is written \(\dfrac{dV}{dt}=-\sqrt{h}\). This negative sign carries through to \(\dfrac{dh}{dt}\).
How do you set up the differential equation for a brine (salt) tank?
Use \(\dfrac{dQ}{dt}=\text{(rate in)}-\text{(rate out)}\). Rate in \(=\) inflow concentration \(\times\) inflow rate; rate out \(=\dfrac{Q}{V}\times\) outflow rate, where \(\dfrac{Q}{V}\) is the current concentration in the well-mixed tank.
What is the difference between a general and a particular solution?
The general solution still contains the constant \(c\) (or \(A\)); it describes every possible solution. The particular solution uses the initial condition to pin down that constant, giving the one curve that fits the problem.
How do you find when a tank is empty?
Solve the differential equation for the depth \(h\) as a function of \(t\), apply the initial depth to find the constant, then set \(h=0\) and solve for \(t\). For an evaporating sphere set the radius \(r=0\) instead.