Related rates
Master related rates for Year 12 Specialist Mathematics in Queensland (QCAA). When two quantities are connected by a formula and both change over time, the chain rule turns a known rate into the one you need — from expanding balloons to filling tanks.
You will learn to write the relation between the quantities, differentiate with respect to time, and substitute the instant to find rates for circles, spheres, cones and sliding ladders — a core calculus application in Unit 4 rates of change and differential equations.
Theory
A related rates problem links the rates at which two connected quantities change with time, using the chain rule, in Year 12 Specialist Mathematics (QCAA, Queensland). When a formula ties the quantities together — a circle's area to its radius, a tank's volume to its depth — differentiating with respect to time turns one known rate into the one you want. This page covers expanding balloons and circles, draining and filling tanks, cones and sliding ladders.
In a related rates problem two or more quantities both change over time and are joined by an equation. You are told the rate of one and asked for the rate of another — for example, a balloon is inflated at a known volume rate and you want how fast its radius grows.
The connection is the chain rule. If a quantity \(Q\) depends on a variable \(x\) and \(x\) depends on time \(t\), then \(\dfrac{dQ}{dt}=\dfrac{dQ}{dx}\times\dfrac{dx}{dt}\). The middle factor \(\dfrac{dQ}{dx}\) comes from the geometric formula; the other factor is the given time rate.
The method is always the same: write the relation between the quantities, differentiate with respect to \(t\), then substitute the known rate and the instant. Differentiate before you substitute numbers, so every changing quantity keeps its rate.
Watch the sign: a quantity that is increasing (filling, inflating) has a positive rate; one that is decreasing (draining, melting, a ladder-top falling) has a negative rate.
The chain rule links any two connected rates:
To make an unknown rate the subject, invert the middle factor:
The middle factor comes from the standard geometric formulas:
How to solve a related rates problem
- Identify the quantities that change with time, the rate you are given, and the rate you want.
- Write the relation connecting the quantities; if two geometric variables both change, use similar triangles to reduce it to a single variable.
- Differentiate both sides with respect to \(t\) using the chain rule, keeping the derivative in terms of the variable.
- Substitute the known rate and the given instant, then solve for the required rate (and note its sign).
Write the chain rule, the area relation, and its derivative:
| \(\dfrac{dA}{dt}\) | \(=\) | \(\dfrac{dA}{dr}\times\dfrac{dr}{dt}\) |
| \(A\) | \(=\) | \(\pi r^2\) |
| \(\dfrac{dA}{dr}\) | \(=\) | \(2\pi r\) |
Substitute \(r=6\) and \(\dfrac{dr}{dt}=4\):
| \(\dfrac{dA}{dt}\) | \(=\) | \(2\pi r\times\dfrac{dr}{dt}\) |
| \(=\) | \(2\pi(6)(4)\) | |
| \(=\) | \(48\pi\) |
\(\dfrac{dA}{dt}=48\pi~\text{cm}^2/\text{s}\).
Differentiate the sphere volume with respect to \(r\):
| \(V\) | \(=\) | \(\dfrac{4}{3}\pi r^3\) |
| \(\dfrac{dV}{dr}\) | \(=\) | \(4\pi r^2\) |
Apply the chain rule with \(\dfrac{dr}{dt}=3\), then \(r=5\):
| \(\dfrac{dV}{dt}\) | \(=\) | \(\dfrac{dV}{dr}\times\dfrac{dr}{dt}\) |
| \(=\) | \(4\pi r^2\times 3\) | |
| \(=\) | \(12\pi(5)^2\) | |
| \(=\) | \(300\pi\) |
\(\dfrac{dV}{dt}=300\pi~\text{cm}^3/\text{s}\).
Use \(r=\dfrac{h}{3}\) to write \(V\) in terms of \(h\) only:
| \(V\) | \(=\) | \(\dfrac{1}{3}\pi r^2 h\) |
| \(=\) | \(\dfrac{1}{3}\pi\left(\dfrac{h}{3}\right)^2 h\) | |
| \(=\) | \(\dfrac{\pi}{27}h^3\) |
Differentiate, then rearrange the chain rule for \(\dfrac{dh}{dt}\):
| \(\dfrac{dV}{dh}\) | \(=\) | \(\dfrac{\pi}{9}h^2\) |
| \(\dfrac{dh}{dt}\) | \(=\) | \(\dfrac{dh}{dV}\times\dfrac{dV}{dt}\) |
| \(=\) | \(\dfrac{9}{\pi h^2}\times 9\) |
Substitute \(h=6\):
| \(=\) | \(\dfrac{81}{\pi(6)^2}\) | |
| \(=\) | \(\dfrac{9}{4\pi}\) |
\(\dfrac{dh}{dt}=\dfrac{9}{4\pi}~\text{cm/s}\).
Differentiate \(x^2+y^2=100\) with respect to \(t\):
| \(x^2+y^2\) | \(=\) | \(100\) |
| \(2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}\) | \(=\) | \(0\) |
At \(x=6\) the height is \(y=8\); substitute with \(\dfrac{dx}{dt}=1\):
| \(\dfrac{dy}{dt}\) | \(=\) | \(-\dfrac{x}{y}\times\dfrac{dx}{dt}\) |
| \(=\) | \(-\dfrac{6}{8}\times 1\) | |
| \(=\) | \(-0.75\) |
The top slides down at \(0.75~\text{m/s}\) (the negative sign shows \(y\) is decreasing).
Common pitfalls
Frequently asked questions
What is a related rates problem?
A problem where two quantities are linked by a formula and both change with time; you know one rate and use the chain rule to find the other.
Which rule links the two rates?
The chain rule: \(\dfrac{dQ}{dt}=\dfrac{dQ}{dx}\times\dfrac{dx}{dt}\). The first factor is the derivative of the geometric formula, the second is the given time rate.
How do I handle a cone where the radius and height both change?
Use similar triangles to write the radius in terms of the height (for example \(r=\dfrac{h}{2}\)), substitute so the volume is in one variable, then differentiate.
Do I substitute the numbers before or after differentiating?
After. Differentiate the relation with respect to \(t\) first so each changing quantity keeps its rate, then substitute the given instant.
Why is my rate negative?
A negative rate means the quantity is decreasing — a tank draining, a snowball melting, or the top of a ladder sliding down. Report the size and state that it is decreasing.
How fast does the top of a sliding ladder move?
Differentiate \(x^2+y^2=L^2\) to get \(\dfrac{dy}{dt}=-\dfrac{x}{y}\dfrac{dx}{dt}\); the negative sign shows the top falls as the base is pulled out.