Related Rates

Nothing in the world changes alone. Inflate a balloon and its radius, its surface area and its volume all grow at once — three different rates, locked together by the geometry of a sphere. A plane cruises overhead at a steady 900 km/h while its shadow races along the ground below it, perfectly in step. Drop a pebble in a pond and the ripple's radius, circumference and enclosed area all swell together, each at its own pace. Pull the foot of a ladder away from a wall and the top must slide down — Pythagoras insists.

In every case the quantities are tied together by an equation, and both are changing with time. Know how fast one changes, and the equation tells you how fast the other must. These are related rates problems, and the engine that powers them is the chain rule.

Here is the deep idea, and it is worth reading twice: you differentiate a relationship, not a formula. Up to now, differentiating has meant taking a formula like f(x) = x^2 and producing its derivative. Related rates asks something subtler: take a true statement connecting two changing quantities — x^2 + y^2 = 100, say — and differentiate the whole statement with respect to time t. Every variable in it picks up a \frac{d}{dt} factor through the chain rule (x^2 becomes 2x\frac{dx}{dt}, because x is secretly a function of t), and a static equation about sizes turns into a live equation about speeds.

Worked example 1 — the expanding balloon

Air is pumped into a spherical balloon at 100 cm³ per second. How fast is the radius growing when the radius is 5 cm?

Name what changes. Two quantities change with time: the volume V and the radius r. We are given \frac{dV}{dt} = 100 cm³/s and we want \frac{dr}{dt} at the instant when r = 5. Notice the phrasing: "when r = 5" is a snapshot — one frozen instant — while the rates describe motion through that instant.

Write the geometric relation. For a sphere, V = \tfrac{4}{3}\pi r^3. This is true at every instant, which is exactly what lets us differentiate it.

Differentiate both sides with respect to t. The chain rule supplies the 3r^2 from the power and then the \frac{dr}{dt} because r itself depends on t:

\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \,\frac{dr}{dt} = 4\pi r^2 \,\frac{dr}{dt}

Substitute the snapshot values — only now. With \frac{dV}{dt} = 100 and r = 5:

100 = 4\pi (5)^2 \frac{dr}{dt} = 100\pi\,\frac{dr}{dt} \qquad\Longrightarrow\qquad \frac{dr}{dt} = \frac{1}{\pi} \approx 0.32 \text{ cm/s}

Interpret. The answer is positive — the balloon is growing, as it should be — and its units are centimetres per second: a rate, not a length. And look at the formula itself: the factor 4\pi r^2 is the sphere's surface area. That's no accident — pumping in air spreads the new volume in a thin shell over the whole surface, so the bigger the balloon, the more air one centimetre of radius costs. The same 100 cm³/s that grows a small balloon briskly barely nudges a big one. Drag the radius below and watch the surface factor climb.

The ritual

Every related-rates problem — every one — yields to the same six moves. Learn the ritual and the scariest exam question becomes a form to fill in:

  1. Draw a diagram. Label every quantity that changes with a letter, and every quantity that stays fixed with its number. (The ladder's length never changes: write 10. The foot's distance from the wall does: write x.)
  2. Name the rates. Translate the words into derivatives: "pumped at 100 cm³/s" means \frac{dV}{dt} = 100; "how fast is the water level falling?" means find \frac{dh}{dt}. Note which rate you're given and which you want.
  3. Relate the variables with an equation that holds at every instant — Pythagoras, similar triangles, an area or volume formula.
  4. Differentiate both sides with respect to t — chain rule on every variable, since each is a hidden function of time.
  5. Substitute the snapshot values last. Only after differentiating do the particular numbers ("when r = 5…") enter the story.
  6. Interpret the answer — check the sign (growing or shrinking?) and state the units (a rate is always something per time).

The order of steps 4 and 5 is the whole game. The equation in step 3 is a statement about all time; the snapshot is one instant. Differentiate the eternal truth first, then freeze the frame.

Worked example 2 — the sliding ladder

A 10 m ladder leans against a wall. Its foot is pulled away from the wall at 1 m/s. How fast is the top sliding down when the foot is 6 m out?

Diagram and names. Let x be the foot's distance from the wall and y the top's height — both change. The ladder's length is constant, so it enters as the number 10, not a letter. We are given \frac{dx}{dt} = 1 and want \frac{dy}{dt} when x = 6.

Relate. Wall, ground and ladder form a right triangle at every instant, so Pythagoras gives x^2 + y^2 = 100.

Differentiate both sides with respect to t. The constant 100 has rate zero; each square picks up a chain-rule factor:

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \qquad\Longrightarrow\qquad \frac{dy}{dt} = -\frac{x}{y}\,\frac{dx}{dt}

Substitute the snapshot. When x = 6, Pythagoras fills in the missing side: y = \sqrt{100 - 36} = 8 (a 6–8–10 triangle). Then

\frac{dy}{dt} = -\frac{6}{8}(1) = -0.75 \text{ m/s}

Interpret. Negative: the top is moving downward at 0.75 m/s. Keep the sign — it's not decoration, it's the direction. And the formula holds a surprise: the factor x/y blows up as y \to 0. The foot slides out at a calm, steady 1 m/s the whole time, yet the top falls faster and faster, without limit, as the ladder nears flat. Watch it happen:

Equal steps of the foot, ever-bigger drops of the top. Two quantities chained by one equation do not have to change at the same pace — that's precisely what \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} quantifies: the exchange rate between the two speeds, and it shifts from instant to instant as x and y shift.

Worked example 3 — the draining conical tank

Water drains from an inverted conical tank (height 4 m, top radius 2 m) at 2 m³ per minute. How fast is the water level falling when the water is 2 m deep?

This one adds a twist the first two didn't have: three quantities change — the volume V, the depth h, and the radius r of the water's surface — but we only know one rate and want one other. The move is to eliminate the extra variable before differentiating. Step through the figure:

Relate — and reduce. The water always forms a cone similar to the tank, so its surface radius and depth stay in the tank's proportions: \frac{r}{h} = \frac{2}{4}, i.e. r = \frac{h}{2} at every instant. Substitute that into the cone volume formula:

V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^{\!2} h = \frac{\pi h^3}{12}

Differentiate both sides with respect to t.

\frac{dV}{dt} = \frac{\pi \cdot 3h^2}{12}\,\frac{dh}{dt} = \frac{\pi h^2}{4}\,\frac{dh}{dt}

Substitute the snapshot — draining means negative. The volume is decreasing, so \frac{dV}{dt} = -2 (the sign carries the physics!). At h = 2:

-2 = \frac{\pi (2)^2}{4}\,\frac{dh}{dt} = \pi\,\frac{dh}{dt} \qquad\Longrightarrow\qquad \frac{dh}{dt} = -\frac{2}{\pi} \approx -0.64 \text{ m/min}

Interpret. The level falls at about 0.64 metres per minute at that instant. And because of the h^2 in the denominator's role, the same outflow drops the level slowly when the tank is full (wide surface) and terrifyingly fast near the bottom (narrow surface) — the last metre of water vanishes far quicker than the first. Anyone who has watched a conical coffee filter drain has seen this equation in action.

A police radar gun doesn't measure a car's speed. It measures the rate of change of the straight-line distance between gun and car — \frac{ds}{dt} — which is a genuinely different number unless the car is driving straight at the gun. Set it up: the gun stands d metres off the road, the car is x metres along it, so s^2 = x^2 + d^2. Differentiate with respect to t: 2s\frac{ds}{dt} = 2x\frac{dx}{dt}, so

\frac{ds}{dt} = \frac{x}{s}\,\frac{dx}{dt}

Since x < s always (the hypotenuse beats the leg), the radar reads less than the true speed \frac{dx}{dt} — the famous "cosine effect". The error always favours the driver, which is why nobody has ever won a court case claiming the geometry overstated their speed. Traffic engineers correct for it with exactly the related-rates equation above.

Now scale up — a lot. In 1929 Edwin Hubble found that every distant galaxy recedes from us at a rate proportional to its distance: \frac{dD}{dt} = H_0 D, the Hubble law. That is a related-rates statement about the entire universe. And the balloon from our first example is the standard picture of it: paint dots on a balloon and inflate it — every dot recedes from every other, and dots twice as far apart separate twice as fast, purely because the space between them is stretching. Cosmologists run the relationship backwards in time (rates and all) to estimate the age of the universe: about 13.8 billion years. Same ritual, bigger balloon.

Watch it on Khan Academy