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Related Rates

The calculus here is easy — a single implicit differentiation. The whole challenge is the setup: seeing the scene, finding the one equation that ties the quantities together, and differentiating it in time so their rates lock. Master the setup and every problem is the same problem.
When two quantities are tied by an equation and both change with time, differentiating that equation with respect to t chains their rates together — know one, get the other. Like meshed gears: turn one and the other must turn, at a rate the linkage fixes. Your job is never the derivative; it's finding the linkage and reading it at the right instant.

The recipe — every related-rates problem

Four steps, always in this order. The one that ends careers is doing step 4 too early.

STEP 1

Name & sort

Label the changing quantities. Mark which rate is given and which you want. Note what's constant.

STEP 2

Find the relation

Write the one equation linking the quantities — geometry, a volume formula, or similar triangles. Eliminate extra variables now.

STEP 3

Differentiate d/dt

Differentiate both sides with respect to t. Chain rule on every variable — each gets its own rate.

STEP 4

Substitute — last

Only now plug in the instant's numbers and solve for the wanted rate. Substitute earlier and you freeze a moving quantity.

The Related-Rates Lab

Pick a scene. Play it — the quantity you control changes at a steady rate you set, and the linked rate is computed live from the equation, so you watch them lock. The four setup steps fill in beside it. Then build the setup yourself in the challenge below.

given rate
given
you set this — steady
found
locked by the equation
Build the setupStep 1 of 4
Four scenes, one method. A ladder, a ripple, a cone, a shadow — every one was the same four steps: name the quantities, find the equation that links them, differentiate in time, substitute last. Learn the steps, not the answers, and there's no related-rates problem you can't set up.

Where it shows up

This isn't a textbook trick. Anywhere two changing quantities obey one law, related rates reads a hidden rate off a measured one — often in real time, often when it matters.

Air-traffic control

A controller can't measure two jets' closing speed directly — she computes it. Their positions are tied by the distance formula, so one differentiation turns each plane's known speed into how fast the gap is shrinking. Safe separation versus near-miss.

Oil-spill response

A slick spreads as a growing circle. Responders need dA/dt — how fast the contaminated area expands — to size booms and crews. It falls straight out of A = πr² and a measured dr/dt: the ripple you dragged, at industrial stakes.

Reservoir management

Operators release water at a controlled dV/dt, but what they watch is the level. The basin's shape links the two — and because it widens with height, the same outflow drops the level far faster when the reservoir runs low.

Tracking a launch

A ground camera holds a rising rocket in frame by matching its angle of elevation to the rocket's height: tan θ = h/d. Differentiate and the climb rate dh/dt sets exactly how fast the camera must pivot — quick near the pad, easing as it climbs away.

Set these up yourself

Three more, in words. For each, write the relation and its t-derivative before revealing — the numbers are the easy part.

Quick reps — name the relation

Recognition drills. For each, decide the equation and its t-derivative in your head, then reveal to check — this is the fast half of every problem.

The capstone challenge

Everything at once. A balloon rises straight up while you watch from 500 ft away — and two different rates hide in the same scene: how fast it pulls away from you, and how fast you must tilt your gaze to follow it. Reveal each part step by step, then let the scene lock to the instant and confirm both answers.

A weather balloon rises straight up from a launch point; you watch from a fixed d = 500 ft away on level ground. It climbs at a steady dh/dt = 30 ft/s. At the instant it reaches h = 500 ft: (a) how fast is its straight-line distance from you increasing? (b) how fast is your angle of elevation increasing?

Part (a) · how fast it pulls away

Part (b) · how fast your gaze tilts

Common mistakes & exam tips

Substituting before differentiating

The cardinal sin. If you plug in x = 3 before taking d/dt, you've frozen x — its rate vanishes and the whole problem collapses. Differentiate first, substitute last.

Forgetting a variable's rate

Every changing variable gets a rate when you differentiate. In x²+y²=L², both x and y change, so both dx/dt and dy/dt appear. Dropping one is the most common slip.

Not eliminating extra variables

In the cone, V = ⅓πr²h has two changers. Use similar triangles (r = h/2) to get V in terms of h alone before differentiating — otherwise you'd need dr/dt too.

Losing the chain rule

d/dt of r² is 2r·(dr/dt), not 2r. Every derivative is with respect to t, so a factor of the variable's rate rides along. Missing it is a silent, fatal error.

Sign confusion

A shrinking quantity has a negative rate. The ladder's height falls, so dy/dt < 0. Let the equation produce the sign; don't force it.

Answering the wrong rate

"How fast is the shadow's tip moving" is not "how fast is the shadow lengthening." Re-read what's asked — tip speed is dx/dt + ds/dt, not ds/dt.

On the AP exam

You can now turn any changing scene into an equation and read its locked rates. Next, we ride a single rate — the slope at a point — to estimate values a calculator would need for: continue to Linear Approximation →, or head back to the Unit 4 Guide.