Related Rates

When two changing quantities are tied together by an equation, their rates of change are tied together too. Differentiate the equation with respect to time — every changing variable picks up its own rate via the chain rule — and a known rate hands you the unknown one.

By the end you'll be able to run the differentiate-then-substitute procedure: relate the variables, differentiate with respect to \(t\), plug in the instant's values last, and solve for the unknown rate with correct sign and units.

Predict: the base slides out at a steady 2 ft/s. Does the top slide down at a steady speed too? Predict what happens to the top's speed as the base nears 13 ft — then drag the base outward and watch \(|dy/dt|\) grow without bound.

A 13 ft ladder leans on a wall. The base distance x and the top height y are chained by \(x^2 + y^2 = 169\), so setting \(x\) forces \(y = \sqrt{169 - x^2}\) — and forcing the rates: \(2x\,x' + 2y\,y' = 0\) gives the top's rate \(y' = -(x/y)\,x'\). Hover or focus either endpoint for exact readouts.

x = 5.00 ft · y = 12.00 ft · top speed |dy/dt| = 0.83 ft/s

x² + y² = 169 — the ladder, its base rate (fixed), and the induced top rate
base distance x (rate x′ = 2 ft/s) top height y top rate y′ = −(x/y)·2

Related-rates problems differentiate an equation relating two or more quantities with respect to time, then use known rates to solve for an unknown rate at a given instant.

Applied

A ladder slips, a balloon inflates, a cone of sand grows — in each, one measurable rate (how fast the base slides, how fast air pumps in) secretly sets another (how fast the top drops, how fast the radius grows). The recipe: draw and label the changing quantities, find one equation relating them (Pythagoras, a volume formula, a trig ratio), differentiate with respect to time, then substitute the instant's values and solve. Attach units at the end — a rate answer without units is half an answer, and a quantity that's shrinking must carry a negative rate.

Formal

Treat every changing quantity as a function of \(t\) and differentiate the relation term by term — this is implicit differentiation with \(t\) as the variable. Just as you treated \(y\) as \(y(x)\) and gave it a \(dy/dx\) factor there, here every variable is a function of time, so each carries its own \(dx/dt\) via the chain rule: \(\frac{d}{dt}\big[x^2\big] = 2x\,\frac{dx}{dt}\). The crucial discipline: substitute numbers only after differentiating. A changing quantity must stay a variable while you differentiate — plug in early and the derivative treats it as a constant, silently deleting its rate term from the equation.

Worked example

A 13 ft ladder leans against a wall; its base slides away at \(dx/dt = 2\) ft/s. How fast does the top slide down when \(x = 5\) ft?
Relate: \(x^2 + y^2 = 169\). Differentiate: \(2x\,\dfrac{dx}{dt} + 2y\,\dfrac{dy}{dt} = 0\). Substitute the instant: at \(x = 5\), \(y = \sqrt{169 - 25} = 12\), so \(2(5)(2) + 2(12)\,\dfrac{dy}{dt} = 0\). Solve: \(\dfrac{dy}{dt} = -\dfrac{20}{24} = -\dfrac{5}{6}\) ft/s — the top descends at 5/6 ft/s. The minus sign is the geometry talking: \(y\) is shrinking.

Your turn

Same 13 ft ladder, same 2 ft/s base rate — but now the base is 12 ft from the wall. Differentiating gave \(2x\,\dfrac{dx}{dt} + 2y\,\dfrac{dy}{dt} = 0\). At \(x = 12\): \(y = \sqrt{169 - 144} = 5\). So \(2(12)(2) + 2(5)\,\dfrac{dy}{dt} = 0\), which means \(\dfrac{dy}{dt} = \) ____

Reveal the answer

\(\dfrac{dy}{dt} = -\dfrac{48}{10} = -4.8\) ft/s — the top drops at 4.8 ft/s, nearly six times faster than at \(x = 5\), with the same base speed. Set the slider to 12 above and check the readout. That's the blow-up you predicted: as \(y \to 0\), the factor \(x/y\) explodes.

More info — this is implicit differentiation wearing a hat

In the implicit differentiation lesson you differentiated a relation like \(x^2 + y^2 = 25\) with respect to \(x\), treating \(y\) as an unknown function \(y(x)\) — so \(y^2\) produced \(2y\,\dfrac{dy}{dx}\) by the chain rule. Related rates is the identical move with \(t\) as the differentiation variable and every quantity treated as a function of time: \(x^2\) produces \(2x\,\dfrac{dx}{dt}\) and \(y^2\) produces \(2y\,\dfrac{dy}{dt}\). Nothing new was invented — you're reusing the differentiate-a-relation machinery, then solving for the one rate you don't know instead of for \(dy/dx\). The Paul's Online Notes link below runs this same parallel across ladder and cone problems.

Check your understanding

Question 1 of 4

A 10 ft ladder leans against a wall. Its base slides away from the wall at 3 ft/s. How fast is the top sliding down the wall when the base is 6 ft out?

Question 2 of 4

In a related-rates problem, why must you substitute the instant's numeric values only AFTER differentiating the relating equation?

Question 3 of 4

Both \(x\) and \(y\) change with time and satisfy \(x^2 + y^2 = 169\). Differentiating both sides with respect to \(t\) — the same move as implicit differentiation — gives which equation?

Question 4 of 4

For the 13 ft sliding ladder with its base moving out at a constant 2 ft/s, what happens to the top's speed \(|dy/dt|\) as the base approaches 13 ft from the wall?

Recap

  • Procedure: draw and label → find an equation relating the variables → differentiate both sides with respect to \(t\) → substitute the instant's values → solve for the unknown rate, with units.
  • Every changing variable carries its own rate factor via the chain rule: \(\frac{d}{dt}[x^2] = 2x\,\frac{dx}{dt}\) — the same move as implicit differentiation, with \(t\) as the variable.
  • Substitute numbers only after differentiating — plugging in early treats a changing quantity as constant and deletes its rate term.
  • Signs carry meaning: a shrinking quantity has a negative rate. And a constant input rate does not mean a constant output rate — the ladder's top speed \((x/y)\,x'\) blows up as \(y \to 0\).

Dive deeper

Sources

  • Related Rates