Related Rates – Solving

Related rates problems ask you to find how fast one quantity is changing given information about how fast another quantity is changing, when both quantities are linked by some geometric or physical relationship. For example, if you know how fast the radius of a circle is growing, you can find how fast the area is growing — even though area and radius change at different speeds. The key insight is that both quantities are functions of time, so their rates of change are connected through differentiation. The technique relies entirely on implicit differentiation with respect to time. You start with an equation that relates the two (or more) quantities — such as the Pythagorean theorem, a volume formula, or a trigonometric relationship — and then differentiate both sides with respect to t. This produces a new equation connecting the rates of change (derivatives like dx/dt and dy/dt), which you can solve algebraically once you substitute the known values. Related rates problems appear throughout physics, engineering, and everyday life: a ladder sliding down a wall, a spherical balloon being inflated, a shadow lengthening as a person walks away from a lamp post, or water draining from a conical tank. Mastering the solution strategy — draw, label, relate, differentiate, substitute, solve — will let you handle all of these systematically.

Key takeaways
  • Draw and label a diagram
  • Identify what is given and what is wanted
  • Write an equation relating the variables
  • Differentiate both sides with respect to time t
  • Substitute known values
  • Solve and interpret

Draw and label a diagram

Sketch the situation at a general moment in time (not just the specific instant you care about). Label all quantities that change with variables (x, y, r, h, etc.). Label any quantities that are constant with their numerical values. A clear diagram prevents sign errors and helps you see which geometric relationship to use.

Identify what is given and what is wanted

Write down explicitly which derivative is given (e.g., dr/dt = 3 cm/s) and which derivative you need to find (e.g., dV/dt = ?). Also note the specific values of the variables at the instant in question (e.g., r = 5 cm at the moment you care about). Do NOT substitute these values yet — they go in only after differentiation.

Write an equation relating the variables

Find a single equation that connects all the variables you labeled. Use geometry (Pythagorean theorem, similar triangles, area/volume formulas) or the physical laws of the problem. The equation must be true at every moment in time, not just the specific instant. For instance, for a sphere: V = (4/3)πr³. For a right triangle with legs x and y: x² + y² = L² (if the hypotenuse L is constant).

Differentiate both sides with respect to time t

Apply d/dt to every term on both sides using the chain rule. Remember that each variable is an implicit function of t, so d/dt[r³] = 3r²(dr/dt), not just 3r². Constants (like π or a fixed length L) have zero derivative. This step produces an equation involving the rates of change alongside the variables themselves.

Substitute known values

Now substitute the specific numerical values of the variables at the instant of interest, as well as the given rate(s). This turns the differentiated equation into an algebraic equation with only one unknown — the rate you are looking for.

Solve and interpret

Solve algebraically for the unknown rate. Include correct units (always a rate, such as cm/s, m²/min, etc.). Check the sign: a positive value means increasing, a negative value means decreasing. State your answer in a full sentence contextualizing it within the original problem.

The Key Equation

dVdt=4πr2drdt\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}
  • \frac{dV}{dt} — Rate of change of volume (cm³/s): How quickly the volume of the sphere is increasing or decreasing with respect to time.
  • r — Radius (cm): The current radius of the sphere at the instant of interest.
  • \frac{dr}{dt} — Rate of change of radius (cm/s): How quickly the radius is increasing or decreasing with respect to time; this is the given rate in balloon-inflation problems.
  • 4\pi r^2 — Surface area of the sphere (cm²): Geometrically, this factor is the surface area of the sphere, showing that volume change equals surface area times radial expansion rate.

Worked Example

Problem. A ladder 10 m long leans against a vertical wall. The bottom of the ladder slides away from the wall at a constant rate of 0.5 m/s. How fast is the top of the ladder sliding down the wall at the instant when the bottom is 6 m from the wall?

dydt=−38 m/s\frac{dy}{dt} = -\frac{3}{8} \text{ m/s}

Answer. The top of the ladder is sliding down the wall at 0.375 m/s when the base is 6 m from the wall.

Figure
Diagram
x = 6 myL = 10 m↔ dx/dt = 0.5 m/s↕ dy/dt = ?●Right-triangle relation: x² + y² = L² at every instant t
Origins · 1 minGottfried Wilhelm Leibniz and how this was discovered
GW
Gottfried Wilhelm Leibniz
1646–1716 · German

Leibniz independently co-invented calculus and introduced the notation dy/dx and d/dt that we still use today. His notation makes implicit differentiation and related rates completely natural: writing dV/dt immediately signals 'the rate of change of V with respect to time', making the chain rule easy to apply symbolically. Leibniz was also a diplomat, philosopher, and librarian, and he built one of the first mechanical calculators capable of multiplication — he called it the 'Step Reckoner' in 1672, more than a decade before he published his calculus.

  • Implicit Differentiation — Related rates problems require differentiating both sides of an equation with respect to t, treating all variables as implicit functions of time — exactly the technique of implicit differentiation.
  • The Chain Rule — When you differentiate x²(t) with respect to t to get 2x(dx/dt), you are applying the chain rule. Every related-rates step of differentiation relies on this rule.
  • Related Rates – Setting Up — Before you can solve a related-rates problem, you must correctly identify variables, draw a diagram, and write the geometric equation that links them — the setup stage that precedes all the solving steps here.
  • Pythagorean Theorem — Many classic related-rates problems (ladders, distance problems) use x² + y² = c² as the linking equation, so comfort with the Pythagorean theorem is essential.