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Daily Math Minute

Applications of Derivatives

Related Rates

Solving problems involving rates of change of related quantities.

Advanced25 min lesson2 min readUpdated August 12, 2026Author not yet attributed

Prerequisites

  • Optimization Problems

When an Equation Ties Two Changing Quantities Together

Definition — Related Rates

Differentiating both sides of an equation with respect to time t — treating each changing quantity as an implicit function of t — relates their rates of change. Numerical values are substituted only after differentiating.

Worked Example — A Sliding Ladder

A 25-foot ladder leans against a wall; its base slides away at 3 ft/sec. How fast is the top sliding down when the base is 15 feet from the wall? Let x = base distance, y = height on the wall: x² + y² = 625. Differentiate: x(dx/dt) + y(dy/dt) = 0. At this instant, x = 15, dx/dt = 3; find y: 15² + y² = 625, so y² = 400, y = 20 (a 15-20-25 right triangle). Substitute: 15(3) + 20(dy/dt) = 0, so 45 + 20(dy/dt) = 0, giving dy/dt = −45/20 = −2.25 ft/sec — the top is sliding down at 2.25 ft/sec.

Worked Example — A Growing Oil Spill

A circular oil spill's radius grows at 2 m/min. How fast is the area growing when the radius is 10 m? A = πr². Differentiate: dA/dt = 2πr(dr/dt). Substitute r = 10, dr/dt = 2: dA/dt = 2π(10)(2) = 40π ≈ 125.7 m²/min.

Tip

Substitute specific numerical values only after differentiating — substituting first treats a still-changing quantity as constant and loses its rate.

Common Mistakes

  • Plugging in the known value of x before differentiating the constraint equation.

    x is still changing at the instant being analyzed — differentiate the general relationship first, then substitute the specific values.

Key Takeaways

  • A related rates problem differentiates a constraint equation with respect to time to connect two rates of change.
  • Numerical substitution always happens after differentiating, never before.

Summary

This closes the study of derivative applications. The next unit turns to the opposite question — recovering accumulated quantities from a rate of change.

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