Unit 4: Contextual Applications of Differentiation

Related Rates

Solving related rates problems involving multiple changing quantities.

Advanced25 min lesson3 min readUpdated August 12, 2026Author not yet attributed

When Two Changing Quantities Are Linked by One Equation

A ladder's base sliding away from a wall and its top sliding down the wall aren't independent — the ladder's fixed length ties the two distances together at every instant. Before reading on: if an equation relates two quantities that both change with time, what would differentiating that equation with respect to time reveal?

Differentiating both sides of an equation with respect to time t — treating every changing quantity as an implicit function of t, the same way implicit differentiation treated y as a function of x — produces an equation relating the quantities' rates of change. That's a related rates problem: set up the equation connecting the quantities, differentiate with respect to t, then substitute the known values at the instant in question.

Worked Example — The Sliding Ladder

A 13-foot ladder leans against a wall. Its base slides away from the wall at 2 ft/sec. How fast is the top sliding down when the base is 5 feet from the wall? Let x = distance from wall to base, y = height of the top on the wall. The ladder's length is fixed: x² + y² = 13² = 169. Differentiate with respect to t: 2x(dx/dt) + 2y(dy/dt) = 0, so x(dx/dt) + y(dy/dt) = 0. At the instant in question, x = 5 and dx/dt = 2; find y from the constraint: 5² + y² = 169, so y² = 144, y = 12 (a 5-12-13 right triangle). Substitute: 5(2) + 12(dy/dt) = 0, so dy/dt = −10/12 = −5/6 ft/sec. The negative sign confirms y is decreasing — the top is sliding down at 5/6 ft/sec at that instant.

Worked Example — An Inflating Balloon

A spherical balloon's volume increases at 100 cm³/sec. How fast is its radius increasing when r = 5 cm? Volume: V = (4/3)πr³. Differentiate with respect to t: dV/dt = 4πr²(dr/dt). Substitute dV/dt = 100 and r = 5: 100 = 4π(25)(dr/dt) = 100π(dr/dt), so dr/dt = 100/(100π) = 1/π ≈ 0.318 cm/sec.

Tip

Only substitute the specific numerical values (like x = 5) after differentiating, never before — differentiating a number instead of a variable that's still changing at that instant silently throws away the related rate you're solving for.

Common Mistakes

Plugging in the known value of x before differentiating the constraint equation.
x is still changing at the moment being analyzed — substituting x = 5 before differentiating treats it as a constant and produces dx/dt = 0 in the derivative, losing the relationship entirely. Differentiate first, substitute second.
Forgetting the chain rule's extra rate-of-change factor when differentiating a squared or cubed quantity with respect to t.
Just like implicit differentiation with respect to x, differentiating r³ with respect to t gives 3r²(dr/dt), not just 3r² — every quantity changing with time picks up its own rate-of-change factor.

Key Takeaways

A related rates problem starts from an equation connecting two or more quantities, then differentiates that equation with respect to time to relate their rates of change.
Specific numerical values are substituted only after differentiating — differentiating first preserves the still-changing variable's rate.
Differentiating with respect to t treats every quantity as an implicit function of time, so each squared or cubed term picks up its own rate-of-change factor via the chain rule.

Summary

Related rates connect two changing quantities at a shared instant. The next lesson uses the derivative differently — to approximate a function's value near a point, using nothing but its tangent line.

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