A streetlight stands 20 feet tall. A person 6 feet tall walks away from it at 5 feet per second. How fast is the person's shadow growing? This is a related rates problem—one of calculus's most practical applications, combining geometric relationships with differentiation through the chain rule to answer questions about how fast things change in the real world.
The Method
Related rates problems follow a consistent procedure. First, draw a diagram and identify all variables. Second, write an equation relating the variables—often from geometry, the Pythagorean theorem, or similar triangles. Third, differentiate both sides with respect to time, using the chain rule for every variable that changes with time. Fourth, substitute known values and solve for the unknown rate. The key insight is that every changing quantity is implicitly a function of time, so every variable in your equation has a time derivative.
The Shadow Problem Solved
Let x be the distance from the person to the streetlight, and s be the length of the shadow. By similar triangles: 20/(x + s) = 6/s. Cross-multiplying: 20s = 6x + 6s, giving 14s = 6x, so s = (3/7)x. Now differentiate both sides with respect to time: ds/dt = (3/7)(dx/dt). With dx/dt = 5 ft/s: ds/dt = (3/7)(5) = 15/7 ≈ 2.14 ft/s. The shadow grows at about 2.14 feet per second—remarkably, at a constant rate regardless of how far the person has walked.
The Balloon Problem
A spherical balloon is inflated at 100 cubic centimeters per second. How fast is the radius growing when the radius is 5 cm? The relationship: V = (4/3)πr³. Differentiating with respect to time: dV/dt = 4πr² × dr/dt. With dV/dt = 100 and r = 5: 100 = 4π(25)(dr/dt), giving dr/dt = 100/(100π) = 1/π ≈ 0.318 cm/s. Notice that as the balloon grows, dr/dt decreases—the same volume added per second produces a smaller radius change when the balloon is larger, because the surface area 4πr² has increased.
The Ladder Problem
A 10-foot ladder leans against a wall. The bottom slides away at 2 ft/s. How fast is the top sliding down when the bottom is 6 feet from the wall? By Pythagoras: x² + y² = 100. Differentiating: 2x(dx/dt) + 2y(dy/dt) = 0. When x = 6: y = sqrt(100 - 36) = 8. So: 2(6)(2) + 2(8)(dy/dt) = 0, giving dy/dt = -24/16 = -1.5 ft/s. The negative sign confirms the top slides downward. Notably, the top speeds up as the ladder becomes more horizontal—an effect clearly captured by the mathematics.
Applications Beyond Textbooks
Related rates appear constantly in applied contexts. In medicine, clinicians use related rates reasoning to understand how drug concentration changes as it's simultaneously absorbed and metabolized. Engineers calculate how quickly a cone-shaped pile of sand grows as material is added at a known rate. In fluid dynamics, the continuity equation connecting fluid velocity to cross-sectional area is a related rates equation. Any situation where multiple changing quantities are geometrically or physically linked requires this kind of reasoning.
Conclusion
Related rates problems are calculus at its most applied—transforming geometric or physical relationships into statements about rates of change. The chain rule is the key that connects rates of different quantities, and the skill lies in correctly identifying and differentiating the underlying relationship between variables. From the shadow growing behind you on a sunny day to the changing depth of water in a draining tank, related rates are the mathematical language for describing how the world's interconnected quantities evolve together through time.