Find the rate of change of one quantity given the rate of another using implicit differentiation.
In related rates problems, multiple quantities change with time. We differentiate a geometric or physical equation with respect to time t using implicit differentiation, then substitute known rates (dX/dt) to find the unknown rate.
Related Rates Strategy
A spherical balloon is inflated at 100 cm³/s. How fast is the radius growing when r = 5 cm? V = (4/3)πr³ → dV/dt = 4πr²(dr/dt). Substitute: 100 = 4π(25)(dr/dt) → dr/dt = 100/(100π) = 1/π ≈ 0.318 cm/s.
A 10-ft ladder leans against a wall. The bottom slides away at 2 ft/s. How fast does the top slide down when the bottom is 6 ft from the wall? x² + y² = 100 → 2x(dx/dt) + 2y(dy/dt) = 0. At x=6: y=8. So 2(6)(2) + 2(8)(dy/dt) = 0 → dy/dt = −24/16 = −3/2 ft/s.
Remember This!
Identify what is given (a rate like dx/dt) and what is asked (another rate like dy/dt). Negative rates mean a quantity is decreasing. Always state units in your final answer.
A square's side grows at 3 cm/s. How fast is its area increasing when the side is 4 cm?