In many problems two quantities change together: as a ladder slides, its top and bottom both move; as a balloon inflates, its radius and volume both grow. If an equation links the quantities, Implicit Differentiation with respect to time links their rates. The Position, Velocity and Acceleration lesson gave us the language of rates in time.
Key idea
Related rates method. Name every changing quantity as a function of time $t$. Write an equation that relates them and holds at every moment. Differentiate both sides with respect to $t$ (chain rule). Only then substitute the values for the moment you care about, and solve for the unknown rate.
Why it works
The equation $x^2 + y^2 = 100$ is true at every instant, so its two sides change at the same rate: their derivatives with respect to $t$ are equal. That is why we may differentiate it. A statement like $x = 6$ is true only at one instant; it does not hold over time, so differentiating it ($\frac{dx}{dt} = 0$) would claim the ladder's foot is not moving, which is false.
Worked example: A sliding ladder
A 10-ft ladder leans on a wall. The foot slides away at 2 ft/s. How fast is the top moving down when the foot is 6 ft from the wall?
- Let $x$ = distance of the foot, $y$ = height of the top. At all times $x^2 + y^2 = 100$.
- Differentiate in $t$: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$.
- Now the moment: $x = 6$, so $y = \sqrt{100 - 36} = 8$; and $\frac{dx}{dt} = 2$.
- $2(6)(2) + 2(8)\frac{dy}{dt} = 0$, so $\frac{dy}{dt} = -\frac{24}{16} = -1.5$.
Answer: The top slides down at 1.5 ft/s ($\frac{dy}{dt} = -1.5$).
Worked example: A growing balloon
Air is pumped into a spherical balloon at 100 cm³/s. How fast is the radius growing when $r = 5$ cm?
- $V = \frac{4}{3}\pi r^3$ at all times.
- Differentiate: $\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$.
- Substitute: $100 = 4\pi(25)\frac{dr}{dt}$, so $\frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi}$.
Answer: $\frac{1}{\pi} \approx 0.32$ cm/s
Worked example: Filling a cone
Water flows at 2 m³/min into a cone-shaped tank (point down) with top radius 3 m and height 6 m. How fast is the water rising when it is 4 m deep?
- Similar triangles: the water's radius is $r = \frac{3}{6}h = \frac{h}{2}$.
- So $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi\frac{h^2}{4}h = \frac{\pi h^3}{12}$, using only $h$.
- Differentiate: $\frac{dV}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt}$.
- Substitute $h = 4$, $\frac{dV}{dt} = 2$: $2 = 4\pi\frac{dh}{dt}$, so $\frac{dh}{dt} = \frac{1}{2\pi}$.
Answer: $\frac{1}{2\pi} \approx 0.16$ m/min
Common mistake
Substituting the given values before differentiating. In the ladder problem, writing $6^2 + y^2 = 100$ first and then differentiating loses $\frac{dx}{dt}$ entirely, because 6 is a constant. Check your order of work: the equation you differentiate should contain only letters for the changing quantities; numbers for one instant come in only after.