Related rates

A ladder slides down a wall. A balloon inflates. A shadow lengthens. In each, two quantities are locked together by geometry, and knowing how fast one changes tells you how fast the other must.

The technique is one idea: differentiate the relationship with respect to time. Every variable becomes a rate via the chain rule, which is why chapter 2.3 was worth the effort.

The method

1. Draw it, and name the changing quantities as variables. Not numbers — variables.
2. Write the equation relating them. Usually Pythagoras, a volume formula, or similar triangles.
3. Differentiate both sides with respect to $t$. Every variable picks up a $\tfrac{d\ }{dt}$ factor.
4. Substitute the numbers — now, and not before.
5. Solve, and check the sign.

Step 4 is where most marks are lost, and it deserves its own warning.

The ladder

A $10$ m ladder leans against a wall. The base slides out at $2$ m/s. How fast is the top falling when the base is $6$ m from the wall?

With $x$ the base distance and $y$ the height, Pythagoras gives $x^2 + y^2 = 100$. The $100$ is a genuine constant — the ladder does not stretch — so it may stay.

Differentiating with respect to $t$:

$$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$$

Now substitute: $x = 6$, so $y = 8$, and $\tfrac{dx}{dt} = 2$:

$$2(6)(2) + 2(8)\frac{dy}{dt} = 0 \;\Longrightarrow\; \frac{dy}{dt} = -\frac{24}{16} = -1.5$$

The top falls at $1.5$ m/s. The minus sign is the answer, not an error — $y$ is decreasing. A related-rates answer whose sign contradicts the physical story means a slip somewhere.

The cone, and why the rate is not constant

Water pours into an inverted cone at $3$ m³/min. The cone is $4$ m tall with a $2$ m top radius. How fast is the level rising when the water is $2$ m deep?

$V = \tfrac13\pi r^2 h$ has two variables, and related rates wants one. Similar triangles give $\tfrac{r}{h} = \tfrac{2}{4}$, so $r = \tfrac{h}{2}$:

$$V = \frac{1}{3}\pi\left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{12}$$

So $\tfrac{dV}{dt} = \tfrac{\pi h^2}{4}\tfrac{dh}{dt}$, and at $h = 2$ with $\tfrac{dV}{dt} = 3$:

$$3 = \frac{\pi(4)}{4}\frac{dh}{dt} \;\Longrightarrow\; \frac{dh}{dt} = \frac{3}{\pi} \approx 0.95\ \text{m/min}$$

Note what the $h^2$ means physically: the level rises more slowly as the cone fills, because each extra centimetre of depth needs more water than the last. Pouring at a constant rate does not raise the level at a constant rate, and the equation says so.

Reading the signs

Write the sign into the setup rather than patching it at the end: a ladder base moving away has $\tfrac{dx}{dt} = +2$, and a balloon losing air has $\tfrac{dV}{dt} < 0$. Getting the signs right at step 1 makes step 5 a check rather than a guess.