A sphere's radius grows at 2 cm/s; how does the chain rule give the rate of volume change?

## What the question is really testing Related rates is the simplest place where the chain rule stops being a symbol-pushing rule and becomes a statement about how a change propagates through a composition. You have a quantity you care about (volume), a quantity that drives it (radius), and a driver of that (time). The chain rule says the rates multiply. ## The setup The radius is a function of time, `r(t)`, and the volume is a function of the radius, `V(r) = (4/3) * pi * r^3`. What you actually have is the composition `V(r(t))` — a function of time. Its derivative is `dV/dt = (dV/dr) * (dr/dt)` read as: how fast V changes per unit of r, times how fast r changes per unit of t. The units make the composition visible: (cm^3 per cm) times (cm per second) = cm^3 per second. ## Doing the arithmetic Differentiate the volume formula with respect to the radius: `dV/dr = 3 * (4/3) * pi * r^2 = 4 * pi * r^2` which is exactly the surface area of the sphere — a nice check, since growing the radius by a sliver dr adds a shell of volume (surface area) times dr. Substitute the given `dr/dt = 2` cm/s: `dV/dt = 4 * pi * r^2 * 2 = 8 * pi * r^2` cm^3/s This is still a function of r, so it must be evaluated at the radius of interest. At r = 5 cm: `8 * pi * 25 = 200 * pi`, about 628 cm^3/s. At r = 10 cm it is 800*pi — four times larger, from the same radius rate. A constant inflation rate in the radius produces an accelerating volume rate. ## The mistake the rule exists to prevent The classic error is to differentiate the volume formula and stop, writing `dV/dt = 4 * pi * r^2`. That silently assumes `dr/dt = 1`. The factor `dr/dt` is the inner derivative, and forgetting it is the single most common chain-rule failure. A second error is to multiply the volume itself by the rate — `(4/3) * pi * r^3 * 2` — which has units of cm^3 per second only by accident and is not a derivative of anything. ## Notation and rigour The Leibniz form `dV/dt = (dV/dr)(dr/dt)` looks like the dr symbols cancel. That mnemonic is harmless in one variable, where every factor is an ordinary number, but it is a mnemonic, not a proof, and it stops working once the intermediate quantity is a vector — there the factors become matrices, which do not cancel and do not commute. Writing the rule in function form makes the honest statement clear: if `F(t) = V(r(t))`, then `F'(t) = V'(r(t)) * r'(t)`, and note that the outer derivative is evaluated at `r(t)`, not at t. ## More than one driver The same idea extends when the target depends on several changing quantities. For a cylinder of radius r and height h, `V = pi * r^2 * h`, and if both change with time the multivariable chain rule adds one term per path: `dV/dt = (dV/dr) * (dr/dt) + (dV/dh) * (dh/dt) = 2 * pi * r * h * (dr/dt) + pi * r^2 * (dh/dt)` Every way time can reach V contributes a product of rates, and the contributions add. That sum-over-paths structure is exactly what the matrix form of the chain rule encodes once inputs and outputs become vectors: each entry of the resulting matrix is one of these sums. ## How to answer under pressure Name the composition out loud (volume through radius through time), write the chain rule symbolically before plugging in numbers, differentiate the geometry formula, substitute the given rate, and finish by checking the units and evaluating at the stated radius. Stating that the answer still depends on r — that there is no single number unless a radius is specified — is what separates a confident answer from a memorised one.