Related Rates and Units

Several quantities move together

A related-rates problem links time-dependent quantities through a relation that remains true as the system changes. The relation might be a volume formula, a distance constraint or a conservation law. The unknown is a rate at a particular instant, not a whole function of time.

Write the relation using variable quantities. Differentiate with respect to time, including a chain-rule factor for every changing variable. Only then insert the values and rates given for the instant of interest. Inserting a temporary measurement too early can make a changing quantity appear constant.

Visual guide

VISUAL GUIDEA sliding ladder links two rates

Ladder

A ladder of length 5 has foot distance x = 3 and height y = 4 at this instant. The constraint x² + y² = 25 gives 2x·x′ + 2y·y′ = 0. Moving the foot outward forces the top downward.

Worked example: a sliding ladder

A $5$ -meter ladder leans against a vertical wall. Its foot is $x$ meters from the wall and its top is at height $y$ . The fixed length gives $x^2+y^2=25$ . Differentiate:

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

When the foot is $3$ meters out, the height is $4$ meters. If the foot moves away at $dx/dt=0.5$ m/s, then

\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}=-\frac38\text{ m/s}.

The negative sign means the top moves downward in our coordinate system. Saying “speed downward” would instead report the positive magnitude $3/8$ m/s with the direction stated separately.

Worked example: a filling conical tank

A cone has height $6$ m and top radius $3$ m. At water depth $h$ , similar triangles give $r/h=1/2$ , hence $r=h/2$ . The water volume is

V=\frac13\pi r^2h=\frac{\pi h^3}{12}.

Thus $dV/dt=(\pi h^2/4)(dh/dt)$ . If water enters at $2$ m³/min and $h=2$ m, then $dh/dt=2/\pi$ m/min. The changing water surface has a changing radius; using the tank's full radius $3$ at every depth would give the wrong model.

Check the answer as a rate

The relation should be dimensionally consistent before and after differentiation. Volume rate divided by surface area has units of length per time, as in the tank example. A denominator approaching zero can produce a large predicted rate; that is a model behavior to interpret, not a reason to discard the calculation automatically.

Remember that some rates may be negative: a draining volume, shortening distance or falling height. Define your coordinate directions before assigning signs.

Practice

A circle's radius grows at $2$ cm/s. Find its area rate when $r=3$ cm.
A sphere expands at $dV/dt=12\pi$ cm³/s. Find $dr/dt$ when $r=2$ cm.
In the ladder example, what goes wrong if $x=3$ is substituted before differentiating?

Show worked solutions

$A'=2\pi rr'=12\pi$ cm²/s.
From $V'=4\pi r^2r'$ , obtain $r'=12\pi/(16\pi)=3/4$ cm/s.
The equation becomes $9+y^2=25$ , falsely treating $x$ as fixed and eliminating its nonzero contribution to the rate.