Related rates

“Observe constantly that all things take place by change…”

— Marcus Aurelius

One equation, several moving quantities

A ladder slides down a wall. The ladder length stays fixed, while its horizontal and vertical reaches change together. Knowing how fast one end moves lets us find the rate of the other end because the geometry constrains both.

In a related-rates problem, quantities depend on a common variable, usually time. We first connect the quantities, then differentiate that relationship to connect their rates.

Prerequisite

Differentiate with respect to time

If \(x=x(t)\), the chain rule gives

\[\frac{d}{dt}(x^2)=2x\frac{dx}{dt}.\]

The symbol \(x\) may denote a length, but it still represents a time-dependent function. By contrast, a length fixed throughout the motion has time derivative zero.

Follow the ladder

Let a \(5\text{ m}\) ladder have its foot \(x(t)\) metres from a wall and its top at height \(y(t)\) metres. We model the ladder as a rigid segment with both ends staying in contact with the wall and floor. Then

\[x^2+y^2=25,\qquad x>0,\quad y>0.\]

Differentiate before inserting the values at a particular instant:

\[2x\frac{dx}{dt}+2y\frac{dy}{dt}=0,\] \[\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}.\]

The factor \(x/y\) depends on the configuration. Near the wall it is small; near the floor it is large. Equal changes of the foot do not produce equal changes of the top.

Example 1

Evaluate a rate at one instant

The foot moves away at \(dx/dt=0.4\text{ m/s}\). At \(x=3\text{ m}\), the geometry gives \(y=\sqrt{25-9}=4\text{ m}\). Therefore

\[\frac{dy}{dt}=-\frac34(0.4)=-0.3\text{ m/s}.\]

The signed rate is negative because height decreases. The speed of the top is the positive magnitude \(0.3\text{ m/s}\).

Extra reading: could the top move faster than light?

The formula predicts it! If the foot keeps moving outward at a fixed speed \(u>0\), then

\[\left\lvert\frac{dy}{dt}\right\rvert=\frac{x}{y}u\longrightarrow\infty \qquad\text{as }y\to0^+.\]

The unbounded speed signals a limit of the model. A real ladder cannot remain rigid, keep both contacts and maintain a fixed foot speed all the way to horizontal. An object with mass cannot be accelerated to light speed.

If the top’s downward speed instead stays finite near the floor, the same geometric relationship requires

\[\frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}\longrightarrow0.\]

The foot must slow down as the ladder becomes horizontal. At \(y=0\) itself, the formula that divides by \(y\) is undefined.

A method that survives a new context

Define each changing quantity and choose a direction for positive motion.
Write an equation valid over the motion, not just at one instant.
Remove extra variables with geometric constraints if useful.
Differentiate with respect to the common time variable.
Substitute the instantaneous data, solve, and interpret units and sign.

Example 2

Filling a cone

A cone has height \(6\text{ m}\) and top radius \(2\text{ m}\). Water enters at \(0.5\text{ m}^3/\text{min}\). Find the height rate when the water is \(3\text{ m}\) deep.

Let \(r\) be the water-surface radius and \(h\) its depth.

A vertical cross-section of a conical tank, six metres high with top radius two metres. Water at half the tank height forms a smaller similar triangle, with depth h and surface radius r measured from the same vertex. A downward arrow shows the inflow. — The water is shown at \(h=3\,\mathrm{m}\). The water and tank share a vertex and side angle, so their radius-to-height ratios agree.

Similar triangles give \(r/h=2/6\), so \(r=h/3\). Substitute this relation into the volume before differentiating:

\[V=\frac13\pi r^2h=\frac{\pi h^3}{27}.\]

Hence

\[\begin{aligned} \frac{dV}{dt}&=\frac{\pi h^2}{9}\frac{dh}{dt},\\ \frac{dh}{dt}&=\frac9{\pi h^2}\frac{dV}{dt}. \end{aligned}\]

At \(h=3\) this is \(1/(2\pi)\text{ m/min}\), about \(0.159\text{ m/min}\). A constant volume inflow produces a decreasing height rate as the water surface widens.

Example 3

A shadow has two possible rates

A \(2\text{ m}\) tall person walks away from a \(6\text{ m}\) lamp at \(1.5\text{ m/s}\). Let \(x\) be the person’s distance from the lamp and \(s\) the shadow length.

A six-metre lamp and two-metre person form similar right triangles with the shadow tip. Distance x runs from lamp to person, s from person to shadow tip, and x plus s from lamp to shadow tip. The person walks right at 1.5 metres per second. — The same light ray forms both triangles. The shadow has length \(s\), while its tip lies \(x+s\) from the lamp.

Similar triangles give

\[\frac6{x+s}=\frac2s\quad\Longrightarrow\quad s=\frac{x}{2}.\]

Thus \(ds/dt=(1/2)(dx/dt)=0.75\text{ m/s}\): this is the rate at which the shadow length grows. The tip is \(x+s\) from the lamp, so it moves at \(1.5+0.75=2.25\text{ m/s}\) relative to the ground. The question’s reference point matters.

Warning

Do not replace a changing length by its instantaneous value before differentiating: \(x=3\) at one instant does not imply \(dx/dt=0\). Also convert rates to a common time unit before combining them.

Quiz

Enable JavaScript for interactive answer checking.A sphere expands with \(dr/dt=2\,\mathrm{cm/s}\). At \(r=3\,\mathrm{cm}\), what is \(dV/dt\)?

\(24\pi\,\mathrm{cm^3/s}\)

\(72\pi\,\mathrm{cm^3/s}\)

\(36\pi\,\mathrm{cm^2/s}\)

Explanation

Use \(dV/dt=4\pi r^2\,dr/dt=4\pi(9)(2)=72\pi\). Volume rate has cubic length units per time.A \(5\,\mathrm{m}\) ladder stays against a wall. Its foot is \(4\,\mathrm{m}\) from the wall and moves outward at \(0.3\,\mathrm{m/s}\). What is the rate of change of the top's height?

\(-0.225\,\mathrm{m/s}\)

\(-0.4\,\mathrm{m/s}\)

\(0.4\,\mathrm{m/s}\)

\(-0.3\,\mathrm{m/s}\)

Explanation

The geometry gives \(y=\sqrt{25-16}=3\,\mathrm{m}\). Then \(\frac{dy}{dt}=-\frac43(0.3)=-0.4\,\mathrm{m/s}\). The negative sign indicates downward motion.A rectangle has \(A=xy\), with both sides changing. Which time derivative is correct?

\(dA/dt=(dx/dt)(dy/dt)\)

\(dA/dt=x\,dy/dt+y\,dx/dt\)

\(dA/dt=x+y\)

Explanation

The product rule is needed because both factors depend on time.Water in a conical tank satisfies \(r=h/3\), and the volume inflow stays constant. When the water depth \(h\) doubles, what happens to \(dh/dt\)?

It is halved

It stays the same

It doubles

It becomes one quarter as large

Explanation

Here \(\frac{dV}{dt}=\frac{\pi h^2}{9}\frac{dh}{dt}\). With \(dV/dt\) fixed, multiplying \(h^2\) by \(4\) divides \(dh/dt\) by \(4\).A person \(2\,\mathrm{m}\) tall walks away from a \(6\,\mathrm{m}\) lamp at \(1.2\,\mathrm{m/s}\). How fast does the shadow tip move away from the lamp?

\(0.6\,\mathrm{m/s}\)

\(1.2\,\mathrm{m/s}\)

\(1.8\,\mathrm{m/s}\)

\(2.4\,\mathrm{m/s}\)

Explanation

Similar triangles give \(s=x/2\), so the shadow length grows at \(0.6\,\mathrm{m/s}\). The tip is \(x+s\) from the lamp, and its rate is \(1.2+0.6=1.8\,\mathrm{m/s}\).Two cyclists move away from a junction along perpendicular paths. Their distances from the junction are \(3\,\mathrm{m}\) and \(4\,\mathrm{m}\), increasing at \(4\,\mathrm{m/s}\) and \(3\,\mathrm{m/s}\), respectively. How fast is their separation increasing?

\(4.8\,\mathrm{m/s}\)

\(5\,\mathrm{m/s}\)

\(7\,\mathrm{m/s}\)

\(24\,\mathrm{m/s}\)

Explanation

Let \(x,y\) be the two distances and \(z\) the separation. Differentiating \(z^2=x^2+y^2\) gives \(\frac{dz}{dt}=\frac{x}{z}\frac{dx}{dt}+\frac{y}{z}\frac{dy}{dt}\). Here \(z=5\,\mathrm{m}\), so \(\frac35(4)+\frac45(3)=4.8\,\mathrm{m/s}\).