Related Rates

A stone hits the water

You are looking straight down at a pond. A ripple spreads out from the splash in a circle, and its radius r grows at a steady 1 meter per second.

Two things are changing at once: the radius, and the area A of water inside the ripple. The radius gains the same amount every second. The question for this whole tutorial is what that does to the other quantity.

The slider under the picture is time. The green band is the new water the ripple will cover in the next quarter of a second.

Rests on A1 D1

Try itPress Play, or drag the ripple outward. The radius grows steadily. Does the area? Watch how the green band changes as the circle gets bigger.

One equation, two rates

At every moment the area and the radius obey A = πr². Both sides change with time, so differentiate both sides with respect to t. The chain rule turns r² into 2r · dr/dt:

dA/dt = 2πr · dr/dt

That is a related rates equation: it ties the rate you know to the rate you want. The picture says the same thing. The green band is a strip as long as the circumference 2πr and as wide as the distance the ripple moves, so a bigger circle sweeps more water for the same step outward.

The panel on the right carries out these steps for the moment you choose. The graph shows the area over time, and the slope of its tangent line is dA/dt.

Try itSet the radius to 2 m, then to 4 m. The radius doubled. What happened to dA/dt? What happened to the area itself?

Change the speed

A new slider sets how fast the ripple spreads, from 0.5 to 2 m/s. When you move it the radius stays where it is, so you can compare speeds at the same moment in the ripple's life.

The equation dA/dt = 2πr · dr/dt has two factors. One is where the ripple is (r), and the other is how fast it moves (dr/dt). A related rates answer always belongs to one particular moment, which is why problems say "when the radius is 3 m".

Try itPut the ripple at r = 3 m. Before you touch the speed slider, predict dA/dt at 2 m/s. Then find a slow, large ripple whose area grows exactly as fast as a fast, small one.

Sand piling up

A hopper drops sand at a steady 2 cubic meters per minute. The sand slides until its slope settles, so the pile is always a cone of the same shape. For this sand the radius of the base is 1.5 times the height.

This time the rate you are given is a volume rate, dV/dt = 2, and the rate you want is how fast the peak rises, dh/dt. With the ripple you knew the simple rate and found the complicated one. Here it runs the other way.

The green coat on the pile is the sand that will land in the next minute.

Rests on A2 A3

Try itPress Play, or drag the peak. The sand arrives at a steady rate. Does the peak rise at a steady rate? Compare the green coat on a small pile and on a big one.

One variable first, then differentiate

The volume of a cone is V = ⅓πr²h, and here both r and h are changing. The shape of the pile ties them together, r = 1.5h, so substitute and get everything in terms of h:

V = ⅓π(1.5h)²h = 0.75πh³

Now differentiate with respect to t and solve for the rate you want:

dV/dt = 2.25πh² · dh/dt

Order matters. Substituting r = 1.5h first is fine, because it is true at every moment. A number like h = 1.5 is true at one moment only. Plug it in before differentiating and V becomes a constant with derivative zero. Numbers go in last.

Rests on A2

Try itSet the height to 1 m, then to 2 m. The pile is twice as tall. How many times slower is the peak rising?

Why the peak slows down

Look at the coefficient: 2.25πh² = π(1.5h)² = πr². That is the area of the pile's base, so

dh/dt = (dV/dt) ÷ (base area)

Here is why. Lift the pile straight up by a small amount. The bigger pile is the old pile sitting on a thin slab as wide as the base. For a small rise the slab is very nearly a disk, so the new sand has the volume of that disk: base area times the rise. The same sand over a wider base makes a thinner disk.

A new slider sets the pour rate. As before, the pile holds its height while you move it.

Try itHold the pile at 1.5 m and double the pour rate from 2 to 4 m³/min. Then let the pile grow until the peak is rising only as fast as it did before you doubled it. How much wider did the base have to get?

A funnel draining

Turn the cone upside down and fill it with water. This funnel is 12 cm deep and 6 cm in radius at the top, and water leaves through the bottom at a steady 10 cubic centimeters per second.

The volume is going down, so its rate is negative: dV/dt = −10. The depth h is falling too, so dh/dt will come out negative. In a related rates problem the sign tells you which way a quantity is heading.

The orange slab is the water that will leave in the next 2 seconds. Every slab holds the same 20 cm³.

Rests on A4 A5 D1

Try itPress Play, or drag the water surface. Where is the level falling slowly, and where is it racing? Watch the thickness of the orange slab.

Similar triangles

The water is a small cone inside the big one, and the two have the same shape. So the water's radius and depth are in the same ratio as the funnel's:

r/h = 6/12,  so  r = h/2

That plays the part r = 1.5h played for the sand. Substitute, then differentiate:

V = ⅓π(h/2)²h = (1/12)πh³  ⇒  dV/dt = (1/4)πh² · dh/dt

Again the coefficient is an area: (1/4)πh² = πr², the water's surface. The level falls at the outflow divided by the surface area. The sand pile slowed down because its base kept widening. The funnel speeds up because its surface keeps shrinking.

Rests on A4

Try itSet the depth to 8 cm, then to 4 cm. Half the depth: how many times faster is the level falling?

Change the funnel

Two new sliders: the outflow, and the radius of the funnel's top. The depth stays where you left it while you move them.

A different funnel means a different ratio in the similar triangles, so every line of the computation changes with it. Watch step 2 as you move the radius slider.

Try itHold the depth at 6 cm. Make the funnel narrow (3 cm), then wide (8 cm). Which one drops faster at the same depth and the same outflow, and what in the picture explains it?

Into a glass

Now the water lands in a glass, a cylinder 5 cm in radius. Call the depth in the glass g.

One flow links three rates. Every cubic centimeter that leaves the funnel arrives in the glass, so the glass gains volume at +10 cm³/s while the funnel loses it at −10. For a cylinder V = 25πg, so

dg/dt = 10 ÷ 25π ≈ 0.127 cm/s

That is the same rule: flow divided by surface area. The glass has the same surface at every depth, so its level rises steadily. The funnel's surface shrinks, so its level does not fall steadily. The green slab in the glass and the orange slab in the funnel hold the same water.

Rests on A5 A6

Try itFind the moment when the funnel's level falls exactly as fast as the glass's level rises. Then look at the two water surfaces. What do they have in common?

The recipe, and what the stories leave out

Every problem here went the same way:

• Name the quantities that change, and the rate you know and the rate you want.
• Write an equation that ties the quantities together at every moment. If it has an extra variable, use the geometry to remove it.
• Differentiate both sides with respect to time.
• Only then plug in the numbers for the moment you were asked about, and solve.

The steady rates are assumptions, and each one leaves something out: an outflow that changes as the funnel empties, sand that arrives in slides and not as a smooth coat, a ripple that slows as it spreads. The related rates equation survives all of that: dh/dt = (dV/dt) ÷ (surface area) holds at each moment, whatever the flow happens to be at that moment.

Everything is unlocked now: all three scenes and every slider. Play as long as you like, then scroll on to finish.

Rests on A1 A3 A5

Try itPick a scene and predict before you look: if the level is twice as high, what happens to the rate? Check each scene. Two of the three answers match. What do those two scenes have in common?