⏳Related Rates

Technician Fluxion keeps a shallow tray of water on the bench at the Tangent Laboratory. Drop a pebble in and a ripple spreads: the radius grows, and so does the area. Nobody measures the area with a ruler — they measure the radius, because the radius is the thing you can catch. And yet the area's speed is completely determined, because the two are tied together by one equation:

A = pi*r²

That is the whole idea of related rates. Two quantities are linked by an equation. Both of them are functions of time, even when the letter t appears nowhere in the equation. So differentiate the equation with respect to t, and the link between the quantities becomes a link between their speeds:

dA/dt = 2*pi*r*(dr/dt)

Read that sentence out loud, because it is the answer to the only question the topic ever asks. *The area's rate is twice pi times the radius, times the radius's rate.* Every problem in this craft has that shape: a known rate on one side, an unknown rate on the other, and a geometric relation between them.

The move that produces the extra factor is the chain rule. r is not the variable of differentiation — t is — so d/dt of r² is not 2r. It is 2r times dr/dt, exactly the way d/dx of (something)^2 is 2(something) times the derivative of that something. This is implicit differentiation, with time as the hidden variable.

d/dt of r² = 2r*(dr/dt) d/dt of h³ = 3h²*(dh/dt) d/dt of sin(x) = cos(x)*(dx/dt) d/dt of 25 = 0

The last line matters as much as the first three. A quantity that never changes has a rate of zero, and that zero is often the whole solution — a ladder's length, a dock's height, a kite's altitude.

Every rate is a derivative with respect to t. "How fast is the area growing" is dA/dt. "The radius grows at 3 cm/s" is dr/dt = 3. A rate that is negative means the quantity is shrinking, and dt/dr — the derivative upside down — is not a rate of anything you were asked about.

Related-rates problems are word problems, and word problems are lost in the setup rather than in the calculus. These five steps in this order will carry every one of them.

1. Name the quantities and write down every rate you are told. Use single letters, and write the units beside each. "The radius grows at 3 cm/s" becomes dr/dt = 3. "How fast is the area growing" becomes dA/dt = ?. Anything you cannot name as a rate is a *value at an instant*, not a rate, and belongs in step 4.

2. Draw the picture, and label the moving lengths with letters, not numbers. This is the step everyone skips and the step that decides the problem. A ladder's foot is x metres from the wall — x, not 5. The moment you write 5 on the drawing you have promised the ladder will never move.

3. Find the equation that ties the quantities together. There are only a handful, and you already know all of them:

• ◆Pythagoras, x² + y² = c², for anything sliding, separating or approaching at a right angle
• ◆area and volume formulas, for anything expanding or filling
• ◆similar triangles, for shadows, cones and troughs — the ratio of two sides is constant
• ◆basic trigonometry, sin, cos and tan, whenever an angle is one of the moving quantities

If the equation has three letters in it and you only have rates for two, use similar triangles to write one of them in terms of another until only two remain.

4. Differentiate with respect to t. Only then substitute the instant. This is the rule the whole craft turns on, so it gets its own paragraph below.

5. Solve for the rate you were asked for, and check the sign and the units. A shrinking quantity has a negative rate. If a question asks "how fast is it falling", the expected answer is the positive speed, and the minus sign in your working is what tells you it is indeed falling. Units are a free check: dV/dt in cm³/s divided by an area in cm² gives cm/s, which is a speed, which is what a rising water level should be.

Why substitution comes last. Suppose the ripple's radius is 4 cm at the moment we care about. Substituting first gives A = pi*4² = 16pi — a constant, whose derivative is 0. The answer would be that the area is not changing at all, which is plainly wrong. The number 4 is only true for one instant, but the equation A = pi*r² is true for every instant, and you can only differentiate something that is true for a stretch of time. Differentiate the relation, then substitute the instant.

A 13 m ladder leans against a wall. Its foot is pulled away from the wall at 2 m/s. How fast is the top sliding down when the foot is 5 m out?

Let x be the distance from the wall to the foot and y the height of the top up the wall. Both change with time. The ladder does not: it is 13 m long for the whole story, which is what makes Pythagoras the right relation.

x² + y² = 13²

Differentiate with respect to t, remembering that the right-hand side is a constant:

2x*(dx/dt) + 2y*(dy/dt) = 0

Divide by 2 and rearrange for the unknown rate:

dy/dt = -x*(dx/dt)/y

Now, and only now, substitute the instant. We are told x = 5 and dx/dt = 2, but we are not told y — so find it from the same relation: y = sqrt(169 - 25) = 12.

dy/dt = -5*2/12 = -5/6 m/s

The minus sign is the physics: the top is going down while the foot goes out. If the question asks for the speed at which the top slides down, the answer is 5/6 m/s and the minus sign is what justified the word "down".

Two things worth noticing, because they turn up in every relation of this shape.

The two ends move at different speeds, and the ratio keeps changing. At x = 5 the top falls at 5/6 m/s, slower than the foot moves. At x = 12 (so y = 5) it falls at 12*2/5 = 24/5 m/s, more than twice the foot's speed. As the ladder flattens, y shrinks towards zero and the top's speed grows without bound. A related rate is a snapshot: it is only true at the instant you substituted.

The same picture answers other questions. If theta is the angle between the ladder and the ground, then cos(theta) = x/13, and differentiating gives -sin(theta)*w = (dx/dt)/13, where w is the angle's rate. Since sin(theta) = y/13, the 13s cancel and w = -(dx/dt)/y = -1/6 rad/s: the angle is closing at 1/6 rad/s. And if you want the rate of the triangle's area, A = xy/2 has two moving factors, so the product rule applies: dA/dt = (x*(dy/dt) + y*(dx/dt))/2.

Air into a balloon. Air is pumped into a spherical balloon at 36 cm³/s. How fast is the radius growing when r = 3 cm?

V = (4/3)*pi*r³ dV/dt = 4*pi*r²*(dr/dt)

(The 3 the power rule brings down cancels the 3 underneath — that is why the neat 4*pi*r², the sphere's surface area, appears.) Rearrange before substituting:

dr/dt = (dV/dt)/(4*pi*r²) = 36/(4*pi*9) = 1/pi cm/s

An exact answer keeps pi as a symbol. 1/pi is a better answer than 0.318, and it is the answer a related-rates question wants unless it says how to round.

The filling cone. Here is the one extra difficulty in the whole topic. Water pours into a cone-shaped tank, point downwards, 10 cm deep and 5 cm in radius at the top, at 8 cm³/s. How fast is the level rising when the water is 4 cm deep?

V = (1/3)*pi*r²*h

Two moving letters, r and h, and only one known rate — so this equation cannot be differentiated usefully yet. Similar triangles fix it. The cone of water is the same shape as the whole tank, so

r/h = 5/10 = 1/2, giving r = h/2

Substitute *that* into the volume before differentiating:

V = (1/3)*pi*(h/2)²*h = pi*h³/12 dV/dt = pi*h²*(dh/dt)/4

Now put the instant in: 8 = pi*16*(dh/dt)/4, so dh/dt = 2/pi cm/s.

Two traps live in that problem. The first is using r = 5, the radius at the top, when the water is only 4 cm deep and the surface is 2 cm across. The second is differentiating with both letters still present and hoping the second rate turns up later. Reduce to one variable first, always.

A trough works the same way without any pi: a triangular cross-section w across the top and H deep is (w/H)h across at depth h, so the volume is length times h²*w/(2H), and dV/dt = length*w*h*(dh/dt)/H.

The streetlight. A lamp holds its light 6 m above the ground; a person 1.5 m tall walks away from it at 1 m/s. Let x be the person's distance from the lamp's foot and s the length of their shadow. The light ray makes two similar triangles — a big one of height 6 on base x + s, a small one of height 1.5 on base s:

s/1.5 = (x + s)/6 => 6s = 1.5x + 1.5s => 4.5s = 1.5x => s = x/3

Differentiate: ds/dt = (dx/dt)/3 = 1/3 m/s. The shadow lengthens at 1/3 m/s.

Now be careful, because two different questions are usually available. The tip of the shadow is at x + s, so its speed is dx/dt + ds/dt = 1 + 1/3 = 4/3 m/s. The tip carries the walker's own speed as well as the shadow's growth; the shadow itself grows more slowly. Read which one is asked.

Notice also what did not appear: how far the person is from the lamp. The relation s = x/3 is a straight line through the origin, so its slope is the same everywhere and both rates are constant. A question may hand you that distance to see whether you can leave it out.

Two objects separating. Two carts leave a crossroads on roads that meet at a right angle, one going east at 40 km/h, the other north at 30 km/h. When they are 30 km and 40 km out, how fast is the gap growing?

z² = x² + y² 2z*(dz/dt) = 2x*(dx/dt) + 2y*(dy/dt), so z*(dz/dt) = x*(dx/dt) + y*(dy/dt)

At that moment z = sqrt(900 + 1600) = 50, so 50*(dz/dt) = 30*40 + 40*30 = 2400 and dz/dt = 48 km/h. Not 70, the sum of the speeds, and not 50, the hypotenuse of the speeds — speeds at right angles never simply add, and the positions do the weighting.

A turning angle. A camera 400 m from a launch pad tracks a rocket rising at 100 m/s. With theta the angle above the ground, tan(theta) = y/400, so differentiating gives w/cos(theta)^2 = (dy/dt)/400. When the rocket is 300 m up the sight line is 500 m, so cos(theta) = 400/500 = 4/5 exactly — a 3-4-5 triangle, no calculator needed — and

w = (16/25)*(100/400) = 4/25 rad/s

The cos(theta)^2 factor is the whole content of the answer: as the rocket climbs, the same vertical speed swings the camera through less and less angle.