Lesson 18: Related Rates

(a) Opening Question

A circular ripple spreads on a pond. At a certain instant the radius is r = 4 ft and the radius is increasing at 0.5 ft/s.

The area enclosed by the ripple is A = πr². Both A and r are changing as time passes, so think of each as a function of time t.

1. The radius is growing at 0.5 ft/s. In derivative notation, what is dr/dt?
2. We want the rate at which the enclosed area is growing — that is, dA/dt. Why can't you simply plug r = 4 into A = πr² and call it done?
3. Differentiate A = πr² with respect to t. (Hint: r is a function of t, so the chain rule introduces a factor of dr/dt.)
4. Now substitute r = 4 and dr/dt = 0.5. What is dA/dt, and what are its units?

Spend two minutes before reading on. The single most important habit in this lesson is hiding in question 2: differentiate first, substitute second. If you plug in r = 4 before differentiating, you freeze the radius at a constant and its rate of change vanishes. Keep the variable alive through the differentiation, then substitute.

(Answers: dr/dt = 0.5 ft/s; you can't plug in early because a constant has derivative zero, killing the rate; dA/dt = 2πr·(dr/dt); dA/dt = 2π(4)(0.5) = 4π ≈ 12.566 ft²/s.)

(b) Core Concepts

A related-rates problem asks: given how fast one quantity is changing, how fast is a related quantity changing? The two quantities are linked by an equation, and because both vary with time, their rates of change are linked too. Calculus turns the geometric relationship into a relationship between rates.

The engine is the chain rule. If A depends on r and r depends on t, then

dA/dt = (dA/dr)·(dr/dt)

Every related-rates computation is this idea applied to an equation that relates the quantities.

The seven-step procedure

Follow these steps every time. The AP readers reward the setup as much as the final number.

Step 1 — Draw and label. Sketch the situation. Label every quantity that changes with a variable (not a number), even if you know its current value.

Step 2 — Identify the given rate and the wanted rate. Write both as derivatives with respect to time t. ("Given: dr/dt = 0.5. Wanted: dA/dt when r = 4.")

Step 3 — Write an equation relating the quantities. Use geometry (area, volume, Pythagorean theorem, similar triangles). The equation should involve the variables whose rates appear in Steps 2.

Step 4 — Differentiate both sides with respect to t. Treat every variable as a function of t and apply the chain rule. Each variable picks up a d(...)/dt factor.

Step 5 — Substitute the known values — only now. Plug in the instantaneous values after differentiating, never before.

Step 6 — Solve for the wanted rate.

Step 7 — Answer with units in a complete sentence, and check the sign (positive = increasing, negative = decreasing).

The critical rule, restated: do not substitute the value of a changing quantity before you differentiate. A number that is plugged in early becomes a constant, and the chain rule will not generate its rate. The only quantities you may substitute before differentiating are ones that are genuinely constant for all time (for example, a fixed ladder length or a tank's fixed dimensions used to eliminate a variable — see the cone below).

A fully worked problem — the sliding ladder

Problem. A 10-foot ladder leans against a vertical wall. The bottom of the ladder is pulled away from the wall along the ground at a constant 2 ft/s. How fast is the top of the ladder sliding down the wall at the instant the bottom is 6 ft from the wall?

Step 1 — Draw and label. The ladder, wall, and ground form a right triangle. Let x be the distance from the wall to the bottom of the ladder, and y the height of the top of the ladder up the wall. The ladder length 10 is the hypotenuse and is constant.

Step 2 — Given and wanted rates.

• Given: dx/dt = 2 ft/s (the bottom moves away from the wall).
• Wanted: dy/dt at the instant x = 6.

Step 3 — Relate the quantities. By the Pythagorean theorem, since the ladder length is constant at 10:

x² + y² = 10² = 100

Step 4 — Differentiate both sides with respect to t. The constant 100 differentiates to 0. Each variable term needs the chain rule:

d/dt[x²] + d/dt[y²] = d/dt[100]
2x·(dx/dt) + 2y·(dy/dt) = 0

Notice the dx/dt and dy/dt factors — those come from the chain rule because x and y are functions of time. Forgetting them is the most common error in the lesson.

Step 5 — Substitute known values, now that we have differentiated. When x = 6, the Pythagorean equation gives the matching y:

6² + y² = 100  ⇒  y² = 64  ⇒  y = 8

Substitute x = 6, y = 8, dx/dt = 2:

2(6)(2) + 2(8)·(dy/dt) = 0
24 + 16·(dy/dt) = 0

Step 6 — Solve for the wanted rate.

16·(dy/dt) = -24
dy/dt = -24/16 = -3/2 = -1.5

Step 7 — Answer with units. dy/dt = -1.5 ft/s. The negative sign means y is decreasing: the top of the ladder is sliding down the wall at a rate of 1.5 feet per second at the instant the bottom is 6 ft from the wall.

Two things to absorb from this model solution. First, we found y = 8 from the relating equation after setting up the derivative — we never assumed it. Second, the sign told the story: a negative dy/dt is exactly what "sliding down" should produce. If your sign disagrees with the physical picture, recheck the setup before trusting the number.

Why "substitute second" matters — a one-line cautionary example

Suppose you wrongly substituted x = 6 into x² + y² = 100 first, getting 36 + y² = 100, then "differentiated." The 36 is now a constant, its derivative is 0, and you have silently discarded the 2x·(dx/dt) term — the very term carrying the given rate. The whole problem collapses. Differentiate while the variables are still variables.

A note on units and the chain rule's extra factor

Every derivative dQ/dt carries units of (units of Q) per second. Keep them with you: dA/dt is ft²/s, dV/dt is cm³/s, dr/dt is cm/s. A mismatch in units (mixing cm and m, or minutes and seconds) is an easy lost point. And the chain-rule factor is not optional decoration: differentiating V = (4/3)πr³ with respect to time gives dV/dt = 4πr²·(dr/dt), not 4πr². The lone dr/dt is what makes it a related-rates equation.

(c) Worked Examples

Example 1 — Expanding circle / ripple (foundation) · [NO CALC]

Problem. A stone dropped in a pond makes a circular ripple whose radius increases at a constant 4 cm/s. How fast is the enclosed area increasing when the radius is 20 cm?

Strategy. Relate area and radius with A = πr², differentiate with respect to t, substitute.

Solution.

• Given: dr/dt = 4 cm/s. Wanted: dA/dt when r = 20.
• Relate: A = πr².
• Differentiate: dA/dt = 2πr·(dr/dt).
• Substitute: dA/dt = 2π(20)(4) = 160π.

dA/dt = 160π ≈ 502.655 cm²/s. The enclosed area is increasing at a rate of 160π ≈ 502.7 square centimeters per second when r = 20 cm.

Justification language. "Differentiating A = πr² with respect to time gives dA/dt = 2πr(dr/dt); substituting r = 20 and dr/dt = 4 yields dA/dt = 160π cm²/s."

Example 2 — Inflating sphere / balloon (procedural) · [NO CALC]

Problem. Air is pumped into a spherical balloon at a rate of 100 cm³/s. How fast is the radius increasing at the instant the radius is 5 cm? (V = (4/3)πr³)

Strategy. Relate volume and radius, differentiate, solve for dr/dt.

Solution.

• Given: dV/dt = 100 cm³/s. Wanted: dr/dt when r = 5.
• Relate: V = (4/3)πr³.
• Differentiate: dV/dt = 4πr²·(dr/dt). (The (4/3)·3 = 4 and the chain-rule factor dr/dt.)
• Substitute: 100 = 4π(5)²·(dr/dt) = 100π·(dr/dt).
• Solve: dr/dt = 100 / (100π) = 1/π.

dr/dt = 1/π ≈ 0.318 cm/s. The radius is increasing at a rate of 1/π ≈ 0.318 centimeters per second when r = 5 cm.

Sanity check on the sign: air is being added, so the radius should be increasing — and 1/π > 0. Good.

Example 3 — Draining cone with similar triangles (AP level) · [CALC for arithmetic]

Problem. Water is poured into an inverted right-circular cone at a rate of 10 ft³/min. The cone has height 12 ft and radius 6 ft at the top. How fast is the water level rising when the water is 4 ft deep?

Strategy. The trap is that the volume formula V = (1/3)πr²h has two changing variables, r and h. Use similar triangles to write r in terms of h, reduce to one variable, then differentiate.

Step 1 — Draw and label. As water fills the cone, the water itself forms a smaller cone similar to the tank.

Step 2 — Given and wanted rates. Given dV/dt = 10 ft³/min. Wanted dh/dt when h = 4.

Step 3 — Relate (and reduce to one variable). Volume of the water cone:

V = (1/3)πr²h

By similar triangles, the full tank has radius/height = 6/12 = 1/2, and the water cone is similar, so

r/h = 1/2  ⇒  r = h/2

Substitute r = h/2 to eliminate r before differentiating (this is legitimate — it is an exact geometric identity true at all times, not a freezing of a value):

V = (1/3)π(h/2)²h = (1/3)π(h²/4)h = πh³/12

Step 4 — Differentiate with respect to t.

dV/dt = π·(3h²)/12·(dh/dt) = (πh²/4)·(dh/dt)

Step 5 — Substitute. dV/dt = 10, h = 4:

10 = (π(4)²/4)·(dh/dt) = (16π/4)·(dh/dt) = 4π·(dh/dt)

Step 6 — Solve.

dh/dt = 10/(4π) = 5/(2π)

Step 7 — Answer. dh/dt = 5/(2π) ≈ 0.796 ft/min. The water level is rising at a rate of 5/(2π) ≈ 0.796 feet per minute when the water is 4 ft deep.

Why the similar-triangles step is mandatory: if you keep both r and h, differentiating V = (1/3)πr²h requires the product rule and introduces an unknown dr/dt you cannot evaluate. Reducing to a single variable removes that unknown.

Example 4 — The walking shadow (AP level) · [NO CALC]

Problem. A street lamp is 15 ft tall. A person 6 ft tall walks away from the base of the lamp at 5 ft/s. (i) How fast is the tip of the person's shadow moving along the ground? (ii) How fast is the length of the shadow increasing?

Step 1 — Draw and label. Let x be the distance from the lamp's base to the person, and s the length of the shadow beyond the person. The lamp, the person, and their shadow tips form two similar triangles (a big one for the lamp, a small one for the person).

Step 2 — Rates. Given dx/dt = 5. Wanted: (i) d(x+s)/dt (the tip), (ii) ds/dt (the shadow length).

Step 3 — Relate via similar triangles. Matching the big and small triangles:

15/(x + s) = 6/s

Cross-multiply and solve for s in terms of x:

15s = 6(x + s)  ⇒  15s = 6x + 6s  ⇒  9s = 6x  ⇒  s = (2/3)x

Step 4 — Differentiate.

ds/dt = (2/3)·(dx/dt)

Steps 5–6 — Substitute and solve.

• (ii) Shadow length: ds/dt = (2/3)(5) = 10/3 ≈ 3.333 ft/s.
• (i) Shadow tip position is x + s, so d(x+s)/dt = dx/dt + ds/dt = 5 + 10/3 = 25/3 ≈ 8.333 ft/s.

Step 7 — Answer. The tip of the shadow moves at 25/3 ≈ 8.33 ft/s, and the shadow length grows at 10/3 ≈ 3.33 ft/s. Note the tip moves faster than the person (25/3 > 5), which makes sense — the tip is the person's motion plus the shadow's growth. A frequent error is to report the same number for both; "tip speed" and "shadow-length rate" are different quantities.

(d) Common Mistakes

Substituting a changing value before differentiating.

What students do: Plug r = 4 (or x = 6, h = 4) into the relating equation, then differentiate.

Why it's wrong: A substituted number is a constant; its derivative is 0, so the term carrying the given rate disappears and the answer is garbage.

Fix: Differentiate first with variables intact, then substitute the instantaneous values (Steps 4 → 5). The only early substitution allowed is an exact identity like r = h/2 that holds for all time.

Forgetting the chain-rule factor (dr/dt, dh/dt, …).

What students do: Differentiate V = (4/3)πr³ to dV/dt = 4πr² and stop.

Why it's wrong: r is a function of t, so the chain rule requires a ·(dr/dt) factor: dV/dt = 4πr²(dr/dt). Without it, the equation no longer relates the two rates.

Fix: When you differentiate with respect to t, attach a d(variable)/dt to every variable term. If a dt-derivative is missing, you have made this error.

Mismatched or missing units.

What students do: Mix centimeters with meters, or seconds with minutes, or omit units entirely.

Why it's wrong: A rate dV/dt in cm³/s cannot be combined with a radius given in meters; and the AP rubric awards the answer point only with correct units.

Fix: Convert everything to one consistent unit system before substituting, and always report (output unit)/(time unit) — ft²/s, cm³/s, ft/min, etc.

Choosing the wrong relating equation.

What students do: Use perimeter when the problem is about area, or use V = πr²h (a cylinder) for a cone.

Why it's wrong: The relating equation must match the geometry, or the rates it produces are meaningless.

Fix: Match the figure to its formula — circle area A = πr², sphere volume V = (4/3)πr³, cone volume V = (1/3)πr²h, right triangle x² + y² = z². Draw the picture first.

Not using similar triangles to reduce variables (the cone trap).

What students do: Keep both r and h in V = (1/3)πr²h and differentiate, producing an unknown dr/dt.

Why it's wrong: You end up with two unknown rates and only one equation — unsolvable.

Fix: Use the fixed proportions of the tank (similar triangles) to write r in terms of h (e.g., r = h/2), substitute to get V in one variable, then differentiate.

(e) Practice Problems

For multiple choice, choose the single best answer. Calculator labels match AP rules; "calculator" here means arithmetic only — all setups are by hand.

(Short answer, full setup)Air leaks from a spherical balloon at 12 in³/min. Set up — but you need not finish solving past the final substitution — the related-rates equation for dr/dt, and find dr/dt when r = 3 in. Show the relating equation, the differentiated equation, and the substitution. (V = (4/3)πr³)

(Short answer, full setup)A 25-ft ladder slides down a wall. The top slides down at 3 ft/s at the instant the top is 15 ft above the ground. Find how fast the bottom is moving away from the wall at that instant. Show all seven steps briefly, including how you find the missing side.

(Justification / interpretation)In the draining-cone problem, a student writes: "Since dV/dt is constant at 10 ft³/min, the water level dh/dt must also be constant." State whether this is correct and justify your answer using the differentiated equation dV/dt = (πh²/4)(dh/dt).

(Justification / interpretation)A circular oil slick's area grows at a constant 8 m²/min. Explain, with reference to dA/dt = 2πr(dr/dt), why the radius grows more slowly as the slick gets larger, even though the area grows at a steady rate.

(f) AP Exam Focus

FRQ — Draining Conical Tank (Section II style). Calculator permitted (arithmetic only). Total: 9 points.

Water is draining from an inverted right-circular conical tank. The tank has a height of 10 ft and a radius of 4 ft at the top. At the instant shown, water is draining out so that the volume of water is decreasing at a constant rate of 2 ft³/min. Let h be the depth of the water and r the radius of the water's circular surface, both measured in feet, at time t minutes.

(a) Using similar triangles, express the radius r of the water's surface in terms of the depth h, and write the volume V of the water as a function of h alone. (2 points)

(b) Find the rate at which the water depth is changing, dh/dt, at the instant when the water is 5 ft deep. Indicate units. (4 points)

(c) At the instant in part (b), is the radius r of the water's surface increasing or decreasing, and how fast? Justify your answer. (2 points)

(d) Interpret the sign of dh/dt found in part (b) in the context of this problem. (1 point)

Model Solution

(a) The full tank has radius/height = 4/10 = 2/5. The water forms a cone similar to the tank, so by similar triangles

r/h = 2/5  ⇒  r = (2/5)h.

Substitute into the cone-volume formula to write V in terms of h alone:

V = (1/3)πr²h = (1/3)π((2/5)h)²h = (1/3)π(4/25)h²·h = (4π/75)h³.

(b) Differentiate V = (4π/75)h³ with respect to t, applying the chain rule:

dV/dt = (4π/75)·3h²·(dh/dt) = (12π/75)h²·(dh/dt) = (4π/25)h²·(dh/dt).

Substitute the known values dV/dt = -2 and h = 5 after differentiating:

-2 = (4π/25)(5)²·(dh/dt) = (4π/25)(25)·(dh/dt) = 4π·(dh/dt).

Solve:

dh/dt = -2/(4π) = -1/(2π) ≈ -0.159 ft/min.

The water depth is changing at dh/dt = -1/(2π) ≈ -0.159 feet per minute when the water is 5 ft deep.

(c) Since r = (2/5)h, differentiate with respect to t:

dr/dt = (2/5)·(dh/dt) = (2/5)(-1/(2π)) = -1/(5π) ≈ -0.064 ft/min.

Because dr/dt = -1/(5π) < 0, the radius of the water's surface is decreasing, at a rate of 1/(5π) ≈ 0.064 feet per minute. This follows from differentiating the similar-triangles relationship r = (2/5)h, which forces dr/dt to have the same sign as dh/dt.

(d) Since dh/dt = -1/(2π) < 0 is negative, the depth h is decreasing: at the instant the water is 5 ft deep, the water level is falling (dropping) at about 0.159 ft/min. This is consistent with water draining out of the tank.

Scoring Commentary — where students lose points

• Part (a) [2 pts]: 1 pt for the correct similar-triangles relation r = (2/5)h; 1 pt for the correct one-variable volume V = (4π/75)h³. The most common loss is using r/h = 4/10 upside down (r = (5/2)h) or never reducing to one variable. Students who keep both r and h set themselves up to fail part (b).
• Part (b) [4 pts]: 1 pt for differentiating correctly to dV/dt = (4π/25)h²(dh/dt) (chain-rule factor present); 1 pt for substituting dV/dt = -2 and h = 5 after differentiating; 1 pt for solving to dh/dt = -1/(2π); 1 pt for correct units (ft/min). The classic zero-credit move is substituting h = 5 into V before differentiating — that freezes the variable and destroys the problem. Forgetting (dh/dt) in the derivative loses the first point.
• Part (c) [2 pts]: 1 pt for dr/dt = (2/5)(dh/dt) = -1/(5π); 1 pt for the justification — stating it is decreasing because dr/dt < 0 and tying it to the differentiated relation r = (2/5)h. A bare "decreasing" with no sign or no derivative reference earns only partial credit.
• Part (d) [1 pt]: 1 pt for connecting the negative sign to the physical meaning (depth decreasing / water level falling / tank draining). Restating the number without interpreting the sign earns 0.

Justification language that earns full credit: "Because dr/dt = -1/(5π) < 0, the surface radius is decreasing at 1/(5π) ft/min." Vague phrasing such as "the radius goes down a little" earns no justification credit.