Lesson 9: Related RatesAP

A related rates problem (CED topics 4.4–4.5) hands you an equation linking several quantities that all change with time, tells you how fast some of them are changing, and asks for the rate of another. Every such problem is the chain rule in disguise: differentiate the linking equation with respect to t, and each variable contributes its own d/dt factor — exactly as y-terms contributed dy/dx in Lesson 8. The exam's related-rates questions are long but formulaic, and this lesson's five-step routine solves all of them.

The five-step routine

• 1. Name the variables and record what is given and what is wanted, with units. Draw and label a diagram; constants get numbers, changing quantities get letters.
• 2. Write the linking equation connecting the variables (Pythagoras, similar triangles, a volume or area formula). If the equation has two changing variables where only one is asked about, use a constraint (like similar triangles) to eliminate one *before* differentiating.
• 3. Differentiate with respect to t. Every variable gets its d/dt factor. Only now — never before differentiating — substitute the instantaneous values.
• 4. Solve for the wanted rate.
• 5. Report with sign and units, and interpret: a negative answer means the quantity is decreasing.

Worked Example 1: the sliding ladder

Problem. A 10 ft ladder leans against a wall. The foot slides away from the wall at 2 ft/s. How fast is the top sliding down when the foot is 6 ft from the wall?

Let x = foot distance, y = top height. Linking equation: x^2 + y^2 = 100.
When x = 6: y = sqrt(100 − 36) = 8.
Differentiate w.r.t. t:  2x (dx/dt) + 2y (dy/dt) = 0
Substitute: 2(6)(2) + 2(8)(dy/dt) = 0  →  dy/dt = −24/16 = −1.5 ft/s

Answer: the top slides down at 1.5 ft/s (dy/dt = −1.5; the negative sign records the downward direction). Note the order of operations: the values x = 6, dx/dt = 2 are substituted *after* differentiation. Substituting x = 6 into the equation first would turn it into 36 + y^2 = 100, whose derivative is just 2y·(dy/dt) = 0 — nonsense. This single discipline separates correct solutions from most incorrect ones.

Worked Example 2: the conical tank and similar triangles

Problem. Water is pumped at 2 m^3/min into an inverted conical tank of height 6 m and top radius 3 m. How fast is the water level rising when the water is 2 m deep?

The water forms a cone similar to the tank: r/h = 3/6, so r = h/2.
Eliminate r:  V = (1/3)π r^2 h = (1/3)π (h/2)^2 h = π h^3/12.
Differentiate:  dV/dt = (π h^2/4) (dh/dt)
Substitute dV/dt = 2, h = 2:  2 = (π·4/4)(dh/dt) = π (dh/dt)
dh/dt = 2/π ≈ 0.637 m/min

Answer: ≈ 0.64 m/min. The similar-triangle elimination in line one is the step students skip — differentiating V = (1/3)π r^2 h with both r and h varying produces a two-unknown equation and a stalled solution. Whenever a cone, a shadow, or a trough appears, hunt for the similar triangles first.

Worked Example 3: estimating a rate from measurements (numerical)

Problem. A spherical balloon is inflated. Its radius is measured: r(0) = 5.0 cm, r(1) = 5.4 cm, r(2) = 5.9 cm. Estimate dV/dt at t = 1 s.

Estimate dr/dt at t = 1 by the symmetric difference: (5.9 − 5.0)/(2 − 0) = 0.45 cm/s.
V = (4/3)π r^3  →  dV/dt = 4π r^2 (dr/dt)
At t = 1: dV/dt ≈ 4π (5.4)^2 (0.45) ≈ 165 cm^3/s

Answer: ≈ 165 cm^3/s. Related rates on the calculator-active section sometimes supply *measured data* rather than a clean rate; the difference quotient of Lesson 5 supplies the missing rate, and the chain-rule structure does the rest.

Real-world application: tracking a rocket

An observer stands 600 m from a launch pad, watching a rocket through a telescope that tilts at angle θ. When the rocket is 800 m up (so its distance from the observer is 1000 m, a 3-4-5 triangle scaled by 200) and climbing at 120 m/s, how fast must the telescope tilt? From tan θ = h/600, differentiate: sec^2 θ · (dθ/dt) = (1/600)(dh/dt). Here cos θ = 600/1000 = 0.6, so sec^2 θ = 1/0.36, giving dθ/dt = (120/600)(0.36) = 0.072 rad/s (about 4 degrees per second). Tracking systems, from telescope mounts to missile guidance, solve exactly this equation continuously.

Common mistakes

• Substituting instantaneous values before differentiating. The routine's step order exists because of this error.
• Forgetting the similar-triangle elimination in cone/shadow problems, or eliminating *after* differentiating.
• Sign blindness: a shrinking quantity has a negative rate; if water is draining, dV/dt < 0. Report "decreasing at 1.5 ft/s" or "−1.5 ft/s", and make the two agree.
• Answering with the wrong quantity — solving for dr/dt when dh/dt was asked, or forgetting that the question may want the rate of change of the *diameter*, the *surface area*, or the *angle*, not the variable you solved for.
• Units missing or mixed (cm with m, minutes with seconds). The exam deducts for a rate with no units on FRQs.

AP Exam Experience

1. (No calculator.) The radius of a sphere increases at 0.5 cm/s. At the instant the radius is 10 cm, the volume is increasing at what rate?

• (A) 20π cm^3/s
• (B) 50π cm^3/s
• (C) 200π cm^3/s
• (D) 400π cm^3/s

Answer: (C). V = (4/3)π r^3, so dV/dt = 4π r^2 (dr/dt) = 4π(100)(0.5) = 200π. (Note 4π r^2 is the surface area — volume's rate always passes through it.)

2. A 13 ft ladder leans against a wall. The foot is 5 ft from the wall and sliding away at 3 ft/s. How fast is the top of the ladder sliding down?

• (A) 5/4 ft/s
• (B) 4/5 ft/s
• (C) 12/5 ft/s
• (D) 3 ft/s

Answer: (A). With x = 5, y = 12 (5-12-13 triangle): 2x(dx/dt) + 2y(dy/dt) = 0 gives dy/dt = −(5·3)/12 = −5/4, so the top descends at 5/4 ft/s. (B) and (C) are the reciprocal-confusion variants.

3. Water drains from a conical tank (height 6 m, top radius 2 m) at 3 m^3/min. When the water is 4 m deep, the depth is changing at

• (A) −27/(16π) m/min
• (B) −16π/27 m/min
• (C) −3/(4π) m/min
• (D) −27π/16 m/min

Answer: (A). Similar triangles: r = h/3, so V = (1/3)π(h/3)^2 h = π h^3/27 and dV/dt = (π h^2/9)(dh/dt). With dV/dt = −3 and h = 4: −3 = (16π/9)(dh/dt), so dh/dt = −27/(16π) ≈ −0.54 m/min. The negative sign is required — the depth is falling.

Free-response question. An inverted conical tank has height 12 ft and top radius 4 ft. Water is pumped into the tank at a constant rate of 3 ft^3/min. Let h be the depth of the water and V its volume at time t.

(a) Express V in terms of h alone. Show the similar-triangles step. (b) Find dh/dt at the instant when h = 6 ft. (c) Find dr/dt at the same instant, where r is the radius of the water's surface.

Scoring guideline.

• (a) 2 points. 1 point for the proportion r/h = 4/12, i.e. r = h/3. 1 point for V = (1/3)π r^2 h = (1/3)π(h/3)^2 h = π h^3/27.
• (b) 2 points. 1 point for differentiating with respect to t: dV/dt = (π h^2/9)(dh/dt) (chain rule factor dh/dt present). 1 point for substitution and the answer: 3 = (π·36/9)(dh/dt) = 4π (dh/dt), so dh/dt = 3/(4π) ≈ 0.239 ft/min.
• (c) 1 point. From r = h/3, dr/dt = (1/3)(dh/dt) = 1/(4π) ≈ 0.0796 ft/min. (Using the consistent value from (b) earns this point even if (b) was numerically wrong; differentiating r = h/3 with the dh/dt factor is what is being tested.)

Additional worked practice

Problem. A point moves around the circle x^2 + y^2 = 25 so that its x-coordinate increases at 3 units/s. Find dy/dt at the moment the point is at (−3, 4), and interpret the motion.

Differentiate w.r.t. t: 2x (dx/dt) + 2y (dy/dt) = 0
Substitute x = −3, y = 4, dx/dt = 3:  2(−3)(3) + 2(4)(dy/dt) = 0 → −18 + 8 (dy/dt) = 0
dy/dt = 18/8 = 9/4 = 2.25 units/s

Answer: dy/dt = 9/4 units/s, positive — at (−3, 4), in the second quadrant, moving right (dx/dt > 0) means moving *toward* the top of the circle, so y increases: consistent with the positive sign. The speed along the circle is sqrt(3^2 + 2.25^2) = 3.75 units/s. Sanity-checking a rate's sign against the geometry — quadrant by quadrant — is a thirty-second habit that catches the majority of related-rates sign errors before they cost points.

Extension: the shadow problem, solved completely — and why two rates differ

A streetlight stands 6 m tall. A person 1.8 m tall walks directly away from the pole at 1.2 m/s. Let x be the person's distance from the pole and s the length of their shadow. The tip of the shadow is at distance x + s from the pole, and similar triangles (large: pole; small: person) give (x + s)/6 = s/1.8, so 1.8(x + s) = 6s, hence 1.8x = 4.2s and s = (3/7)x. Differentiate: ds/dt = (3/7)(1.2) ≈ 0.514 m/s — the shadow lengthens at about half a metre per second, regardless of where the person is (the rate is constant because the geometry is self-similar at every position). The tip, however, sits at x + s = (10/7)x, so it moves at (10/7)(1.2) ≈ 1.714 m/s — faster than the person, since the tip gains both from the walking and from the lengthening. Two different questions, two different rates: read FRQs carefully for *which* point or length is being tracked.

A companion classic: a circle's radius grows at 0.3 cm/s; its area grows at dA/dt = 2πr (dr/dt), which at r = 5 is 2π(5)(0.3) = 3π ≈ 9.42 cm^2/s — and grows *faster* as r grows, because the same radial gain adds a longer ring. Related-rates answers that themselves change with the state of the system are the norm, not the exception; if your computed rate does not depend on the given instant's values at all, re-read the problem — a substitution was probably made too early.