Applications of Differentiation

Related Rates

When two quantities are linked by an equation, their rates of change are linked too, and implicit differentiation with respect to time tells you exactly how.

10 worked examples 5-step strategy Ladders · tanks · balloons
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§ 01What Is a Related Rates Problem?

You know how fast one quantity is changing. The question is: how fast is some other quantity changing at the same moment?

Here is the key idea. Two quantities (call them x and y) are related by some equation, for example the Pythagorean theorem, the volume formula for a cone, or the area formula for a circle. Because they are related by an equation, their rates of change with respect to time are also related. Differentiating both sides of the equation implicitly with respect to time t creates a new equation connecting dx/dt and dy/dt.

Related rates problems appear constantly in physics, engineering, and economics: a ladder sliding down a wall, water draining from a tank, a shadow lengthening across the ground, a balloon expanding as air is pumped in. The calculus is always the same; only the geometry changes.

The Core Principle If F(x, y) = 0 holds for all time, then differentiating both sides with respect to t gives a relationship between dx/dt and dy/dt that holds at every instant.

The symbol dx/dt is read as "the rate of change of x with respect to time": in plain English, how fast x is growing or shrinking at a particular moment. A positive rate means the quantity is increasing; a negative rate means it is decreasing.

§ 02The 5-Step Strategy

Every related rates problem follows the same logical skeleton. Internalise these five steps and no scenario can surprise you.

The 5-Step Method for Related Rates

  1. Draw and label a diagram. Sketch the situation. Assign variable names to every quantity that changes with time. Record any given numerical values: but keep changing quantities as variables for now.
  2. Identify the rates. Write down which rate(s) you are given (e.g. dV/dt = −3 m³/min) and which rate you must find (e.g. dh/dt = ?).
  3. Write an equation relating the variables. Use geometry (Pythagoras, area, volume formulas) or another relevant relationship to write a single equation that connects all the changing quantities. This equation must be always true, not just at the moment you care about.
  4. Differentiate both sides with respect to time t. Use implicit differentiation. Every variable gets a d/dt; apply the chain rule wherever a variable appears.
  5. Substitute the known values and solve. Now (and only now) substitute in the given numerical values (rates and measurements at the instant of interest). Solve for the unknown rate.
⚠ Common Mistake: Substituting Too Early Do not substitute numerical values for changing quantities until after you have differentiated. If you plug in, say, x = 6 before differentiating, you lose the variable and get 0 for its derivative. Always differentiate first, substitute second.

§ 03Worked Examples

Ten fully solved problems, from the classic ladder problem to a shadow lengthening across the ground.

Example 1 Expanding circle — ripple on a pond

A stone is dropped into a still pond and creates a circular ripple. The radius of the ripple is increasing at 4 cm/s. How fast is the area of the circle increasing when the radius is 10 cm?

Step 1 — Label variables

Let r = radius (cm), A = area (cm²). Given: dr/dt = 4. Find: dA/dt when r = 10.

Step 2 — Equation
A = πr²
Step 3 — Differentiate w.r.t. t
dA/dt = 2πr · dr/dt
Step 4 — Substitute r = 10, dr/dt = 4
dA/dt = 2π(10)(4) = 80π ≈ 251.3 cm²/s
dA/dt = 80π cm²/s ≈ 251.3 cm²/s
Example 2 Ladder sliding down a wall

A 5 m ladder leans against a vertical wall. The base is sliding away from the wall at 0.5 m/s. How fast is the top of the ladder sliding down the wall when the base is 3 m from the wall?

y x 5 m

x = distance from wall (base), y = height on wall (top)

Step 1 — Label variables

Let x = distance of base from wall, y = height of top on wall. Given: dx/dt = 0.5, ladder length = 5. Find: dy/dt when x = 3.

Step 2 — Equation (Pythagoras)
x² + y² = 25
Step 3 — Differentiate w.r.t. t
2x · dx/dt + 2y · dy/dt = 0
Step 4 — Find y when x = 3
y = √(25 − 9) = √16 = 4
Step 5 — Substitute and solve
2(3)(0.5) + 2(4)(dy/dt) = 0 3 + 8 · dy/dt = 0 dy/dt = −3/8 = −0.375 m/s
dy/dt = −3/8 m/s (top slides down at 0.375 m/s)

The negative sign confirms the top is moving downward: exactly what we expect.

Example 3 Conical water tank draining

Water drains from an inverted conical tank at 2 m³/min. The tank has height 6 m and top radius 3 m. How fast is the water level falling when the water is 4 m deep?

Step 1 — Set up similar triangles

Let h = depth of water, r = radius of water surface. The cone's proportions give r/h = 3/6, so r = h/2.

Step 2 — Volume formula (with r eliminated)
V = (1/3)πr²h = (1/3)π(h/2)²h = πh³/12
Step 3 — Differentiate w.r.t. t
dV/dt = (π/4)h² · dh/dt
Step 4 — Substitute dV/dt = −2, h = 4
−2 = (π/4)(16) · dh/dt = 4π · dh/dt dh/dt = −2/(4π) = −1/(2π) ≈ −0.159 m/min
dh/dt = −1/(2π) ≈ −0.159 m/min
Example 4 Expanding sphere — balloon inflation

Air is pumped into a spherical balloon at 100 cm³/s. How fast is the radius increasing when the radius is 5 cm?

Equation
V = (4/3)πr³
Differentiate
dV/dt = 4πr² · dr/dt
Substitute dV/dt = 100, r = 5
100 = 4π(25) · dr/dt = 100π · dr/dt dr/dt = 1/π ≈ 0.318 cm/s
dr/dt = 1/π ≈ 0.318 cm/s
Example 5 Two cars — approaching intersection

Car A travels north at 60 km/h and car B travels east at 80 km/h, both heading toward the same intersection. When A is 0.3 km away and B is 0.4 km away, how fast is the distance between them decreasing?

Step 1 — Let x = B's distance, y = A's distance, z = distance between cars
z² = x² + y²
Step 2 — At the given instant
z = √(0.4² + 0.3²) = √(0.16 + 0.09) = √0.25 = 0.5 km
Step 3 — Differentiate
2z · dz/dt = 2x · dx/dt + 2y · dy/dt
Step 4 — Cars approach → rates are negative: dx/dt = −80, dy/dt = −60
2(0.5) · dz/dt = 2(0.4)(−80) + 2(0.3)(−60) dz/dt = (−64 − 36) / 1 = −100 km/h
dz/dt = −100 km/h (closing at 100 km/h)
Example 6 Angle of elevation — aircraft tracking

An aircraft flies horizontally at an altitude of 2 km and a speed of 400 km/h directly over an observer. How fast is the angle of elevation increasing when the aircraft is 2 km away horizontally?

Setup — let x = horizontal distance, θ = angle of elevation
tan θ = 2/x
Differentiate w.r.t. t (implicit)
sec²θ · dθ/dt = −2/x² · dx/dt
At x = 2: tan θ = 1 → θ = π/4 → sec²θ = 2. Aircraft moves toward observer so dx/dt = −400.
2 · dθ/dt = −2/(4) · (−400) = 200 dθ/dt = 100 rad/h = 100/3600 rad/s ≈ 0.0278 rad/s
dθ/dt = 100 rad/h ≈ 1.67 rad/min
Example 7 Rectangular box — volume and surface area

A cube's side length is increasing at 3 cm/s. How fast is the volume increasing when the side is 8 cm?

Equation
V = s³
Differentiate
dV/dt = 3s² · ds/dt
Substitute s = 8, ds/dt = 3
dV/dt = 3(64)(3) = 576 cm³/s
dV/dt = 576 cm³/s
Example 8 Shadow lengthening — street lamp

A person 1.8 m tall walks away from a 4.5 m street lamp at 1.2 m/s. How fast is the tip of their shadow moving along the ground?

Step 1 — Let x = person's distance from lamp, s = length of shadow

The tip of the shadow is at distance x + s from the lamp. By similar triangles:

4.5/(x + s) = 1.8/s
Step 2 — Solve for s in terms of x
4.5s = 1.8(x + s) = 1.8x + 1.8s 2.7s = 1.8x → s = (2/3)x
Step 3 — Differentiate
ds/dt = (2/3) · dx/dt = (2/3)(1.2) = 0.8 m/s
Step 4 — Speed of shadow tip = d(x+s)/dt
d(x+s)/dt = dx/dt + ds/dt = 1.2 + 0.8 = 2.0 m/s
Shadow tip moves at 2.0 m/s
Example 9 Cylindrical tank filling

A cylindrical tank of radius 2 m is being filled with water at 5 m³/min. How fast is the water level rising?

Volume of cylinder
V = πr²h = π(4)h = 4πh

(radius is constant at 2 m)

Differentiate
dV/dt = 4π · dh/dt
Substitute dV/dt = 5
5 = 4π · dh/dt → dh/dt = 5/(4π) ≈ 0.398 m/min
dh/dt = 5/(4π) ≈ 0.398 m/min
Example 10 Kite flying — string length and angle

A kite is at a constant altitude of 80 m and drifts horizontally at 5 m/s. How fast is the string (from a fixed point on the ground directly below the launch) being let out when 100 m of string has been released?

Let x = horizontal drift, L = string length
L² = x² + 80² → L² = x² + 6400
When L = 100: x = √(10000 − 6400) = √3600 = 60 m
Differentiate
2L · dL/dt = 2x · dx/dt dL/dt = (x/L) · dx/dt = (60/100)(5) = 3 m/s
dL/dt = 3 m/s

§ 04Common Mistakes

Mistake 1 — Substituting values before differentiating If the ladder base is 3 m from the wall, do not write x = 3 in your equation before you differentiate. Once substituted, the variable is gone and its derivative disappears. Differentiate first — substitute after.
Mistake 2 — Forgetting the chain rule When differentiating a term like r² with respect to t, the answer is 2r · dr/dt, not 2r. Every differentiation of a variable produces a corresponding rate factor.
Mistake 3 — Wrong signs If a quantity is decreasing (water level falling, ladder top sliding down), its rate of change must be negative. Always check whether your rates should be positive or negative based on the physical situation before substituting.
Mistake 4 — Not eliminating extra variables In the conical tank example, you had both r and h as unknowns. Use the geometric similarity relationship to reduce to one variable before differentiating. If your equation still has two changing quantities after differentiation, check whether you missed a constraint.
Mistake 5 — Units mismatch Check that all quantities use consistent units throughout. If the radius is in metres but the volume rate is in litres per minute, convert before substituting.

§ 05Practice Quiz — 10 Questions

Apply the 5-step method. Enter numerical answers (accept ±0.01). Show working to reveal full solutions.

Score: 0 / 0
Question 1 — Expanding circle

The radius of a circle grows at 3 cm/s. How fast is the area growing (cm²/s) when r = 5 cm? Enter your answer in terms of π, enter the coefficient of π only (e.g. for 60π enter 60).

Use A = πr². Differentiate to get dA/dt = 2πr · dr/dt.
  1. A = πr²
  2. dA/dt = 2πr · dr/dt = 2π(5)(3) = 30π cm²/s
Question 2 — Ladder

A 10 m ladder has its base sliding away at 2 m/s. How fast is the top sliding down (m/s) when the base is 6 m from the wall? (Give a positive number for the speed.)

x² + y² = 100. When x = 6, y = 8. Differentiate and solve.
  1. x² + y² = 100 → y = 8 when x = 6
  2. 2x·dx/dt + 2y·dy/dt = 0
  3. 2(6)(2) + 2(8)(dy/dt) = 0 → dy/dt = −24/16 = −3/2
  4. Speed = 1.5 m/s downward
Question 3 — Spherical balloon

Air is pumped into a sphere at 50 cm³/s. How fast is the radius increasing (cm/s, to 3 d.p.) when r = 3 cm? Use π ≈ 3.14159.

V = (4/3)πr³ → dV/dt = 4πr²·dr/dt. Isolate dr/dt.
  1. dV/dt = 4πr²·dr/dt
  2. 50 = 4π(9)·dr/dt = 36π·dr/dt
  3. dr/dt = 50/(36π) = 25/(18π) ≈ 0.442 cm/s
Question 4 — Conical tank

A conical tank has height 9 m and top radius 3 m. Water drains at 6 m³/min. How fast is the depth falling (m/min, 3 d.p.) when h = 3 m?

Similar triangles: r/h = 3/9 = 1/3, so r = h/3. Volume V = πh³/27.
  1. r = h/3, so V = (1/3)π(h/3)²h = πh³/27
  2. dV/dt = (π/9)h²·dh/dt
  3. At h = 3: −6 = (π/9)(3²)·dh/dt = π·dh/dt
  4. dh/dt = −6/π
  5. Speed ≈ 1.910 m/min
Question 5 — Cylindrical tank

A cylindrical tank with radius 4 m is filled at 12 m³/min. How fast does the water level rise? Give answer to 2 d.p. (m/min).

V = π(4²)h = 16πh. Differentiate and solve.
  1. V = 16πh → dV/dt = 16π·dh/dt
  2. 12 = 16π·dh/dt → dh/dt = 12/(16π) = 3/(4π) ≈ 0.239 m/min
Question 6 — Cube

A cube's side grows at 2 cm/s. How fast is the surface area growing (cm²/s) when the side is 6 cm?

Surface area S = 6s². Differentiate: dS/dt = 12s · ds/dt.
  1. S = 6s² → dS/dt = 12s·ds/dt
  2. dS/dt = 12(6)(2) = 144 cm²/s
Question 7 — Concept (Multiple Choice)

A ladder problem gives the equation x² + y² = L² where L is the constant ladder length. After differentiating with respect to t, what do you get?

Question 8 — Concept (Multiple Choice)

In a related rates problem you are told to find dA/dt. Which of these is the correct first step?

Question 9 — Shadow tip

A 2 m tall person walks at 1 m/s away from a 5 m lamp. Using similar triangles, the shadow length s = (2/3)x where x is distance from lamp. How fast is the shadow tip moving (m/s)?

Tip position = x + s = x + (2/3)x = (5/3)x. Differentiate.
  1. Tip = x + s = (5/3)x
  2. d(tip)/dt = (5/3)·dx/dt = (5/3)(1) = 5/3 ≈ 1.667 m/s
Question 10 — Kite string

A kite flies at fixed altitude 60 m and drifts horizontally at 4 m/s. When 100 m of string is out, how fast is string being paid out (m/s, 2 d.p.)?

L² = x² + 60². When L = 100, x = 80. Then 2L·dL/dt = 2x·dx/dt.
  1. L² = x² + 3600. When L = 100: x = √(10000 − 3600) = √6400 = 80.
  2. 2(100)·dL/dt = 2(80)(4) = 640
  3. dL/dt = 640/200 = 3.2 m/s

§ 06What to Study Next

Related rates sit at the intersection of implicit differentiation, geometry, and real-world modelling. To consolidate and extend this topic, work through these closely related pages:

  • Implicit Differentiation, the core technique underlying every related rates problem.
  • Optimisation, another powerful application: finding the maximum or minimum of a quantity subject to constraints.
  • Linearisation and Differentials, differentials (dx, dy) let you estimate how much one quantity changes when another changes by a small amount.
  • Newton's Method, uses derivatives to solve equations numerically, another key application of differentiation.

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