Difficulty: Intermediate | Prerequisites: Sections 2.1–2.7 (limits, definition of derivative, differentiation rules, chain rule, implicit differentiation)
Related rates sit at the intersection of geometry and differentiation. The core idea: two or more quantities are changing over time, linked by some geometric or physical equation. You know how fast one quantity is changing, and you want to find how fast another is changing at a particular instant. This is one of the first places in calculus where you apply the chain rule to a real physical scenario, and it is a staple of Calc I exams.
You should already be comfortable with implicit differentiation and the chain rule before tackling this section. If those feel shaky, revisit Sections 2.5–2.6 first.
Related rates problems ask you to find how fast one quantity changes, given how fast a related quantity changes. You set up a geometric equation linking the variables, differentiate both sides with respect to time (using the chain rule), then plug in the known values at the specific instant. The critical rule: never substitute numerical values until after you differentiate.
Related rates
A class of problem where two or more quantities, each changing over time, are connected by an equation. You know the rate of change of one and want to find the rate of change of another at a specific instant.
In simple terms: "this thing is moving, and that thing is connected to it, so how fast is that thing moving?"
Rate of change
The derivative of a quantity with respect to time, written as dy/dt or y'(t). It tells you how quickly a variable is increasing or decreasing at a given moment.
Think of it as the speedometer reading for any changing quantity, not just position.
Target rate
The unknown rate of change you are solving for in a related rates problem. This is what the question asks you to find.
Implicit differentiation (with respect to time)
Differentiating both sides of an equation with respect to t, treating every variable as a function of t. The chain rule produces a dy/dt term for each variable y.
In simple terms: instead of solving for one variable first, you differentiate the whole equation as-is and let the chain rule do the work.
Chain rule (in context)
When you differentiate a function of a function, you multiply by the derivative of the inner function. In related rates, this means d/dt[f(y(t))] = f'(y) * dy/dt.
Think of it as the engine that links the rates together.
Draw and label a picture. Mark constant values with numbers. Label changing quantities with letters (as functions of t). Draw arrows for known rates (derivatives) and mark the unknown target rate with a question mark.
Write an equation relating the variables, based on the geometry or physics of the situation. Common sources: Pythagorean theorem, area/volume formulas, trigonometric definitions.
Differentiate both sides with respect to t. Every variable is a function of time, so the chain rule applies. If necessary, solve the resulting equation for the target rate.
Plug in the known current values of the variables and their rates. Compute the target rate.
A weight hangs from a rope that goes up to a pulley 10 ft above the floor, then down to your hand, which is 3 ft above the floor and 15 ft horizontally from the pulley. You walk away at 2 ft/sec. How fast does the weight rise?
Constants: the vertical gap between the pulley and your hand is 10 – 3 = 7 ft.
Variables: h(t) = horizontal distance from pulley to hand; r(t) = length of rope from pulley to hand.
Known: h(0) = 15, h'(0) = 2 ft/sec.
Target: r'(0), since the weight rises at the same rate the rope shortens on the other side.
Setting up the equation:
By the Pythagorean theorem, r² = h² + 7².
Differentiating implicitly:
2r * r' = 2h * h', which gives r' = (h * h') / r.
Plugging in at t = 0:
r(0) = √(15² + 49) = √274.
r'(0) = (15)(2) / √274 = 30/√274 ≈ 1.8 ft/sec.
Water spreads as a circular puddle. The puddle is 1 metre across (radius = 0.5 m) and the area increases at 2 sq m/min. How fast is the radius growing?
Variables: r(t) = radius, A(t) = area.
Known: r(0) = 1/2, A'(0) = 2.
Target: r'(0).
Equation: A = πr².
Differentiating: A' = 2πr * r'.
Solving for the target: r' = A' / (2πr).
Plugging in: r'(0) = 2 / (2π * 1/2) = 2/π ≈ 0.64 m/min.
Plausibility check: the radius is growing at roughly half a metre per minute, which makes sense for a puddle fed by a stream.
A searchlight is 20 metres from a wall. The beam is at 30° from perpendicular and rotates at 5°/sec. How fast does the light spot move along the wall?
Constant: distance to wall = 20 m.
Variables: θ(t) = angle from perpendicular, s(t) = distance of the light spot along the wall from the nearest point.
Known: θ(0) = 30°, θ'(0) = 5°/sec.
Target: s'(0).
Equation: tan(θ) = s/20, so s = 20 tan(θ).
Differentiating: ds/dt = 20 sec²(θ) * dθ/dt.
Converting to radians (essential):
θ(0) = 30° = π/6 rad.
θ'(0) = 5°/sec = π/36 rad/sec.
Plugging in: s'(0) = 20 sec²(π/6) * π/36 = (20/27)π ≈ 2.3 m/sec.
Note: the problem gives speed, not velocity, so the answer is |s'(0)| = (20/27)π, regardless of which direction the light rotates.
Pythagorean theorem (right-triangle problems):
r² = a² + b², which differentiates to 2r * r' = 2a * a' + 2b * b'.
Area of a circle (expanding/contracting circles):
A = πr², which differentiates to A' = 2πr * r'.
Tangent relationship (spotlight/shadow problems):
tan(θ) = opposite/adjacent, which differentiates to sec²(θ) * θ' = (1/adjacent) * s'.
Always convert angles to radians before plugging into any trig derivative.
Volume of a sphere (balloon problems):
V = (4/3)πr³, which differentiates to V' = 4πr² * r'.
Volume of a cone (draining/filling tank problems):
V = (1/3)πr²h. Note: if r and h are related by the cone's geometry (similar triangles), substitute to get V in terms of one variable before differentiating.
Related rates turn up whenever two changing quantities are linked by geometry or physics. Engineers use them to work out how fast a shadow lengthens as the sun moves, how quickly pressure changes in a piston as the cylinder compresses, or how fast the water level drops in a conical tank being drained. Radar tracking systems rely on the same mathematics: a known rotational speed of an antenna, combined with the geometry of the target, gives the target's actual velocity.
Students often plug in numerical values before differentiating. This collapses a variable into a constant, and the derivative of a constant is zero. You lose the rate you are trying to find. Always differentiate first, substitute second.
Students sometimes forget that every variable is a function of time. When you differentiate h², you must apply the chain rule and get 2h * dh/dt, not just 2h.
In trig-based problems, students frequently leave angles in degrees when differentiating. The standard derivative formulas (d/dx tan(x) = sec²(x), etc.) only hold when x is in radians. Forgetting to convert gives wildly wrong answers.
Students occasionally confuse speed and velocity in the answer. If a problem asks for speed, the answer is the absolute value of the rate. If it asks for velocity, the sign matters (positive or negative indicates direction).
⚠️ Related rates are a near-certainty on Calc I exams. Expect at least one problem involving a geometric setup (triangle, circle, cone, or sphere).
⚠️ The most commonly tested mistake: substituting values before differentiating. Exam graders look for this.
⚠️ If trig is involved, you will likely need to convert degrees to radians. Forgetting this conversion is one of the most penalised errors.
⚠️ Always check whether the problem asks for a rate of change (signed, with direction) or a speed (absolute value). Losing a mark over a missing absolute value is common.
True or false: You should plug in known values before differentiating.
False. Always differentiate first, then substitute.
Fill in the blank: Differentiating A = πr² with respect to t gives A' = ________.
2πr * r'.
True or false: d/dt[tan(θ)] = sec²(θ) whether θ is in degrees or radians.
False. The formula only works when θ is in radians.
Fill in the blank: In a related rates problem, the chain rule turns d/dt[h²] into ________.
2h * dh/dt.
True or false: If a problem says "speed is 5°/sec," the velocity could be +5 or −5.
True. Speed is the magnitude; velocity has a sign.
Q: A circular oil slick has a radius of 10 m and the radius is increasing at 0.5 m/min. How fast is the area increasing?
A: A = πr², so A' = 2πr * r' = 2π(10)(0.5) = 10π ≈ 31.4 sq m/min.
Q: A 13-ft ladder leans against a wall. The bottom slides away at 2 ft/sec. When the bottom is 5 ft from the wall, how fast is the top sliding down?
A: x² + y² = 13². Differentiating: 2x * x' + 2y * y' = 0. When x = 5, y = 12. So y' = –(x * x')/(y) = –(5)(2)/12 = –5/6 ≈ –0.83 ft/sec (the negative sign means the top is sliding downward).
Q: A conical tank (radius 3 m, height 6 m) is being filled with water at 2 cu m/min. How fast is the water level rising when the water is 4 m deep?
A: By similar triangles, r = h/2. So V = (1/3)π(h/2)²h = πh³/12. Differentiating: V' = πh²/4 * h'. Solving: h' = 4V'/(πh²) = 4(2)/(π16) = 1/(2π) ≈ 0.16 m/min.
Q: Why is it wrong to substitute h(0) = 15 into r = √(h² + 49) before differentiating?
A: Because substituting makes r a fixed number (√274), and the derivative of a constant is zero. You lose the relationship between h' and r', which is the entire point.
Related rates build directly on the chain rule (Section 2.5) and implicit differentiation (Section 2.6). If either of those felt rough, revisiting them will make related rates problems considerably easier.
This section also connects forward to optimisation problems (Section 3.7), where you again set up equations from geometry and differentiate, but solve for a maximum or minimum rather than a rate. The skill of translating a word problem into an equation, then applying calculus, is shared across both.
In physics, related rates are the foundation of kinematics problems where multiple quantities (position, angle, length) change simultaneously. If you go on to take multivariable calculus, the same idea generalises to partial derivatives and the multivariable chain rule.
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