Difficulty: Intermediate | Prerequisites: Differentiation rules (chain rule, product rule, implicit differentiation), basic geometry and trigonometry.
Related rates sit at the intersection of implicit differentiation and real-world modelling. The core idea: when two or more quantities are linked by an equation and each changes over time, their rates of change are linked too. You already know how to differentiate y = f(x) with respect to x; here you differentiate every variable with respect to t (time), which turns a static geometric or algebraic relationship into one that describes motion, growth, or change. This topic appears in nearly every introductory calculus course and is a staple of exam problem sets because it forces you to combine differentiation technique with spatial reasoning.
Related rates problems give you a relationship between quantities (distance, angle, volume, etc.) and tell you how some of them change over time. Your job is to differentiate the relationship implicitly with respect to time, plug in the known rates and values at the specific instant described, and solve for the unknown rate. The entire technique boils down to: draw a picture, write an equation, differentiate with respect to t, substitute, solve.
Related rates
A class of problems in which two or more quantities, related by an equation, change with respect to time, and you use implicit differentiation to find how one rate of change depends on another.
Think of it as: everything in the equation is moving, and the chain rule is the thread that ties their speeds together.
Implicit differentiation (with respect to t)
Differentiating both sides of an equation with respect to time, treating every variable as a function of t. Each variable x picks up a dx/dt factor via the chain rule.
In simple terms, this means you differentiate the shape of the equation as usual, but every variable gets a "speed tag" (its derivative with respect to t) attached.
Instantaneous rate of change
The derivative of a quantity at one specific moment in time, not an average over an interval.
Think of it as: the speedometer reading at the exact second you glance at it, rather than your average speed over the whole journey.
Chain rule
If y = f(g(t)), then dy/dt = f'(g(t)) · g'(t). In related rates, this is why differentiating x² with respect to t gives 2x · (dx/dt) rather than just 2x.
In simple terms, every time you differentiate a variable that itself depends on t, you multiply by that variable's own rate of change.
Step 1: Draw and label a diagram. Assign variables to every quantity that changes. Use fixed numbers only for things that never change in the problem.
Step 2: Write an equation relating the variables. This is usually a geometric formula (Pythagorean theorem, area, volume, trig identity) or an algebraic constraint given in the problem.
Step 3: Differentiate both sides with respect to t. Apply the chain rule to every term. Each variable x produces a dx/dt factor.
Step 4: Substitute known values. Plug in the specific numerical values of the variables and their rates at the instant described. Do this after differentiating, not before.
Step 5: Solve for the unknown rate.
Pythagorean theorem: a² + b² = c². Used when a right triangle is involved (ladders, moving objects, distances).
Differentiated: 2a(da/dt) + 2b(db/dt) = 2c(dc/dt)
Trigonometric relationships: sin θ = y/r, tan θ = y/x, etc. Used when angles change (lighthouses, rotating beams, elevation problems).
Example: tan θ = x/3 differentiates to sec²θ · (dθ/dt) = (1/3)(dx/dt)
Multi-variable constraints: x² + y² + z² = constant. Used when three or more linked quantities change simultaneously.
Differentiated: 2x(dx/dt) + 2y(dy/dt) + 2z(dz/dt) = 0
Similar triangles: Ratios between corresponding sides stay constant as the triangle scales. Used in shadow and spotlight problems.
Example: (object height)/(shadow length) = (light height)/(total distance)
Never plug in numerical values for changing quantities before you differentiate. If you replace a variable with a number before differentiating, its derivative vanishes and you lose the rate term you need. Constants (fixed lengths, fixed speeds given as rates) can be substituted at any stage, but variables that change must remain as letters until after the d/dt step.
\text{Pythagorean: } a^2 + b^2 = c^2 \quad\Rightarrow\quad 2a\frac{da}{dt} + 2b\frac{db}{dt} = 2c\frac{dc}{dt}\text{Sphere surface/constraint: } x^2 + y^2 + z^2 = r^2 \quad\Rightarrow\quad 2x\frac{dx}{dt} + 2y\frac{dy}{dt} + 2z\frac{dz}{dt} = 0\tan\theta = \frac{x}{k} \quad\Rightarrow\quad \sec^2\theta \cdot \frac{d\theta}{dt} = \frac{1}{k}\frac{dx}{dt}\sin\theta = \frac{y}{r} \quad\Rightarrow\quad \cos\theta \cdot \frac{d\theta}{dt} = \frac{1}{r}\frac{dy}{dt}\text{Similar triangles: } \frac{a}{b} = \frac{c}{d} \quad\Rightarrow\quad ad = bc \quad\text{(cross-multiply, then differentiate)}Setup: x² + y² + z² = 9. Given dx/dt = 5, dy/dt = 4. Find dz/dt when x = 2, y = 2, z = 1.
Solution:
Differentiate implicitly with respect to t:
2x(dx/dt) + 2y(dy/dt) + 2z(dz/dt) = 0
Substitute the known values:
2(2)(5) + 2(2)(4) + 2(1)(dz/dt) = 0
20 + 16 + 2(dz/dt) = 0
36 + 2(dz/dt) = 0
dz/dt = -18
The negative sign means z is decreasing at that instant.
Setup: A spotlight is on the ground, 8 m from a wall. A 2 m tall person walks toward the wall. The total distance from spotlight to wall is 12 m. The person's shadow falls on the wall. Given dx/dt = 1.6 m/s (the person's walking speed), find dy/dt (the rate the shadow moves on the wall) when y = 3 m.
Solution:
By similar triangles:
(12 - x)/12 = 2/y
Cross-multiplying and differentiating:
1 - (1/12)x = 2(1/y)
Differentiate with respect to t:
-(1/12)(dx/dt) = 2 · (-1/y²)(dy/dt)
-(1/12)(dx/dt) = -2/y² · (dy/dt)
Substitute dx/dt = 1.6, y = 3:
-(1/12)(1.6) = -(2/9)(dy/dt)
dy/dt = 0.6 m/s
The shadow is moving down the wall at 0.6 m/s.
Setup: A right triangle with one fixed leg a = 1, the other leg b = 8. The hypotenuse c satisfies a² + b² = c². Given db/dt = -1 m/s (the horizontal leg is shrinking), find dx/dt when x = 8.
Solution:
First find c: c = √(1 + 64) = √65.
Differentiate a² + b² = c²:
2a(da/dt) + 2b(db/dt) = 2c(dc/dt)
Since a is constant (da/dt = 0):
0 + 2(8)(dx/dt) = 2√65(-1)
16(dx/dt) = -2√65
dx/dt = -√65 / 8 ≈ -1.01 m/s
Setup: A lighthouse is 3 km from a straight shoreline. Its beam rotates at dθ/dt = 4 rev/min = 8π rad/min. Find dx/dt (the speed of the light spot along the shore) when x = 1 km.
Solution:
The relationship is tan θ = x/3, so x = 3 tan θ.
Differentiate with respect to t:
dx/dt = 3 sec²θ · (dθ/dt)
When x = 1: tan θ = 1/3, so sec²θ = 1 + tan²θ = 1 + 1/9 = 10/9.
Alternatively, sec θ = √10/3, so sec²θ = 10/9.
Substitute:
dx/dt = 3(10/9)(8π) = (10/3)(8π) = 80π/3
dx/dt = 80π/3 km/min ≈ 83.8 km/min
The spot of light races along the shore, and it moves faster the further from the closest point it gets (because sec²θ grows).
Setup: An object moves around a circle of radius 10 m. It completes one revolution every 2 minutes, so dθ/dt = 2π/2 = π rad/min. Find dy/dt when y = 6 m.
Solution:
The vertical position is given by sin θ = y/10, so y = 10 sin θ.
Differentiate with respect to t:
dy/dt = 10 cos θ · (dθ/dt)
When y = 6: sin θ = 6/10 = 0.6, so cos θ = √(1 - 0.36) = √0.64 = 0.8.
Substitute:
dy/dt = 10(0.8)(π) = 8π
dy/dt = 8π m/min ≈ 25.1 m/min
The object is rising at 8π m/min at that instant. Notice the vertical speed is greatest when the object is at the leftmost or rightmost point of the circle (cos θ = ±1) and zero at the top and bottom (cos θ = 0).
Related rates are not just exam exercises. Engineers use them when designing systems where multiple dimensions change simultaneously: how fast oil fills a conical tank, how quickly a radar dish must rotate to track a plane, how the shadow of a building moves across a park as the sun changes angle. In physics, nearly every kinematics problem with coupled motion (a pulley system, a piston in a cylinder, an orbiting satellite's projected position) is a related rates problem in disguise.
Substituting before differentiating. Students often plug in the numerical value of a changing variable before taking the derivative. This kills the rate term. If you set x = 2 before differentiating, x looks like a constant and dx/dt disappears from your equation. Always differentiate first, then substitute.
Forgetting the chain rule factor. When differentiating x² with respect to t, the result is 2x(dx/dt), not 2x. Every variable that depends on t picks up its own derivative. Leaving off dx/dt is the single most common algebraic error in these problems.
Confusing angular speed units. Revolutions per minute and radians per minute are different. One revolution = 2π radians. If a problem says "4 rev/min," you need 8π rad/min before using trig derivatives.
Using the wrong triangle or angle. In trig-based problems, students sometimes set up the angle from the wrong reference line, or confuse adjacent and opposite sides. A quick sketch with the angle clearly marked prevents this.
⚠️ Related rates problems appear on virtually every Calculus I and II exam. Expect at least one.
⚠️ The most commonly tested setups: sliding ladder (Pythagorean), expanding/contracting shapes (area or volume), shadow problems (similar triangles), and rotating beacon (trig).
⚠️ Professors often test whether you can identify the correct geometric relationship before any calculus begins. If you pick the wrong equation, the rest of the work is wasted.
⚠️ Sign matters. A negative rate means the quantity is decreasing. Exams frequently ask you to interpret the sign of your answer, not just find the number.
⚠️ Units carry through. If distances are in km and time is in minutes, your rate is in km/min. Some problems mix units deliberately to test whether you convert.
True or False: You should substitute known numerical values into your equation before differentiating with respect to t.
False. Differentiate first, then substitute.
Fill in the blank: When differentiating x² with respect to t, the chain rule gives ____.
2x(dx/dt)
True or False: 1 revolution per minute equals π radians per minute.
False. 1 rev/min = 2π rad/min.
Fill in the blank: In a right triangle where a² + b² = c², differentiating with respect to t gives ____.
2a(da/dt) + 2b(db/dt) = 2c(dc/dt)
True or False: A negative value of dy/dt means y is increasing.
False. Negative dy/dt means y is decreasing.
Q: A 10-foot ladder leans against a wall. The bottom slides away at 2 ft/s. How fast is the top sliding down when the bottom is 6 feet from the wall?
A: Using x² + y² = 100, differentiate: 2x(dx/dt) + 2y(dy/dt) = 0. When x = 6, y = 8. So 2(6)(2) + 2(8)(dy/dt) = 0, giving dy/dt = -3/2 ft/s. The top slides down at 1.5 ft/s.
Q: A conical tank has height equal to its diameter. Water pours in at 3 m³/min. How fast is the water level rising when the depth is 4 m?
A: With h = 2r, so r = h/2. Volume V = (1/3)πr²h = (1/3)π(h/2)²h = πh³/12. Differentiate: dV/dt = (πh²/4)(dh/dt). Substitute: 3 = (π·16/4)(dh/dt), so dh/dt = 3/(4π) m/min.
Q: A light on the ground is 20 m from a building. A person 2 m tall walks toward the building at 1.5 m/s. How fast is the person's shadow on the building shrinking when the person is 8 m from the building?
A: Set up similar triangles with the person's distance from the light as the key variable. The shadow height s satisfies 2/(20 - d) = s/20, where d is distance walked from the light. Differentiate and substitute to find ds/dt.
Q: In Problem 4 above, what happens to dx/dt as the beam rotates so that θ approaches π/2 (the beam is nearly parallel to the shore)?
A: sec²θ grows without bound as θ approaches π/2, so dx/dt increases without limit. The spot of light accelerates along the shore and its speed tends to infinity. This is physically reasonable: a small angular change near 90° sweeps the beam across a huge stretch of coastline.
Q: If x² + y² + z² = 25, dx/dt = 3, dy/dt = -2, and at a certain instant x = 3, y = 0, z = 4, find dz/dt.
A: 2(3)(3) + 2(0)(-2) + 2(4)(dz/dt) = 0. So 18 + 0 + 8(dz/dt) = 0, giving dz/dt = -9/4.
Related rates build directly on the chain rule from single-variable differentiation. If the chain rule feels shaky, revisit it before tackling these problems.
This topic also connects forward to optimisation problems (finding max/min values), which use similar setup steps: draw a picture, write an equation, then differentiate. The difference is that optimisation sets the derivative equal to zero, while related rates sets up a relationship between derivatives.
In multivariable calculus, related rates generalise into partial derivatives and the multivariable chain rule, where you track how changes in several independent variables propagate through a function simultaneously.
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