Source: MTH 132, Michigan State University, Sections 2.6/2.8
Difficulty: Intermediate to Challenging | Prerequisites: Sections 2.1 through 2.6 (all differentiation rules, implicit differentiation, basic geometry formulas for triangles, circles, cones, and Pythagorean theorem).
Related rates problems ask you to find how fast one quantity is changing when you know how fast another quantity is changing, and the two quantities are linked by an equation. This is implicit differentiation with respect to time instead of x. These problems are word problems, so the main challenge is translating a physical scenario into a mathematical relationship before you differentiate. You should be fully comfortable with implicit differentiation (Section 2.6) and with basic geometry (Pythagorean theorem, similar triangles, volume of a cone) before tackling this section.
Find an equation connecting the changing quantities, differentiate both sides with respect to time t (every variable picks up a d/dt via the chain rule), plug in the known rates and the known values at the instant in question, and solve for the unknown rate. The hard part is drawing the picture and choosing the right equation, not the calculus itself.
Related rates
A class of problems where two or more quantities change with time and are connected by an equation, so that knowing one rate of change lets you find another.
Think of it as: "If I know how fast this is changing, how fast is that changing?"
Rate of change
The derivative of a quantity with respect to time, written as dx/dt, dy/dt, dV/dt, etc.
In simple terms, it tells you how many units that quantity gains or loses per second (or per minute, etc.).
Differentiate with respect to t
Apply d/dt to both sides of an equation, treating every variable as a function of time. Each variable's derivative picks up a d(variable)/dt factor, the same way y picked up dy/dx in implicit differentiation.
Similar triangles
A geometric tool used to eliminate a variable from a related rates equation. When two triangles share the same angles, the ratios of their corresponding sides are equal.
Think of it as: the relationship that lets you express the radius of a cone's liquid surface in terms of the liquid's height, so you can work with one variable instead of two.
Draw a picture and label every quantity that changes with a variable (not a number).
Write down what you know (given rates) and what you want (the unknown rate).
Find an equation that relates the variables. This is the modelling step, and it is where most mistakes happen.
If you have too many variables, use a geometric constraint (e.g. similar triangles) to eliminate one.
Differentiate both sides with respect to t.
Plug in all known values at the specific instant.
Solve for the unknown rate.
A rope runs from a boat to a pulley on a dock that sits 2 m above the bow. The rope is pulled in at 1/2 m/s. How fast is the boat approaching the dock when it is 10 m away?
Setup: Let x = horizontal distance from boat to dock, and L = length of rope from bow to pulley. The pulley is 2 m above the bow, so the Pythagorean theorem gives:
L² = x² + 2² = x² + 4
Given: dL/dt = −1/2 (the rope is getting shorter). Want: dx/dt when x = 10.
Find L when x = 10: L² = 100 + 4 = 104, so L = √104 = 2√26.
Differentiate with respect to t:
2L · dL/dt = 2x · dx/dt
Plug in and solve:
2(2√26)(−1/2) = 2(10) · dx/dt
−2√26 = 20 · dx/dt
dx/dt = −√26/10 ≈ −0.51 m/s
The boat approaches the dock at √26/10 m/s (roughly 0.51 m/s). The negative sign confirms the distance is decreasing.
A cone-shaped filter has height 15 in and top diameter 10 in (radius 5 in). Liquid drips out at 4 in³/sec. How fast is the level dropping when the liquid is 1 in deep?
Setup: Let h = depth of liquid, r = radius of the liquid surface. The volume of a cone is:
V = (1/3)πr²h
Eliminate r using similar triangles: The full cone has radius 5 at height 15, so r/h = 5/15 = 1/3, giving r = h/3.
Substitute:
V = (1/3)π(h/3)²h = (1/3)π(h²/9)h = πh³/27
Given: dV/dt = −4 (liquid is leaving). Want: dh/dt when h = 1.
Differentiate:
dV/dt = (3πh²/27) · dh/dt = (πh²/9) · dh/dt
Plug in h = 1:
−4 = (π/9) · dh/dt
dh/dt = −36/π ≈ −11.46 in/sec
The level drops at 36/π in/sec. That is fast because at h = 1 the cone is very narrow, so a small volume change produces a large height change.
A 10 ft ladder leans against a wall. The angle θ between the ladder and the wall is π/4 and increasing at π/12 rad/sec. How fast is the top sliding down?
Setup: Let y = height of the top of the ladder on the wall. The relationship between the angle and y is:
cos(θ) = y/10, so y = 10 cos(θ)
Differentiate:
dy/dt = −10 sin(θ) · dθ/dt
Plug in θ = π/4, dθ/dt = π/12:
dy/dt = −10 sin(π/4) · (π/12) = −10 · (√2/2) · (π/12) = −5π√2/12 ≈ −1.85 ft/sec
The top slides down at 5π√2/12 ft/sec.
A 10 m statue and a 50 m pole are 20 m apart. A light rises up the pole. When the light is at 40 m and rising at 5 m/min, how fast is the shadow shrinking?
Setup: Let h = height of the light on the pole, and s = length of the shadow cast by the statue (measured from the base of the statue, away from the pole). The tip of the shadow, the top of the statue, and the light source form a line. Using similar triangles:
10/(20 + s) = h/... is one approach, but more directly: the light at height h on the pole, the statue top at height 10 and 20 m from the pole, and the shadow tip on the ground at distance (20 + s) from the pole all lie along one ray. By similar triangles:
10/h = (20 + s − 20)/(50 − 0)... Let us set up coordinates. Place the pole at x = 0 and the statue at x = 20. The shadow extends from x = 20 to x = 20 + s. The light is at (0, h). The line from (0, h) through (20, 10) hits the ground at (20 + s, 0).
Slope of the line: (10 − h)/(20 − 0) = (0 − h)/((20 + s) − 0)
So: (10 − h)/20 = −h/(20 + s)
Cross-multiply: (10 − h)(20 + s) = −20h
200 + 10s − 20h − hs = −20h
200 + 10s − hs = 0
s(10 − h) = −200
s = 200/(h − 10)
Differentiate:
ds/dt = −200/(h − 10)² · dh/dt
Plug in h = 40, dh/dt = 5:
ds/dt = −200/(30)² · 5 = −200/900 · 5 = −1000/900 = −10/9 m/min
The shadow shrinks at 10/9 ≈ 1.11 m/min.
Pythagorean theorem: a² + b² = c² (used whenever a right triangle connects the changing quantities, e.g. boat/dock, ladder problems).
Volume of a cone: V = (1/3)πr²h
Similar triangles: r/h = R/H (where R and H are the full dimensions of the cone), so r = (R/H)h. This lets you rewrite V in terms of h alone.
Trig relationships for ladder/angle problems: If θ is the angle between the ladder and the wall and L is the ladder length, then y = L cos(θ) and x = L sin(θ).
Shadow problems (coordinate geometry): Set up coordinates, write the equation of the line from the light source through the top of the object to the ground, and solve for the shadow tip's position.
General procedure:
Draw and label.
Identify all rates (given and wanted), using d/dt notation.
Write the geometric equation connecting the variables.
Eliminate extra variables using constraints (similar triangles, etc.).
Differentiate with respect to t.
Substitute the known values at the instant in question.
Solve for the unknown rate and interpret the sign.
Related rates problems model real situations in engineering, physics, and everyday life. The boat-and-dock problem is a simplified version of winch-and-cable calculations in marine engineering. The conical filter problem mirrors how chemical engineers track liquid levels in hoppers and funnels. Ladder problems appear in structural safety analysis (how fast a scaffold or ladder moves under loading). Shadow problems are the basis for sundial calculations, stage-lighting design, and even satellite shadow tracking.
Students often plug in the known values before differentiating. You must differentiate first and substitute after, because substituting a constant kills the variable and its rate disappears from the equation.
Students forget to use the chain rule when differentiating with respect to t. Every variable is a function of time, so d/dt[x²] = 2x · dx/dt, not just 2x.
Students sometimes confuse the sign of a rate. A quantity that is decreasing has a negative derivative. If the rope is getting shorter, dL/dt is negative.
In cone problems, students often forget to eliminate the extra variable (r) before differentiating. If you differentiate V = (1/3)πr²h with both r and h, you get a product rule in two unknowns, and you will not be able to solve for dh/dt unless you also know dr/dt separately.
⚠️ Related rates problems are a near-certainty on a Calc I exam. They typically account for one full problem (multi-part) worth significant marks.
⚠️ The most commonly tested setups are: Pythagorean theorem (ladder, boat), cone/tank draining or filling, expanding circle or sphere, and shadow problems.
⚠️ Professors often test whether you differentiate before substituting. If your answer looks suspiciously clean or you get 0 for a rate that should not be 0, you likely substituted too early.
True or False: You should substitute known values into your equation before differentiating with respect to t. Answer: False. Always differentiate first.
Fill in the blank: In a cone with height H and radius R, if the liquid depth is h, then the radius of the liquid surface is r = ______. Answer: (R/H) · h
True or False: If a quantity is decreasing, its rate of change is positive. Answer: False. A decreasing quantity has a negative rate.
Fill in the blank: d/dt[x²] = ______. Answer: 2x · dx/dt
True or False: In a ladder problem, the length of the ladder changes over time. Answer: False. The ladder length is constant; its position changes.
Q: A spherical balloon is being inflated at a rate of 3 ft³/min. How fast is the radius increasing when the radius is 2 ft?
A: V = (4/3)πr³. Differentiate: dV/dt = 4πr² · dr/dt. Plug in: 3 = 4π(4) · dr/dt. Solve: dr/dt = 3/(16π) ft/min.
Q: A 13 ft ladder leans against a wall. The base slides away at 2 ft/sec. How fast is the top sliding down when the base is 5 ft from the wall?
A: x² + y² = 169. When x = 5, y = 12. Differentiate: 2x · dx/dt + 2y · dy/dt = 0. Plug in: 2(5)(2) + 2(12) · dy/dt = 0. Solve: dy/dt = −20/24 = −5/6 ft/sec. The top slides down at 5/6 ft/sec.
Q: Water is poured into a cylindrical tank (radius 3 m) at 2 m³/min. How fast is the water level rising?
A: V = πr²h = 9πh. Differentiate: dV/dt = 9π · dh/dt. So dh/dt = 2/(9π) m/min. This one is simpler because the radius is constant.
Q: In a related rates problem, why do you differentiate before substituting numerical values?
A: If you substitute before differentiating, you replace a variable with a constant. The derivative of a constant is zero, so the rate of change of that quantity vanishes from your equation, and you lose the information you need.
Related rates is implicit differentiation (Section 2.6) with respect to time. If implicit differentiation felt mechanical, related rates is where that mechanics meets physical intuition.
It also connects forward to optimisation (Sections 4.1 and 4.7), where you again build an equation from a word problem and use calculus to find a quantity of interest. The modelling step (drawing the picture, choosing the equation) is the same skill in both topics.
The geometry refresher here (Pythagorean theorem, similar triangles, cone volume) will reappear in volumes of revolution later in Calculus II.
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