A related rates problem asks: if one quantity is changing at a known rate, how fast is a related quantity changing? The chain rule applied to both sides of a geometric equation connects the two rates.

If y = f(x) and both change with time t: dy/dt = f'(x) · dx/dt

The 4-Step Method

📋 General Method
Step 1Draw a diagram and label all variables
Step 2Write an equation relating the variables
Step 3Differentiate both sides with respect to time t
Step 4Substitute known values and solve

Example — Inflating Balloon

📋 Balloon inflated at 10 cm³/s — find dr/dt when r = 5 cm
SetupV = (4/3)πr³ → dV/dt = 4πr² · dr/dt
Substitute10 = 4π(25) · dr/dt
Answerdr/dt = 10/(100π) = 1/(10π) ≈ 0.032 cm/s

Example — Sliding Ladder

📋 10m ladder, bottom slides at 2 m/s — find rate top slides when bottom is 6m out
Setupx² + y² = 100 → 2x·dx/dt + 2y·dy/dt = 0
Valuesx=6, y=8, dx/dt=2
Answerdy/dt = −3/2 m/s
⚠ Critical Rule

Never substitute specific values before differentiating. Differentiate the general equation first, then substitute. Substituting early destroys the variables you need.

Frequently Asked Questions

What geometry do I need?▾
Pythagorean theorem, area/volume formulas (circles, spheres, cones, cylinders), and similar triangles cover 90% of related rates problems.
Why is the answer sometimes negative?▾
A negative rate means the quantity is decreasing. In the ladder example, dy/dt < 0 means the top of the ladder is sliding down — physically correct.

Why Related Rates Feel Difficult

Most students find related rates harder than straightforward differentiation. The reason is that these problems require you to set up the equation yourself — you have to recognise the geometric relationship before you can differentiate it. The calculus part (implicit differentiation with respect to time) is actually the easy step. The hard part is the setup.

The fix: always draw a diagram first, always label every quantity that changes with a variable (not a number), and never substitute specific values until after you have differentiated.

Example — Draining Water Tank

Water drains from a conical tank (vertex down) at 2 m³/min. The tank has radius 3m and height 5m. How fast is the water level falling when the water is 3m deep?

📋 Conical tank solution
SetupBy similar triangles: r/h = 3/5, so r = 3h/5. Volume: V = (1/3)πr²h = (1/3)π(3h/5)²h = 3πh³/25
DifferentiatedV/dt = (9πh²/25) · dh/dt
Substitute−2 = (9π·9/25) · dh/dt → dh/dt = −50/(81π) ≈ −0.197 m/min

Example — Lengthening Shadow

A 1.8m person walks away from a 6m lamppost at 1.2 m/s. How fast is their shadow lengthening?

📋 Shadow problem solution
Variablesx = person's distance from post, s = shadow length. dx/dt = 1.2
Similar triangles6/(x+s) = 1.8/s → 6s = 1.8(x+s) → 4.2s = 1.8x → s = 3x/7
Differentiateds/dt = (3/7)·dx/dt = (3/7)·1.2 = 0.514 m/s

The Universal Strategy

The 4-step approach that works every time

1. Draw and label. Every changing quantity gets a variable. Write given rates as derivatives (e.g. dV/dt = −2).

2. Write one equation connecting the variables (Pythagoras, volume formula, similar triangles, etc.).

3. Differentiate implicitly with respect to t. Every variable gets a d/dt applied to it via chain rule.

4. Substitute and solve. Only now plug in the specific values given in the problem.

AM
Dr. Aisha Malik, PhD Mathematics
Senior Lecturer in Applied Mathematics · 12 years teaching calculus
Dr. Malik holds a PhD in Applied Mathematics from the University of Edinburgh and has taught calculus to over 4,000 students. She has reviewed this article for mathematical accuracy and pedagogical clarity.
Technically reviewed by: Prof. James Chen, Stanford Mathematics Department · April 2026