Author: Rishabh Kumar (IIT + ISI | Global Math Mentor)
Published: October 2025
Category: Calculus | Applications
🔹 What Are Related Rates?
In many real-world situations, two or more quantities change with time — and their rates of change are connected.
For example:
- The radius and volume of a balloon expanding,
- The height and shadow length of a moving lamppost,
- The water level and radius in a conical tank being filled.
These problems belong to a special category of calculus known as Related Rates, where we find how one rate of change affects another using implicit differentiation and the chain rule.
🔹 The Core Idea
If two quantities xxx and yyy are related by an equation f(x,y)=0f(x, y) = 0f(x,y)=0,
then by differentiating with respect to time ttt, we can connect dxdt\frac{dx}{dt}dtdx and dydt\frac{dy}{dt}dtdy: ddtf(x,y)=0\frac{d}{dt}f(x, y) = 0dtdf(x,y)=0
and hence, ∂f∂xdxdt+∂f∂ydydt=0\frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} = 0∂x∂fdtdx+∂y∂fdtdy=0
This allows us to find one rate given another.
🔹 Step-by-Step Example
🧩 Example 1: Expanding Circle
A circle’s radius increases at a rate of 2 cm/s2 \, \text{cm/s}2cm/s.
How fast is the area increasing when r=5 cm?r = 5 \, \text{cm}?r=5cm?
Given: A=πr2,drdt=2A = \pi r^2, \quad \frac{dr}{dt} = 2A=πr2,dtdr=2
Differentiate both sides with respect to ttt: dAdt=2πrdrdt\frac{dA}{dt} = 2\pi r \frac{dr}{dt}dtdA=2πrdtdr
Substitute r=5r = 5r=5: dAdt=2π(5)(2)=20π\frac{dA}{dt} = 2\pi(5)(2) = 20\pidtdA=2π(5)(2)=20π
✅ The area increases at a rate of 20π cm2/s20\pi \, \text{cm}^2/\text{s}20πcm2/s.
🔹 Example 2: Sliding Ladder Problem (Classic Exam Favorite)
A 5 m ladder leans against a wall. The bottom slides away from the wall at 1 m/s1 \, \text{m/s}1m/s.
How fast is the top sliding down when the bottom is 3 m from the wall?
Let the height on the wall be yyy and the distance from the wall be xxx. x2+y2=25x^2 + y^2 = 25×2+y2=25
Differentiate w.r.t time ttt: 2xdxdt+2ydydt=02x\frac{dx}{dt} + 2y\frac{dy}{dt} = 02xdtdx+2ydtdy=0
Substitute values: x=3,dxdt=1,y=4x = 3, \quad \frac{dx}{dt} = 1, \quad y = 4x=3,dtdx=1,y=4 3(1)+4dydt=0⇒dydt=−343(1) + 4\frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = -\frac{3}{4}3(1)+4dtdy=0⇒dtdy=−43
✅ The top slides down at 0.75 m/s0.75 \, \text{m/s}0.75m/s.
🔹 Common Mistakes to Avoid
- ❌ Forgetting to differentiate implicitly with respect to time.
- Always use the chain rule: dydt\frac{dy}{dt}dtdy appears whenever yyy changes with ttt.
- ❌ Plugging in numerical values too early.
- Keep the variables symbolic until after differentiation.
- ❌ Mixing up signs.
- A decreasing quantity has a negative rate of change.
🔹 Real-World Applications
- Physics: velocity, acceleration, angular motion
- Engineering: fluid flow, expansion, optimization
- Economics: rate of cost or demand changes
- Mathematics exams: IB AA HL, AP Calculus BC, STEP, MAT — frequent topic for applied differentiation
🔹 Advanced Challenge
A cone has a height three times its radius. Water is poured into it at a rate of 100 cm3/s100 \, \text{cm}^3/s100cm3/s.
Find how fast the water level is rising when the radius is 5 cm5 \, \text{cm}5cm.
👉 Try solving it using V=13πr2hV = \frac{1}{3}\pi r^2hV=31πr2h and h=3rh = 3rh=3r.
(Full solution discussed in Math by Rishabh’s Mentorship Classes under Applications of Differentiation module.)
🌟 Why This Concept Matters
Related rates sharpen your intuition about how quantities influence each other — a vital skill in higher mathematics and physics.
Understanding how and why change propagates through a system transforms your problem-solving from mechanical to conceptual — exactly what top universities like Oxford, Cambridge, and MIT value.
📘 Learn Beyond Formulas
In the Mathematics Elevate Mentorship, we go beyond just applying formulas:
✅ Connect related rates to implicit differentiation and optimization
✅ Learn structured problem patterns for STEP / MAT
✅ Build mathematical intuition — not just exam tricks
🚀 Ready to elevate your calculus mastery?
👉 Book your personalized mentorship session now at MathByRishabh.com
