Application of Derivatives: The Rate Question Pattern
Every application-of-derivatives question is one of three shapes, and all three are answered by the same six-step checklist.
A strategy note identifying the three shapes of rate-application question in JEE — related rates, optimisation and curve behaviour — and the six-step checklist that solves all three.
How this actually works
Three question shapes, one checklist
Every application-of-derivatives question in JEE is one of three shapes: related rates, where two variables move together; optimisation, where you find a maximum or minimum; or curve behaviour, where you find where a function is increasing. The checklist that solves all three is: name the variable you want, find the relation that links it to the given data, differentiate, set the relevant rate or the derivative to zero, solve, and check the boundary. Every one of those six steps is short; the only way to lose the mark is to skip one, and step two is the one that gets skipped.
Related rates and the lost chain factor
The trap in related rates is the forgotten chain rule factor. If the radius of a sphere is growing at dr/dt, the volume grows at dV/dt = 4πr² dr/dt, and the r² factor is exactly the part that vanishes from most candidates' work. Write dV/dt on one side and 4πr² dr/dt on the other, then substitute the radius at the instant the question names. That instant is almost always t = 0 or the moment the variable reaches a stated value, and using the general expression instead of the value at that instant is the second most common error.
Optimisation in one variable
The optimisation shape is identical except that the quantity being maximised is set to a variable, differentiated, and set to zero. The JEE version usually wraps the shape in a geometry problem — the area of a rectangle inscribed in a circle, the volume of a box made by cutting corners off a cube. The constraint reduces it to one variable before differentiation, and skipping that reduction is what makes the second derivative test fail.
The curve-behaviour shape
Here there is no extremum to find. The question asks where a function increases, decreases, or has a local maximum, and the answer is the sign of the derivative. The JEE twist is the point where the derivative does not exist — a corner, a cusp, or a discontinuity — which is why the domain has to be read before differentiating.
Application of Derivatives: The Rate Question Pattern: the facts to keep
| The three shapes | Related rates, optimisation, and curve behaviour - all answered by the same six-step checklist |
|---|---|
| The checklist | Name the variable, find the linking relation, differentiate, set the rate or derivative to zero, solve, check the boundary |
| The chain factor | If the radius of a sphere grows at dr/dt, the volume grows at dV/dt = 4 pi r squared dr/dt - the r squared is the lost part |
| The instant | Use the radius at the instant the question names, not the general expression |
| Optimisation | Reduce the constraint to one variable before differentiating, or the second derivative test fails |
| Curve behaviour | The answer is the sign of the derivative, and the domain must be read first for the points where the derivative does not exist |
Questions students ask about this
Why do I keep losing the r² in related rates?
Because you differentiate the relationship between the two variables and then forget to multiply by the derivative of the inner variable. Write the full chain on the page every time — dV/dt = (dV/dr)(dr/dt) — and the factor cannot go missing.
Do I need to find a critical point if the question only asks for the maximum?
Yes, unless the maximum is at a boundary. On a closed interval the maximum is either at a critical point or at an endpoint, so you must check both to know which one the answer is.