Applications of Derivatives
Subject: Mathematics
Program: SRMJEEE
Exam: JEE (Main & Advanced)
Why it matters
Derivatives are not just a tool for finding slopes; they are the engine behind all the “why” questions in physics, economics, and engineering. In JEE, you’ll be asked to apply derivative knowledge to optimisation problems, related‑rate questions, and curve‑sketching. These notes give you a quick reference and a worked example that captures the typical style you’ll see on the exam.
1. Simple Explanation
| Concept | What it is | Why you need it |
|---|---|---|
| Derivative as rate of change | (f'(x) = \displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}{h}) | Determines how a quantity changes instant‑by‑instant. |
| Critical points | (f'(x)=0) or (f') undefined | Possible maxima/minima or points of inflection. |
| First‑derivative test | Sign of (f') before/after a critical point | Classifies the critical point. |
| Second‑derivative test | (f''(x)) at a critical point | Quick confirmation of max/min. |
| Related rates | Using ( \frac{d}{dt}g(x(t),y(t)) = g_x,x'(t)+g_y,y'(t) ) | When two or more variables change together. |
| Optimization | Find extreme values of a function or a combination of functions | Minimising cost, maximising area, etc. |
| Inverse functions | ( (f^{-1})'(y) = \frac{1}{f'(x)} ) | When you need the derivative of an inverse. |
2. Key Points (What You Must Master)
| Sub‑Topic | Core Idea | Common JEE Question Types |
|---|---|---|
| A. Finding max/min | Set (f'(x)=0), test with 1st/2nd derivative | Area of triangle with fixed perimeter, Length of a rectangle with given sum of sides |
| B. Related Rates | Differentiate constraint equations w.r.t. time | Water draining from a tank, Distance between moving objects |
| C. Curve Sketching | Use first & second derivative to identify increasing/decreasing, concavity, inflection | Sketch (y=x^3-3x^2+2) |
| D. Inverse Function Derivative | Use ( (f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}) | Derivative of (\sin^{-1}x), (\tan^{-1}x) |
| E. Optimization with Constraints | Lagrange multipliers or substitution | Maximum volume of a right‑cylindrical tank with fixed surface area |
| F. Rate of Change of a Function of a Function | Chain rule | Speed of an object when velocity is a function of position |
3. Worked Example
Problem (Typical JEE style)
A cylindrical can (open top) is to be made from a fixed amount of material (A = 200 \text{ cm}^2) (total surface area). Find the dimensions that maximise the volume (V).
3.1 Setup
- Let radius = (r), height = (h).
- Surface area (open top) :
[ A = \underbrace{2\pi r h}{\text{lateral}} + \underbrace{\pi r^2}{\text{bottom}} = 200 ] - Volume:
[ V = \pi r^2 h ]
3.2 Express (h) in terms of (r)
From the area constraint: [ 2\pi r h + \pi r^2 = 200 \ \Rightarrow h = \frac{200 - \pi r^2}{2\pi r} ]
3.3 Write (V(r))
[ V(r) = \pi r^2 \left( \frac{200 - \pi r^2}{2\pi r} \right) = \frac{r(200 - \pi r^2)}{2} ]
Simplify: [ V(r) = 100r - \frac{\pi}{2} r^3 ]
3.4 Differentiate & Find Critical Point
[ V'(r) = 100 - \frac{3\pi}{2} r^2 ] Set (V'(r)=0): [ 100 - \frac{3\pi}{2} r^2 = 0 \ \Rightarrow r^2 = \frac{200}{3\pi} \ \Rightarrow r = \sqrt{\frac{200}{3\pi}} \approx 4.62 \text{ cm} ]
3.5 Verify Maximum
Second derivative: [ V''(r) = -3\pi r ] At (r>0), (V''(r) < 0) → maximum.
3.6 Find Height
[ h = \frac{200 - \pi r^2}{2\pi r} = \frac{200 - \pi \left( \frac{200}{3\pi} \right)}{2\pi r} = \frac{200 - \frac{200}{3}}{2\pi r} = \frac{\frac{400}{3}}{2\pi r} = \frac{200}{3\pi r} ] Plug (r = \sqrt{\frac{200}{3\pi}}):
[ h = \frac{200}{3\pi} \cdot \frac{1}{\sqrt{\frac{200}{3\pi}}} = \frac{200}{3\pi} \cdot \sqrt{\frac{3\pi}{200}} = \sqrt{\frac{200}{3\pi}} = r ]
Result: For this particular material budget, the optimal cylinder has (h = r).
- **Radius