Rate of Change of Quantities
The derivative $\frac{dy}{dx}$ represents the rate of change of $y$ with respect to $x$. In physical applications, if $y = f(t)$ where $t$ is time, then $\frac{dy}{dt}$ is the rate of change of $y$ with respect to time. This concept is widely used to solve "related rates" problems.
Most Repeated PYQ Topic — Volume/Surface Area Expansion
If two variables $x$ and $y$ are related, say $y = f(x)$, and both vary with time $t$, we use the Chain Rule:
For example, if the radius of a sphere is changing at rate $\frac{dr}{dt}$, the rate of change of volume $V = \frac{4}{3}\pi r^3$ is $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
1.1 Common Geometrical Rates
- Sphere: $V = \frac{4}{3}\pi r^3 \implies \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$. Surface Area $S = 4\pi r^2 \implies \frac{dS}{dt} = 8\pi r \frac{dr}{dt}$.
- Cube: $V = x^3 \implies \frac{dV}{dt} = 3x^2 \frac{dx}{dt}$. $S = 6x^2 \implies \frac{dS}{dt} = 12x \frac{dx}{dt}$.
- Cylinder: $V = \pi r^2 h$. If $h$ constant, $\frac{dV}{dt} = 2\pi r h \frac{dr}{dt}$. If $r$ constant, $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$.
- Cone: $V = \frac{1}{3}\pi r^2 h$. Often involves similar triangles to relate $r$ and $h$.
PYQ — JEE Main 2024 (Jan 31 Shift 1)
Question: Water is being filled in a right circular cone at a constant rate of $5 \text{ cm}^3/\text{s}$. The height of the cone is $10 \text{ cm}$ and the semi-vertical angle is $30^\circ$. Find the rate at which the height of the water is increasing when the water level is $5 \text{ cm}$.
Solution:
Volume of cone $V = \frac{1}{3}\pi r^2 h$.
Given semi-vertical angle $\alpha = 30^\circ$. We know $\tan 30^\circ = \frac{r}{h} \implies \frac{1}{\sqrt{3}} = \frac{r}{h} \implies r = \frac{h}{\sqrt{3}}$.
Substitute $r$ in $V$:
Differentiate w.r.t time $t$:
We are given $\frac{dV}{dt} = 5$ and $h = 5$. Substitute to find $\frac{dh}{dt}$:
Answer: $\frac{3}{5\pi} \text{ cm/s}$
Tangents and Normals
The derivative $\frac{dy}{dx}$ at a point gives the slope of the tangent to the curve $y = f(x)$ at that point. The normal is the line perpendicular to the tangent at that point.
Most Repeated PYQ Topic — Slope & Equations of Tangent/Normal
For a curve $y = f(x)$ at point $(x_1, y_1)$:
- Slope of Tangent: $m = \left(\frac{dy}{dx}\right)_{(x_1, y_1)}$
- Slope of Normal: $m_N = -\frac{1}{m} = -\frac{dx}{dy}$ (if $m \neq 0$)
- Equation of Tangent: $y - y_1 = m(x - x_1)$
- Equation of Normal: $y - y_1 = -\frac{1}{m}(x - x_1)$
2.1 Subtangent and Subnormal
For point $P(x_1, y_1)$ on curve, lengths of subtangent and subnormal are projections on the x-axis:
- Subtangent: $T_m = \left| \frac{y_1}{dy/dx} \right|$
- Subnormal: $N_m = \left| y_1 \cdot \frac{dy}{dx} \right|$
- Length of Tangent: $\left| \frac{y_1 \sqrt{1+(dy/dx)^2}}{dy/dx} \right|$
- Length of Normal: $\left| y_1 \sqrt{1+(dy/dx)^2} \right|$
PYQ — JEE Main 2023 (Apr 10 Shift 2)
Question: If the tangent to the curve $y = x^3 - x$ at a point $P$ is parallel to the line $2x - y + 1 = 0$, then the coordinates of $P$ are:
(a) $(1, 0)$
(b) $(-1, 0)$
(c) $(2, 6)$
(d) Both (a) and (b)
Solution:
Slope of given line $2x - y + 1 = 0 \implies y = 2x + 1$ is $m = 2$.
Curve: $y = x^3 - x \implies \frac{dy}{dx} = 3x^2 - 1$.
Since tangent at $P(x_1, y_1)$ is parallel to the line, their slopes are equal:
If $x_1 = 1$, $y_1 = (1)^3 - 1 = 0$. Point is $(1, 0)$.
If $x_1 = -1$, $y_1 = (-1)^3 - (-1) = -1 + 1 = 0$. Point is $(-1, 0)$.
Answer: (d) Both (a) and (b)
Approximations and Errors
Differentials are used to estimate small changes in the dependent variable $y$ corresponding to small changes in the independent variable $x$. This is highly useful for approximating values of complex functions and calculating relative/percentage errors.
Most Repeated PYQ Topic — Approximating Values
If $y = f(x)$, and $\Delta x$ is a small increment in $x$, then the corresponding increment in $y$ is approximated by:
Thus, $f(x + \Delta x) \approx f(x) + f'(x)\Delta x$.
3.1 Error Calculation Formulas
- Absolute Error: $\Delta x$ (or $dx$)
- Relative Error: $\frac{\Delta x}{x}$
- Percentage Error: $\frac{\Delta x}{x} \times 100\%$
PYQ — JEE Main 2024 (Apr 5 Shift 2)
Question: If the radius of a sphere is measured as $9 \text{ cm}$ with an error of $0.03 \text{ cm}$, then calculate the approximate error in calculating its surface area. (Use $\pi \approx 3.14$)
Solution:
Surface Area of sphere $S = 4\pi r^2$.
Given: $r = 9 \text{ cm}$, Error $\Delta r = 0.03 \text{ cm}$.
Differentiate $S$ w.r.t $r$:
Approximate error $\Delta S \approx dS = \frac{dS}{dr} \Delta r$:
Answer: $6.78 \text{ cm}^2$
Increasing and Decreasing Functions
A function is said to be monotonic in an interval if it is either entirely increasing or entirely decreasing in that interval. The sign of the first derivative $\frac{dy}{dx}$ determines the monotonicity.
Most Repeated PYQ Topic — Intervals of Monotonicity
Let $f(x)$ be differentiable on $(a, b)$.
- Strictly Increasing: $f'(x) > 0$ for all $x \in (a, b)$
- Strictly Decreasing: $f'(x) < 0$ for all $x \in (a, b)$
- Monotonically Increasing (Non-decreasing): $f'(x) \geq 0$
- Monotonically Decreasing (Non-increasing): $f'(x) \leq 0$
4.1 Critical Points
Points where $f'(x) = 0$ or $f'(x)$ is undefined are called critical points. These points divide the domain into intervals of increasing and decreasing behavior.
PYQ — JEE Main 2024 (Jan 30 Shift 2)
Question: The function $f(x) = \frac{x}{\ln x}$ is strictly decreasing in the interval:
(a) $(0, 1) \cup (1, e)$
(b) $(e, \infty)$
(c) $(1, e)$
(d) $(0, e)$
Solution:
$f(x) = \frac{x}{\ln x}$. Domain: $x > 0, x \neq 1$.
Find $f'(x)$ using quotient rule:
For $f(x)$ to be strictly decreasing, $f'(x) < 0$:
Since denominator $(\ln x)^2 > 0$ (for $x \neq 1$), we need $\ln x - 1 < 0$:
Also $x>0$ and $x \neq 1$. So the intervals are $(0, 1) \cup (1, e)$.
Answer: (a) $(0, 1) \cup (1, e)$
Maxima and Minima
Maxima and minima are the largest and smallest values of a function within a given range. We study local (relative) and absolute (global) extrema.
Most Repeated PYQ Topic — Second Derivative Test
Let $f(x)$ be differentiable. Find $x$ where $f'(x) = 0$ (critical point $c$). Then check $f''(c)$:
- Local Maximum: $f'(c) = 0$ and $f''(c) < 0$
- Local Minimum: $f'(c) = 0$ and $f''(c) > 0$
- Point of Inflection / Test Fails: $f'(c) = 0$ and $f''(c) = 0$ (use first derivative test)
5.1 First Derivative Test
At critical point $c$:
- If $f'(x)$ changes sign from positive to negative as $x$ passes through $c$, then $c$ is a point of Local Maximum.
- If $f'(x)$ changes sign from negative to positive as $x$ passes through $c$, then $c$ is a point of Local Minimum.
- If $f'(x)$ does not change sign, $c$ is neither max nor min.
5.2 Absolute Extrema in Closed Interval
How to find Absolute Max/Min on $[a, b]$
- Step 1: Find all critical points $f'(x) = 0$ inside $(a, b)$.
- Step 2: Evaluate $f(x)$ at these critical points.
- Step 3: Evaluate $f(a)$ and $f(b)$.
- Step 4: Largest value is Absolute Max, smallest is Absolute Min.
PYQ — JEE Main 2023 (Apr 13 Shift 1)
Question: The sum of the absolute maximum and absolute minimum values of the function $f(x) = x^3 - 3x + 2$ in the interval $[0, 2]$ is:
Solution:
Given $f(x) = x^3 - 3x + 2$ on $[0, 2]$.
Find $f'(x) = 3x^2 - 3$.
Set $f'(x) = 0 \implies 3x^2 - 3 = 0 \implies x^2 = 1 \implies x = 1, -1$.
Critical point in $(0, 2)$ is $x = 1$. (We discard $x = -1$).
Evaluate $f(x)$ at critical point and endpoints:
Absolute Maximum = $4$ (at $x=2$)
Absolute Minimum = $0$ (at $x=1$)
Sum = $4 + 0 = 4$.
Answer: 4
Rolle's & Mean Value Theorem
These theorems establish a relationship between the values of a function at its endpoints and the value of its derivative inside the interval.
Most Repeated PYQ Topic — Verifying Conditions of Rolle's Theorem
Rolle's Theorem: If a function $f(x)$ is:
- Continuous on closed interval $[a, b]$
- Differentiable on open interval $(a, b)$
- $f(a) = f(b)$
Then there exists at least one point $c \in (a, b)$ such that $f'(c) = 0$.
6.1 Lagrange's Mean Value Theorem (MVT)
If $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists at least one $c \in (a, b)$ such that:
Geometrically, there is a tangent at $c$ that is parallel to the chord joining $(a, f(a))$ and $(b, f(b))$.
PYQ — JEE Main 2021 (Aug 27 Shift 1)
Question: If Rolle's theorem holds for the function $f(x) = x^3 - ax^2 + bx - 4$ in $[1, 2]$ with $f'\left(\frac{4}{3}\right) = 0$, then find the values of $a$ and $b$.
Solution:
Since Rolle's Theorem holds, $f(1) = f(2)$.
Equating $f(1)$ and $f(2)$:
Given $f'\left(\frac{4}{3}\right) = 0$. Find $f'(x)$:
Substitute $b = 3a - 7$:
Answer: $a = 5, b = 8$
Optimization & Word Problems
These involve finding the absolute maximum or minimum values of a function representing a physical quantity (like area, volume, cost, distance) to optimize a real-world scenario.
Most Repeated PYQ Topic — Geometry Based Optimization
Steps to solve:
- Step 1: Draw a figure and label variables.
- Step 2: Write the quantity to be minimized/maximized as a function of one variable.
- Step 3: Differentiate, set to $0$, and verify max/min using second derivative test.
PYQ — JEE Main 2022 (June 28 Shift 1)
Question: A wire of length $36 \text{ m}$ is cut into two pieces. One of the pieces is bent in the form of a square and the other in the form of an equilateral triangle. Find the length of the side of the square that minimizes the total area.
Solution:
Let the side of square be $x$ and side of triangle be $y$.
Perimeter of square = $4x$, Perimeter of triangle = $3y$.
Total wire: $4x + 3y = 36 \implies 3y = 36 - 4x \implies y = 12 - \frac{4x}{3}$.
Total Area $A = x^2 + \frac{\sqrt{3}}{4}y^2$.
Differentiate $A$ w.r.t $x$:
Set $A'(x) = 0$ for minima:
Rationalizing the denominator:
Wait, checking calculations. Let's simplify $A'(x) = 2x - 8\sqrt{3} + \frac{8\sqrt{3}x}{9} = 0$.
Answer: $\frac{36\sqrt{3}}{9 + 4\sqrt{3}} \text{ m}$
Previous Year Questions Analysis
JEE Main (Last 10 Years)
- ✓ Tangents & Normals (Equations/Slopes): 20%
- ✓ Maxima & Minima (Values & Word Problems): 25%
- ✓ Monotonicity (Intervals of Inc/Dec): 18%
- ✓ Rate of Change (Volume/Area): 12%
- ✓ Approximations & Errors: 10%
- ✓ Rolle's & Mean Value Theorem: 15%
JEE Advanced (Last 10 Years)
- ✓ Optimization Word Problems: 30%
- ✓ Roots of Equations using Rolle's: 20%
- ✓ Curve Sketching & Extrema: 20%
- ✓ Multiple Tangents/ Normals: 15%
- ✓ Advanced Monotonicity & Inequalities: 15%
Practice Problem Set
Level 1: Basic
- Find the slope of the tangent to $y = x^2 - 4x + 3$ at $x=2$.
- Find the interval in which $f(x) = x^2 - 4x + 6$ is strictly increasing.
- Find the approximate value of $\sqrt{25.3}$ using differentials.
> FAQ
Frequently Asked Questions
What is the slope of the tangent to a curve at a point?
The slope of the tangent to the curve y = f(x) at point (x1, y1) is given by the value of the derivative at that point, i.e., m = f'(x1) = dy/dx evaluated at x = x1.
What is the condition for a function to be strictly increasing?
A function f(x) is strictly increasing in an interval if its derivative f'(x) > 0 for all x in that interval. If f'(x) ≥ 0, it is monotonically increasing (non-decreasing).
What are the conditions for Rolle's Theorem?
For Rolle's Theorem to hold, the function must be continuous on [a, b], differentiable on (a, b), and f(a) must equal f(b). Then there is at least one c in (a, b) where f'(c) = 0.