Rate of Change PYQs
Practice Class 12 CBSE Board Previous Year Questions (2014-2026)
Q1
2026 Board (Sample)
00:00
The distance $s$ (in meters) travelled by a particle in $t$ seconds is given by $s = t^3 - 6t^2 + 9t + 5$. The time when its velocity is zero is: 1 Mark
Velocity $v = ds/dt = 3t^2 - 12t + 9$.
Set $v = 0$: $3(t^2 - 4t + 3) = 0$.
$(t-1)(t-3) = 0 \Rightarrow t = 1, 3$.
1s and 3s
Q2
2024 Board
00:00
A spherical balloon is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm. 2 Marks
$V = \frac{4}{3}\pi r^3 \Rightarrow \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
Given $dV/dt = 900$ and $r = 15$.
$900 = 4\pi (15)^2 \frac{dr}{dt} \Rightarrow 900 = 900\pi \frac{dr}{dt}$.
$\frac{dr}{dt} = \frac{1}{\pi}$ cm/s.
1/π cm/s
Q3
2023 Board
00:00
Water is dripping out from a conical funnel at a uniform rate of 5 cm³/s through a tiny hole at the vertex in the bottom. When the slanting height of water is 4 cm, find the rate of decrease of the slant height, given that the vertical angle of the funnel is 120°. 3 Marks
Semi-vertical angle $\alpha = 60^\circ$.
$r = l \sin 60^\circ = l\frac{\sqrt{3}}{2}$, $h = l \cos 60^\circ = l/2$.
$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (\frac{3}{4}l^2)(\frac{1}{2}l) = \frac{\pi}{8}l^3$.
$\frac{dV}{dt} = \frac{3\pi}{8}l^2 \frac{dl}{dt}$.
$5 = \frac{3\pi}{8}(16) \frac{dl}{dt} \Rightarrow 5 = 6\pi \frac{dl}{dt} \Rightarrow \frac{dl}{dt} = \frac{5}{6\pi}$ cm/s.
5/(6π) cm/s
Q4
2020
00:00
A particle moves along the curve $y = x^3$. Find the points on the curve at which the y-coordinate is changing twice as fast as the x-coordinate. 2 Marks
Given $dy/dt = 2(dx/dt)$.
Differentiating $y=x^3$ w.r.t $t$: $dy/dt = 3x^2 (dx/dt)$.
Substituting: $2(dx/dt) = 3x^2 (dx/dt) \Rightarrow 3x^2 = 2 \Rightarrow x^2 = 2/3$.
$x = \pm \sqrt{2/3}$.
When $x = \sqrt{2/3}, y = (2/3)^{3/2} = \frac{2\sqrt{2}}{3\sqrt{3}}$.
When $x = -\sqrt{2/3}, y = -\frac{2\sqrt{2}}{3\sqrt{3}}$.
(±√(2/3), ±(2√2)/(3√3))
Q5
2016
00:00
A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 1 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall? 4 Marks
$x^2 + y^2 = 25$ (where $x$ is distance from wall, $y$ is height).
$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \Rightarrow x \frac{dx}{dt} + y \frac{dy}{dt} = 0$.
Given $dx/dt = 0.01$ m/s, $x = 4$ m.
$4^2 + y^2 = 25 \Rightarrow y = 3$ m.
$4(0.01) + 3(dy/dt) = 0 \Rightarrow 3(dy/dt) = -0.04 \Rightarrow dy/dt = -4/300$ m/s.
Rate of decrease = $4/3$ cm/s.
4/3 cm/s
Q6
2019
00:00
The total revenue received from the sale of $x$ units of a product is given by $R(x) = 13x^2 + 26x + 15$. Find the marginal revenue when $x = 7$. 2 Marks
Marginal Revenue $MR = dR/dx = 26x + 26$.
When $x = 7$: $MR = 26(7) + 26 = 182 + 26 = 208$.
208
Q7
2018
00:00
The total cost $C(x)$ associated with the production of $x$ units of an item is given by $C(x) = 0.005x^3 - 0.02x^2 + 30x + 5000$. Find the marginal cost when $x = 3$. 2 Marks
$MC = dC/dx = 0.015x^2 - 0.04x + 30$.
When $x = 3$: $MC = 0.015(9) - 0.04(3) + 30 = 0.135 - 0.12 + 30 = 30.015$.
30.015
Q8
2022 (Term 1 MCQ)
00:00
The rate of change of area of a circle with respect to its radius $r$ when $r = 3$ cm is: 1 Mark
$A = \pi r^2$.
$\frac{dA}{dr} = 2\pi r$.
When $r = 3$, $\frac{dA}{dr} = 2\pi(3) = 6\pi$.
6π
Q9
2017C
00:00
The volume of a sphere is increasing at the rate of 8 cm³/s. Find the rate of increase of its surface area when the radius is 12 cm. 4 Marks
$V = \frac{4}{3}\pi r^3 \Rightarrow \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
Given $dV/dt = 8, r = 12 \Rightarrow 8 = 4\pi(144)\frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{1}{72\pi}$.
$S = 4\pi r^2 \Rightarrow \frac{dS}{dt} = 8\pi r \frac{dr}{dt}$.
$\frac{dS}{dt} = 8\pi(12) \cdot \frac{1}{72\pi} = \frac{96\pi}{72\pi} = \frac{4}{3}$ cm²/s.
4/3 cm²/s
Q10
2015
00:00
The side of an equilateral triangle is increasing at the rate of 2 cm/s. At what rate is its area increasing when the side of the triangle is 20 cm? 2 Marks
$A = \frac{\sqrt{3}}{4}x^2$.
$\frac{dA}{dt} = \frac{\sqrt{3}}{4}(2x) \frac{dx}{dt}$.
Given $dx/dt = 2, x = 20 \Rightarrow \frac{dA}{dt} = \frac{\sqrt{3}}{2}(20)(2) = 20\sqrt{3}$ cm²/s.
20√3 cm²/s