Rate of Change
NCERT EXERCISE 6.1 • FULL SOLUTIONS
💡 Core Principle
The derivative $\frac{dy}{dx}$ represents the rate of change of $y$ with respect to $x$. If both $x$ and $y$ depend on time $t$, we use the chain rule: $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$.
Question 01
Find the rate of change of the area of a circle w.r.t. its radius $r$ when (a) $r=3$ cm (b) $r=4$ cm.
$A = \pi r^2 \implies \frac{dA}{dr} = 2\pi r$.
Question 02
Volume of a cube increases at $8 \text{ cm}^3/\text{s}$. How fast is the surface area increasing when edge is $12$ cm?
$V = x^3 \implies \frac{dV}{dt} = 3x^2 \frac{dx}{dt}$. Given $\frac{dV}{dt} = 8, x = 12$.
$8 = 3(144)\frac{dx}{dt} \implies \frac{dx}{dt} = \frac{1}{54} \text{ cm/s}$.
$S = 6x^2 \implies \frac{dS}{dt} = 12x \frac{dx}{dt} = 12(12)(\frac{1}{54}) = \frac{8}{3} \text{ cm}^2/\text{s}$.
$8 = 3(144)\frac{dx}{dt} \implies \frac{dx}{dt} = \frac{1}{54} \text{ cm/s}$.
$S = 6x^2 \implies \frac{dS}{dt} = 12x \frac{dx}{dt} = 12(12)(\frac{1}{54}) = \frac{8}{3} \text{ cm}^2/\text{s}$.
Result: 8/3 cm²/s
Question 03
The radius of a circle is increasing at $3 \text{ cm/s}$. Find the rate at which area is increasing when $r=10$ cm.
Given $\frac{dr}{dt} = 3$. $A = \pi r^2 \implies \frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
$\frac{dA}{dt} = 2\pi(10)(3) = 60\pi \text{ cm}^2/\text{s}$.
$\frac{dA}{dt} = 2\pi(10)(3) = 60\pi \text{ cm}^2/\text{s}$.
Result: 60π cm²/s
Question 04
An edge of a variable cube is increasing at $3 \text{ cm/s}$. How fast is the volume increasing when edge is $10$ cm?
Let edge be $x$. Given $\frac{dx}{dt} = 3, x = 10$.
$V = x^3 \implies \frac{dV}{dt} = 3x^2 \frac{dx}{dt}$.
$\frac{dV}{dt} = 3(10)^2(3) = 900 \text{ cm}^3/\text{s}$.
$V = x^3 \implies \frac{dV}{dt} = 3x^2 \frac{dx}{dt}$.
$\frac{dV}{dt} = 3(10)^2(3) = 900 \text{ cm}^3/\text{s}$.
Result: 900 cm³/s
Question 05
A stone is dropped into a lake and waves move in circles at $5 \text{ cm/s}$. When $r=8$ cm, how fast is area increasing?
Given $\frac{dr}{dt} = 5, r = 8$.
$\frac{dA}{dt} = 2\pi r \frac{dr}{dt} = 2\pi(8)(5) = 80\pi \text{ cm}^2/\text{s}$.
$\frac{dA}{dt} = 2\pi r \frac{dr}{dt} = 2\pi(8)(5) = 80\pi \text{ cm}^2/\text{s}$.
Result: 80π cm²/s
Question 06
Radius of a circle increases at $0.7 \text{ cm/s}$. Find rate of increase of its circumference.
$C = 2\pi r \implies \frac{dC}{dt} = 2\pi \frac{dr}{dt}$.
$\frac{dC}{dt} = 2\pi(0.7) = 1.4\pi \text{ cm/s}$.
$\frac{dC}{dt} = 2\pi(0.7) = 1.4\pi \text{ cm/s}$.
Result: 1.4π cm/s
Question 07
Length $x$ decreases at $5 \text{ cm/min}$, width $y$ increases at $4 \text{ cm/min}$. Find rate of change of Perimeter and Area when $x=8, y=6$.
$\frac{dP}{dt} = 2(-5 + 4) = -2 \text{ cm/min}$.
$\frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt} = 8(4) + 6(-5) = 2 \text{ cm}^2/\text{min}$.
$\frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt} = 8(4) + 6(-5) = 2 \text{ cm}^2/\text{min}$.
Question 08
A balloon is inflated at $900 \text{ cm}^3/\text{s}$. Find rate of change of radius when $r=15$ cm.
$V = \frac{4}{3}\pi r^3 \implies \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
$900 = 4\pi(15)^2 \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{900}{900\pi} = \frac{1}{\pi} \text{ cm/s}$.
$900 = 4\pi(15)^2 \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{900}{900\pi} = \frac{1}{\pi} \text{ cm/s}$.
Result: 1/π cm/s
Question 09
A balloon which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.
$V = \frac{4}{3}\pi r^3 \implies \frac{dV}{dr} = 4\pi r^2$.
At $r = 10$: $\frac{dV}{dr} = 4\pi(10)^2 = 400\pi \text{ cm}^3/\text{cm}$.
At $r = 10$: $\frac{dV}{dr} = 4\pi(10)^2 = 400\pi \text{ cm}^3/\text{cm}$.
Result: 400π cm³/cm
Question 10
A 5m ladder leans against a wall. Bottom is pulled at $2 \text{ cm/s}$. Find rate of height decrease when foot is 4m away.
$x^2 + y^2 = 25 \implies \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}$.
At $x=4, y=3$: $\frac{dy}{dt} = -\frac{4}{3}(2) = -8/3 \text{ cm/s}$.
At $x=4, y=3$: $\frac{dy}{dt} = -\frac{4}{3}(2) = -8/3 \text{ cm/s}$.
Result: 8/3 cm/s (decreasing)
Question 11
A particle moves along $6y = x^3 + 2$. Find points where $y$-coordinate changes 8 times as fast as $x$-coordinate.
$6\frac{dy}{dt} = 3x^2 \frac{dx}{dt}$. Given $\frac{dy}{dt} = 8\frac{dx}{dt}$.
$6(8\frac{dx}{dt}) = 3x^2 \frac{dx}{dt} \implies 48 = 3x^2 \implies x^2 = 16 \implies x = \pm 4$.
If $x=4, y = \frac{4^3+2}{6} = 11$. If $x=-4, y = \frac{-64+2}{6} = -31/3$.
$6(8\frac{dx}{dt}) = 3x^2 \frac{dx}{dt} \implies 48 = 3x^2 \implies x^2 = 16 \implies x = \pm 4$.
If $x=4, y = \frac{4^3+2}{6} = 11$. If $x=-4, y = \frac{-64+2}{6} = -31/3$.
Points: (4, 11) and (-4, -31/3)
Question 12
Radius of air bubble increases at $0.5 \text{ cm/s}$. Find rate of volume increase when $r=1$ cm.
$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} = 4\pi(1)^2(0.5) = 2\pi \text{ cm}^3/\text{s}$.
Result: 2π cm³/s
Question 13
A balloon has diameter $\frac{3}{2}(2x+1)$. Find rate of change of volume w.r.t $x$.
$r = \frac{3}{4}(2x+1)$. $V = \frac{4}{3}\pi [\frac{3}{4}(2x+1)]^3 = \frac{9\pi}{16}(2x+1)^3$.
$\frac{dV}{dx} = \frac{9\pi}{16} \cdot 3(2x+1)^2 \cdot 2 = \frac{27\pi}{8}(2x+1)^2$.
$\frac{dV}{dx} = \frac{9\pi}{16} \cdot 3(2x+1)^2 \cdot 2 = \frac{27\pi}{8}(2x+1)^2$.
Result: $\frac{27\pi}{8}(2x+1)^2$
Question 14
Sand forms a cone where $h = \frac{1}{6}r$. Find rate of height increase when $h=4$ cm.
$V = 12\pi h^3 \implies \frac{dV}{dt} = 36\pi h^2 \frac{dh}{dt} \implies 12 = 36\pi(16)\frac{dh}{dt}$.
$\frac{dh}{dt} = \frac{1}{48\pi} \text{ cm/s}$.
$\frac{dh}{dt} = \frac{1}{48\pi} \text{ cm/s}$.
Question 15 — 18
Marginal Costs, Revenue and MCQs.
Q15: $C'(17) = 20.967 \approx 20.97$.
Q16: $R'(7) = 26(7)+26 = 208$.
Q17: $2\pi(6) = 12\pi$. Option B
Q18: $6(15)+36 = 126$. Option D