Exercise 6.1 Practice
Rate of Change of Quantities – NCERT Solutions
Similar to Q1
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Find the rate of change of the area of a circle with respect to its radius $r$ when
(a) $r = 5 \text{ cm}$
(b) $r = 6 \text{ cm}$
(a) $r = 5 \text{ cm}$
(b) $r = 6 \text{ cm}$
Given: Area of a circle $A = \pi r^2$.
Differentiation: Differentiating w.r.t radius $r$:
$\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r$.
$\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r$.
(a) At $r = 5 \text{ cm}$:
$\frac{dA}{dr} = 2\pi(5) = 10\pi \text{ cm}^2/\text{cm}$.
$\frac{dA}{dr} = 2\pi(5) = 10\pi \text{ cm}^2/\text{cm}$.
(b) At $r = 6 \text{ cm}$:
$\frac{dA}{dr} = 2\pi(6) = 12\pi \text{ cm}^2/\text{cm}$.
$\frac{dA}{dr} = 2\pi(6) = 12\pi \text{ cm}^2/\text{cm}$.
$\boxed{\text{(a) } 10\pi, \text{ (b) } 12\pi}$
Similar to Q2
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The volume of a cube is increasing at the rate of $9 \text{ cm}^3/\text{s}$. How fast is the surface area increasing when the length of an edge is $10 \text{ cm}$?
Given: Rate of change of volume $\frac{dV}{dt} = 9 \text{ cm}^3/\text{s}$. Edge length $x = 10 \text{ cm}$.
Formulae: Volume $V = x^3$, Surface Area $S = 6x^2$.
Step 1: Find $\frac{dx}{dt}$
$\frac{dV}{dt} = \frac{d}{dt}(x^3) = 3x^2 \frac{dx}{dt}$
$9 = 3x^2 \frac{dx}{dt} \Rightarrow \frac{dx}{dt} = \frac{3}{x^2}$.
$\frac{dV}{dt} = \frac{d}{dt}(x^3) = 3x^2 \frac{dx}{dt}$
$9 = 3x^2 \frac{dx}{dt} \Rightarrow \frac{dx}{dt} = \frac{3}{x^2}$.
Step 2: Find $\frac{dS}{dt}$
$\frac{dS}{dt} = \frac{d}{dt}(6x^2) = 12x \frac{dx}{dt}$
Substitute $\frac{dx}{dt}$: $\frac{dS}{dt} = 12x \left(\frac{3}{x^2}\right) = \frac{36}{x}$.
$\frac{dS}{dt} = \frac{d}{dt}(6x^2) = 12x \frac{dx}{dt}$
Substitute $\frac{dx}{dt}$: $\frac{dS}{dt} = 12x \left(\frac{3}{x^2}\right) = \frac{36}{x}$.
Step 3: At $x = 10$
$\frac{dS}{dt} = \frac{36}{10} = 3.6 \text{ cm}^2/\text{s}$.
$\frac{dS}{dt} = \frac{36}{10} = 3.6 \text{ cm}^2/\text{s}$.
$\boxed{3.6 \text{ cm}^2/\text{s}}$
Similar to Q3
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The radius of a circle is increasing uniformly at the rate of $4 \text{ cm/s}$. Find the rate at which the area of the circle is increasing when the radius is $8 \text{ cm}$.
Given: $\frac{dr}{dt} = 4 \text{ cm/s}$, $r = 8 \text{ cm}$.
Formula: Area $A = \pi r^2$.
Differentiation:
$\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt}$.
$\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt}$.
Substitution:
$\frac{dA}{dt} = 2\pi(8)(4) = 64\pi \text{ cm}^2/\text{s}$.
$\frac{dA}{dt} = 2\pi(8)(4) = 64\pi \text{ cm}^2/\text{s}$.
$\boxed{64\pi \text{ cm}^2/\text{s}}$
Similar to Q4
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An edge of a variable cube is increasing at the rate of $5 \text{ cm/s}$. How fast is the volume of the cube increasing when the edge is $12 \text{ cm}$ long?
Given: Edge $x$, $\frac{dx}{dt} = 5 \text{ cm/s}$. Find $\frac{dV}{dt}$ at $x = 12 \text{ cm}$.
Formula: Volume $V = x^3$.
Differentiation:
$\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$.
$\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$.
Calculation:
$\frac{dV}{dt} = 3(12)^2(5) = 15(144) = 2160 \text{ cm}^3/\text{s}$.
$\frac{dV}{dt} = 3(12)^2(5) = 15(144) = 2160 \text{ cm}^3/\text{s}$.
$\boxed{2160 \text{ cm}^3/\text{s}}$
Similar to Q5
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A stone is dropped into a quiet lake and waves move in circles at the speed of $6 \text{ cm/s}$. At the instant when the radius of the circular wave is $12 \text{ cm}$, how fast is the enclosed area increasing?
Given: $\frac{dr}{dt} = 6 \text{ cm/s}$. Find $\frac{dA}{dt}$ at $r = 12 \text{ cm}$.
Step 1: Area $A = \pi r^2$.
Step 2: Differentiate w.r.t time $t$:
$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
Step 3: Substitute values:
$\frac{dA}{dt} = 2\pi(12)(6) = 144\pi \text{ cm}^2/\text{s}$.
$\frac{dA}{dt} = 2\pi(12)(6) = 144\pi \text{ cm}^2/\text{s}$.
$\boxed{144\pi \text{ cm}^2/\text{s}}$
Similar to Q6
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The radius of a circle is increasing at the rate of $0.5 \text{ cm/s}$. What is the rate of increase of its circumference?
Given: $\frac{dr}{dt} = 0.5 \text{ cm/s}$.
Formula: Circumference $C = 2\pi r$.
Differentiation:
$\frac{dC}{dt} = 2\pi \frac{dr}{dt}$.
$\frac{dC}{dt} = 2\pi \frac{dr}{dt}$.
Calculation:
$\frac{dC}{dt} = 2\pi(0.5) = \pi \text{ cm/s}$.
$\frac{dC}{dt} = 2\pi(0.5) = \pi \text{ cm/s}$.
$\boxed{\pi \text{ cm/s}}$
Similar to Q7
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The length $x$ of a rectangle is decreasing at the rate of $3 \text{ cm/min}$ and the width $y$ is increasing at the rate of $2 \text{ cm/min}$. When $x = 10 \text{ cm}$ and $y = 5 \text{ cm}$, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.
Given: $\frac{dx}{dt} = -3 \text{ cm/min}$, $\frac{dy}{dt} = 2 \text{ cm/min}$. Values: $x=10, y=5$.
(a) Perimeter $P = 2(x+y)$:
$\frac{dP}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right) = 2(-3 + 2) = 2(-1) = -2 \text{ cm/min}$.
$\frac{dP}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right) = 2(-3 + 2) = 2(-1) = -2 \text{ cm/min}$.
(b) Area $A = xy$:
$\frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt}$.
$\frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt}$.
Substitute values:
$\frac{dA}{dt} = 10(2) + 5(-3) = 20 - 15 = 5 \text{ cm}^2/\text{min}$.
$\frac{dA}{dt} = 10(2) + 5(-3) = 20 - 15 = 5 \text{ cm}^2/\text{min}$.
$\boxed{\text{Perimeter: } -2 \text{ cm/min}, \text{ Area: } 5 \text{ cm}^2/\text{min}}$
Similar to Q8
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A balloon, which always remains spherical on inflation, is being inflated by pumping in $800$ cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is $10 \text{ cm}$.
Given: $\frac{dV}{dt} = 800 \text{ cm}^3/\text{s}$. Find $\frac{dr}{dt}$ at $r=10$.
Formula: Volume $V = \frac{4}{3}\pi r^3$.
Differentiation:
$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
Substitution:
$800 = 4\pi(10)^2 \frac{dr}{dt} = 400\pi \frac{dr}{dt}$.
$800 = 4\pi(10)^2 \frac{dr}{dt} = 400\pi \frac{dr}{dt}$.
$\frac{dr}{dt} = \frac{800}{400\pi} = \frac{2}{\pi}$.
$\boxed{\frac{2}{\pi} \text{ cm/s}}$
Similar to Q9
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A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the radius is $20 \text{ cm}$.
Objective: Find $\frac{dV}{dr}$ at $r=20$.
Formula: $V = \frac{4}{3}\pi r^3$.
Differentiation:
$\frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) = 4\pi r^2$.
$\frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) = 4\pi r^2$.
Substitution:
At $r=20$: $\frac{dV}{dr} = 4\pi(20)^2 = 4\pi(400) = 1600\pi$.
At $r=20$: $\frac{dV}{dr} = 4\pi(20)^2 = 4\pi(400) = 1600\pi$.
$\boxed{1600\pi \text{ cm}^3/\text{cm}}$
Similar to Q10
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A ladder $10 \text{ m}$ long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of $3 \text{ cm/s}$. How fast is its height on the wall decreasing when the foot of the ladder is $6 \text{ m}$ away from the wall?
Given: Ladder length $L=10\text{m}$. $\frac{dx}{dt} = 3 \text{ cm/s}$. Find $\frac{dy}{dt}$ when $x=6\text{m}$.
Relation: $x^2 + y^2 = 10^2 = 100$.
Find y: When $x=6$, $36 + y^2 = 100 \Rightarrow y^2 = 64 \Rightarrow y = 8\text{m}$.
Differentiation:
$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \Rightarrow x\frac{dx}{dt} + y\frac{dy}{dt} = 0$.
$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \Rightarrow x\frac{dx}{dt} + y\frac{dy}{dt} = 0$.
Substitution:
$6(3) + 8\frac{dy}{dt} = 0 \Rightarrow 18 + 8\frac{dy}{dt} = 0$.
$\frac{dy}{dt} = -\frac{18}{8} = -2.25 \text{ cm/s}$.
$6(3) + 8\frac{dy}{dt} = 0 \Rightarrow 18 + 8\frac{dy}{dt} = 0$.
$\frac{dy}{dt} = -\frac{18}{8} = -2.25 \text{ cm/s}$.
$\boxed{\text{Decreasing at } 2.25 \text{ cm/s}}$
Similar to Q11
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A particle moves along the curve $6y = x^3 + 5$. Find the points on the curve at which the y-coordinate is changing 2 times as fast as the x-coordinate.
Given: Curve $6y = x^3 + 5$ and $\frac{dy}{dt} = 2\frac{dx}{dt}$.
Differentiation: Differentiate w.r.t $t$:
$6\frac{dy}{dt} = 3x^2\frac{dx}{dt}$.
$6\frac{dy}{dt} = 3x^2\frac{dx}{dt}$.
Substitution: Substitute $\frac{dy}{dt} = 2\frac{dx}{dt}$:
$6(2\frac{dx}{dt}) = 3x^2\frac{dx}{dt} \Rightarrow 12 = 3x^2 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$.
$6(2\frac{dx}{dt}) = 3x^2\frac{dx}{dt} \Rightarrow 12 = 3x^2 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$.
Find Points:
If $x=2$: $6y = 2^3 + 5 = 13 \Rightarrow y = 13/6$. Point $(2, 13/6)$.
If $x=-2$: $6y = (-2)^3 + 5 = -3 \Rightarrow y = -1/2$. Point $(-2, -1/2)$.
If $x=2$: $6y = 2^3 + 5 = 13 \Rightarrow y = 13/6$. Point $(2, 13/6)$.
If $x=-2$: $6y = (-2)^3 + 5 = -3 \Rightarrow y = -1/2$. Point $(-2, -1/2)$.
$\boxed{(2, 13/6) \text{ and } (-2, -1/2)}$
Similar to Q12
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The radius of an air bubble is increasing at the rate of $\frac{1}{4} \text{ cm/s}$. At what rate is the volume of the bubble increasing when the radius is $2 \text{ cm}$?
Given: $\frac{dr}{dt} = \frac{1}{4} \text{ cm/s}$. Find $\frac{dV}{dt}$ at $r=2$.
Formula: $V = \frac{4}{3}\pi r^3$.
Differentiation:
$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
Substitution:
$\frac{dV}{dt} = 4\pi(2)^2 \left(\frac{1}{4}\right) = 4\pi(4)\left(\frac{1}{4}\right) = 4\pi$.
$\frac{dV}{dt} = 4\pi(2)^2 \left(\frac{1}{4}\right) = 4\pi(4)\left(\frac{1}{4}\right) = 4\pi$.
$\boxed{4\pi \text{ cm}^3/\text{s}}$
Similar to Q13
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A balloon, which always remains spherical, has a variable diameter $\frac{3}{4}(2x + 3)$. Find the rate of change of its volume with respect to $x$.
Given: Diameter $D = \frac{3}{4}(2x+3)$. Radius $r = \frac{D}{2} = \frac{3}{8}(2x+3)$.
Volume: $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \left[\frac{3}{8}(2x+3)\right]^3$.
Simplify V: $V = \frac{4}{3}\pi \cdot \frac{27}{512}(2x+3)^3 = \frac{9\pi}{128}(2x+3)^3$.
Differentiation:
$\frac{dV}{dx} = \frac{9\pi}{128} \cdot 3(2x+3)^2 \cdot \frac{d}{dx}(2x+3)$
$= \frac{27\pi}{128}(2x+3)^2 \cdot 2 = \frac{27\pi}{64}(2x+3)^2$.
$\frac{dV}{dx} = \frac{9\pi}{128} \cdot 3(2x+3)^2 \cdot \frac{d}{dx}(2x+3)$
$= \frac{27\pi}{128}(2x+3)^2 \cdot 2 = \frac{27\pi}{64}(2x+3)^2$.
$\boxed{\frac{27\pi}{64}(2x+3)^2}$
Similar to Q14
00:00
Sand is pouring from a pipe at the rate of $10 \text{ cm}^3/\text{s}$. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-fourth of the radius of the base. How fast is the height of the sand cone increasing when the height is $5 \text{ cm}$?
Given: $\frac{dV}{dt} = 10 \text{ cm}^3/\text{s}$. Condition $h = \frac{r}{4} \Rightarrow r = 4h$.
Volume Formula: $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (4h)^2 h = \frac{16\pi}{3}h^3$.
Differentiation:
$\frac{dV}{dt} = \frac{16\pi}{3} \cdot 3h^2 \frac{dh}{dt} = 16\pi h^2 \frac{dh}{dt}$.
$\frac{dV}{dt} = \frac{16\pi}{3} \cdot 3h^2 \frac{dh}{dt} = 16\pi h^2 \frac{dh}{dt}$.
Substitution: At $h=5$:
$10 = 16\pi(5)^2 \frac{dh}{dt} = 16\pi(25) \frac{dh}{dt} = 400\pi \frac{dh}{dt}$.
$10 = 16\pi(5)^2 \frac{dh}{dt} = 16\pi(25) \frac{dh}{dt} = 400\pi \frac{dh}{dt}$.
$\frac{dh}{dt} = \frac{10}{400\pi} = \frac{1}{40\pi}$.
$\boxed{\frac{1}{40\pi} \text{ cm/s}}$
Similar to Q15
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The total cost $C(x)$ in Rupees associated with the production of $x$ units of an item is given by $C(x) = 0.005x^3 - 0.02x^2 + 30x + 6000$. Find the marginal cost when $5$ units are produced.
Marginal Cost (MC): $MC = \frac{dC}{dx}$.
Differentiation:
$MC = \frac{d}{dx}(0.005x^3 - 0.02x^2 + 30x + 6000)$
$MC = 0.015x^2 - 0.04x + 30$.
$MC = \frac{d}{dx}(0.005x^3 - 0.02x^2 + 30x + 6000)$
$MC = 0.015x^2 - 0.04x + 30$.
Substitution: At $x=5$:
$MC = 0.015(5)^2 - 0.04(5) + 30$
$= 0.015(25) - 0.2 + 30 = 0.375 - 0.2 + 30 = 30.175$.
$MC = 0.015(5)^2 - 0.04(5) + 30$
$= 0.015(25) - 0.2 + 30 = 0.375 - 0.2 + 30 = 30.175$.
$\boxed{₹ 30.175}$
Similar to Q16
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The total revenue in Rupees received from the sale of $x$ units of a product is given by $R(x) = 20x^2 + 40x + 10$. Find the marginal revenue when $x = 8$.
Marginal Revenue (MR): $MR = \frac{dR}{dx}$.
Differentiation:
$MR = \frac{d}{dx}(20x^2 + 40x + 10) = 40x + 40$.
$MR = \frac{d}{dx}(20x^2 + 40x + 10) = 40x + 40$.
Substitution: At $x=8$:
$MR = 40(8) + 40 = 320 + 40 = 360$.
$MR = 40(8) + 40 = 320 + 40 = 360$.
$\boxed{₹ 360}$
Similar to Q17 (MCQ)
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The rate of change of the area of a circle with respect to its radius $r$ at $r = 10 \text{ cm}$ is:
(A) $10\pi$
(B) $20\pi$
(C) $30\pi$
(D) $40\pi$
(A) $10\pi$
(B) $20\pi$
(C) $30\pi$
(D) $40\pi$
Formula: $\frac{dA}{dr} = 2\pi r$.
Calculation: At $r=10$, $\frac{dA}{dr} = 2\pi(10) = 20\pi$.
$\boxed{\text{(B) } 20\pi}$
Similar to Q18 (MCQ)
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The total revenue in Rupees received from the sale of $x$ units of a product is given by $R(x) = 5x^2 + 20x + 10$. The marginal revenue, when $x = 10$ is:
(A) 100
(B) 110
(C) 120
(D) 130
(A) 100
(B) 110
(C) 120
(D) 130
Formula: $MR = \frac{dR}{dx} = 10x + 20$.
Calculation: At $x=10$, $MR = 10(10) + 20 = 100 + 20 = 120$.
$\boxed{\text{(C) } 120}$