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NCERT Solutions for Exercise 6.1 Class 12 Maths Chapter 6 - Application of Derivatives

NCERT Solutions for Exercise 6.1 Class 12 Maths Chapter 6 - Application of Derivatives

#CBSE Class 12th
Updated on 22 Apr 2025, 11:25 PM IST

We have learned that the change in distance per unit time is called speed. Similarly, the change in velocity per unit time is called acceleration. This change is also known as an Instantaneous change. In chapter 6, exercise 6.1, we will learn the concept of differential change. All these NCERT solutions given in Class 12 Maths chapter 6 exercise 6.1 provide a deeper knowledge of the concept of rate of change of quantities that is discussed in the NCERT book. These NCERT solutions are created by a subject matter expert at Careers360. These solutions will give you concept clarification and help you to know the proper methods to solve the questions.

This Story also Contains

  1. Access NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1
  2. Application of Derivatives Class 12 Chapter 6 Exercise 6.1
  3. Topics covered in Chapter 6 Application of Derivatives: Exercise 6.1
  4. Subject-wise NCERT Exemplar solutions

Access NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1

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Application of Derivatives Class 12 Chapter 6 Exercise 6.1

Question 1: a) Find the rate of change of the area of a circle with respect to its radius r when

r = 3 cm

Answer:

Given: Length $x$ of a rectangle is decreasing at the rate $\frac{dx}{dt} = -5\ \text{cm/min}$ (negative sign indicates decrease),
the width $y$ is increasing at the rate $\frac{dy}{dt} = 4\ \text{cm/min}$.

To find: $\frac{dP}{dt}$ at $x = 8\ \text{cm}$ and $y = 6\ \text{cm}$, where $P$ is the perimeter.

Perimeter of rectangle: $P = 2(x + y)$

$\frac{dP}{dt} = \frac{d}{dt}[2(x + y)] = 2\left( \frac{dx}{dt} + \frac{dy}{dt} \right) = 2(-5 + 4) = -2\ \text{cm/min}$

Hence, the perimeter is decreasing at the rate of $2\ \text{cm/min}$.

Question 1: b) Find the rate of change of the area of a circle with respect to its radius r when
r = 4 cm

Answer:

Area of the circle (A) = $\pi r^{2}$
Rate of change of the area of a circle with respect to its radius r = $\frac{dA}{dr}$ = $\frac{d(\pi r^{2})}{dr}$ = $2 \pi r$
So, when r = 4, Rate of change of the area of a circle = $2 \pi (4)$ = $8 \pi$
Hence, Rate of change of the area of a circle with respect to its radius r when r = 4 is $8 \pi$

Question 2: The volume of a cube is increasing at the rate of $8\ \text{cm}^3/\text{s}$ . How fast is the surface area increasing when the length of an edge is 12 cm?

Answer:

The volume of the cube(V) = $x^{3}$ where x is the edge length of the cube
It is given that the volume of a cube is increasing at the rate of $8\ \text{cm}^3/\text{s}$

we can write $\frac{dV}{dt} = \frac{dV}{dx}.\frac{dx}{dt}$ ( By chain rule)

$\frac{dV}{dt} = 8 = \frac{dV}{dx}.\frac{dx}{dt}$

$\frac{dx^{3}}{dx}.\frac{dx}{dt} = 8$ $\Rightarrow 3x^{2}.\frac{dx}{dt} = 8$

$\frac{dx}{dt} = \frac{8}{3x^{2}}$ - (i)
Now, we know that the surface area of the cube(A) is $6x^{2}$

$\frac{dA}{dt} = \frac{dA}{dx}.\frac{dx}{dt} = \frac{d6x^{2}}{dx}.\frac{dx}{dt} = 12x. \frac{dx}{dt}$ - (ii)

from equation (i) we know that $\frac{dx}{dt} = \frac{8}{3x^{2}}$

put this value in equation (i)
We get,
$\frac{dA}{dt} = 12x. \frac{8}{3x^{2}} = \frac{32}{x}$
It is given in the question that the value of edge length(x) = 12cm
So,

$\frac{dA}{dt} = \frac{32}{12} = \frac{8}{3} \ \text{cm}^2/\text{s}$

Question 3: The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.

Answer:

Radius of a circle is increasing uniformly at the rate $\left ( \frac{dr}{dt} \right )$ = 3 cm/s
Area of circle(A) = $\pi r^{2}$
$\frac{dA}{dt} =\frac{dA}{dr}.\frac{dr}{dt}$ (by chain rule)
$\frac{dA}{dt} =\frac{d \pi r^{2}}{dr}.\frac{dr}{dt} = 2\pi r \times 3 = 6\pi r$
It is given that the value of r = 10 cm
So,

$\frac{dA}{dt} = 6\pi \times 10 = 60\pi\ \text{cm}^2/\text{s}$

Hence, the rate at which the area of the circle is increasing when the radius is 10 cm is $60\pi\ \text{cm}^2/\text{s}$

Question 4: An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?

Answer:

It is given that the rate at which edge of cube increase $\left ( \frac{dx}{dt} \right )$ = 3 cm/s
The volume of cube = $x^{3}$
$\frac{dV}{dt} = \frac{dV}{dx}.\frac{dx}{dt}$ (By chain rule)

$\frac{dV}{dt} = \frac{d{x^3}}{dx} \cdot \frac{dx}{dt} = 3x^2 \cdot \frac{dx}{dt} = 3x^2 \times 3 = 9x^2\ \text{cm}^3/\text{s}$

It is given that the value of x is 10 cm
So,
$\frac{dV}{dt} = 9(10)^{2} = 9\times 100 = 900\ \text{cm}^3/\text{s}$
Hence, the rate at which the volume of the cube increasing when the edge is 10 cm long is $900\ \text{cm}^3/\text{s}$

Question 5: A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?

Answer:

Given = $\frac{dr}{dt} = 5\ \text{cm/s}$

To find = $\frac{dA}{dt}$ at r = 8 cm

Area of the circle (A) = $\pi r^{2}$
$\frac{dA}{dt} = \frac{dA}{dr}.\frac{dr}{dt}$ (by chain rule)

$\frac{dA}{dt} = \frac{d(\pi r^2)}{dr} \cdot \frac{dr}{dt} = 2\pi r \times 5 = 10\pi r = 10\pi \times 8 = 80\pi \ \text{cm}^2/\text{s}$

Hence, the rate at which the area increases when the radius of the circular wave is 8 cm is $80\pi \ \text{cm}^2/\text{s}$

Question 6: The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?

Answer:

Given = $\frac{dr}{dt} = 0.7\ \text{cm/s}$
To find = $\frac{dC}{dt}$, where C is circumference

We know that the circumference of the circle (C) = $2\pi r$
$\frac{dC}{dt} = \frac{dC}{dr} \cdot \frac{dr}{dt}$ (by chain rule)
$\frac{dC}{dt} = \frac{d(2\pi r)}{dr} \cdot \frac{dr}{dt} = 2\pi \times 0.7 = 1.4\pi\ \text{cm/s}$

Hence, the rate of increase of its circumference is $1.4\pi\ \text{cm/s}$

Question 7: (a) The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8cm and y = 6cm, find the rate of change of the perimeter of rectangle

Answer:

Given: Length $x$ of a rectangle is decreasing at the rate $\frac{dx}{dt} = -5$ cm/min (negative sign indicates decrease),
the width $y$ is increasing at the rate $\frac{dy}{dt} = 4$ cm/min.

To find: $\frac{dP}{dt}$ at $x = 8$ cm and $y = 6$ cm, where $P$ is the perimeter.

Perimeter of rectangle: $P = 2(x + y)$

$\frac{dP}{dt} = \frac{d}{dt}[2(x + y)]$
$= 2\left( \frac{dx}{dt} + \frac{dy}{dt} \right)$
$= 2(-5 + 4)$
$= -2\ \text{cm/min}$

Hence, the perimeter decreases at the rate of $2$ cm/min.

Question 7: (b) The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8cm and y = 6cm, find the rates of change of the area of the rectangle.

Answer:

Given = Length x of a rectangle is decreasing at the rate $(\frac{dx}{dt})$ = -5 cm/minute (−ve sign indicates decrease in rate)
The width y is increasing at the rate $(\frac{dy}{dt})$ = 4 cm/minute

Area of rectangle = $xy$
$\frac{dA}{dt} = \frac{d(xy)}{dt} = \left( x\frac{dy}{dt} + y\frac{dx}{dt} \right) = \left( 8 \times 4 + 6 \times (-5) \right) = (32 - 30) = 2\ \text{cm}^2/\text{minute}$
Hence, the rate of change of area is $2\ \text{cm}^2/\text{minute}$

Question 8: A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Answer:

Given = $\frac{dV}{dt} = 900\ \text{cm}^3/\text{s}$
To find = $\frac{dr}{dt}$ at $r = 15$ cm

Volume of sphere (V) = $\frac{4}{3}\pi r^{3}$
$\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} = \frac{d\left(\frac{4}{3}\pi r^{3}\right)}{dr} \cdot \frac{dr}{dt} = \frac{4}{3}\pi \times 3r^{2} \cdot \frac{dr}{dt}$

$\frac{dV}{dt} = 4\pi r^{2} \cdot \frac{dr}{dt}$
$\frac{dr}{dt} = \frac{\frac{dV}{dt}}{4\pi r^{2}} = \frac{900}{4\pi \times (15)^2} = \frac{900}{900\pi} = \frac{1}{\pi}\ \text{cm}/\text{s}$
Hence, the rate at which the radius of the balloon increases when the radius is 15 cm is $\frac{1}{\pi}\ \text{cm}/\text{s}$

Question 9: 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.

Answer:

We need to find the value of $\frac{dV}{dr}$ at $r = 10$ cm.
The volume of the sphere (V) = $\frac{4}{3}\pi r^{3}$
$\frac{dV}{dr} = \frac{d\left(\frac{4}{3}\pi r^{3}\right)}{dr} = \frac{4}{3}\pi \times 3r^{2} = 4\pi r^{2} = 4\pi (10)^{2} = 4\pi \times 100 = 400\pi\ \text{cm}^3/\text{s}$
Hence, the rate at which its volume is increasing with the radius when the latter is 10 cm is $400\pi\ \text{cm}^3/\text{s}$

Question 10: 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 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

Answer:

Let $h$ be the height of the ladder and $x$ be the distance of the foot of the ladder from the wall.
It is given that $\frac{dx}{dt} = 2\ \text{cm/s}$.
We need to find the rate at which the height of the ladder decreases $\left(\frac{dh}{dt}\right)$.
Length of ladder ($L$) = 5 m and $x = 4$ m (given).

By Pythagoras theorem, we can say that
$h^{2} + x^{2} = L^{2}$
$h^{2} = L^{2} - x^{2}$
$h = \sqrt{L^{2} - x^{2}}$

Differentiate on both sides with respect to $t$:
$\frac{dh}{dt} = \frac{d(\sqrt{L^{2} - x^{2}})}{dx} \cdot \frac{dx}{dt} = \frac{1}{2} \cdot \frac{-2x}{\sqrt{5^{2} - x^{2}}} \cdot \frac{dx}{dt} = \frac{-x}{\sqrt{25 - x^{2}}} \cdot \frac{dx}{dt}$

At $x = 4$:
$\frac{dh}{dt} = \frac{-4}{\sqrt{25 - 16}} \cdot 2 = \frac{-4}{3} \cdot 2 = \frac{-8}{3}\ \text{cm/s}$

Hence, the rate at which the height of the ladder decreases is $\frac{8}{3}\ \text{cm/s}$.

Question 11: A particle moves along the curve $6y = x^3 + 2$ Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

Answer:

We need to find the point at which $\frac{dy}{dt} = 8\frac{dx}{dt}$
Given the equation of curve = $6y = x^3 + 2$
Differentiate both sides w.r.t. t
$6\frac{dy}{dt} = \frac{d(x^{3})}{dx}.\frac{dx}{dt} +0$
$= 3x^{2}.\frac{dx}{dt}$
$\frac{dy}{dt} = 8\frac{dx}{dt}$ (required condition)
$6\times 8\frac{dx}{dt}= 3x^{2}.\frac{dx}{dt}$
$3x^{2}.\frac{dx}{dt} =48\frac{dx}{dt}$ $\Rightarrow x^{2} = \frac{48}{3} = 16$
$x = \pm 4$
when x = 4 , $y = \frac{4^{3}+2}{6} = \frac{64+2}{6} = \frac{66}{6} = 11$
and
when x = -4 , $y = \frac{(-4)^{3}+2}{6} = \frac{-64+2}{6} = \frac{-62}{6} = \frac{-31}{3}$
So , the coordinates are

$(4,11)$ and $(-4,\frac{-31}{3})$

Question 12: The radius of an air bubble is increasing at the rate of 1 /2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Answer:

It is given that $\frac{dr}{dt} = \frac{1}{2} \ \text{cm/s}$
We know that the shape of the air bubble is spherical
So, volume(V) = $\frac{4}{3}\pi r^{3}$
$\frac{dV}{dt} = \frac{dV}{dr}.\frac{dr}{dt} = \frac{d(\frac{4}{3}\pi r^{3})}{dr}.\frac{dr}{dt} = 4\pi r^{2}\times\frac{1}{2} = 2\pi r^{2} = 2\pi \times (1)^{2} = 2\pi \ \text{cm}^{3}/\text{s}$
Hence, the rate of change in volume is $2\pi \ \text{cm}^{3}/\text{s}$

Question 13: A balloon, which always remains spherical, has a variable diameter $\frac{3}{2}( 2x +1)$ Find the rate of change of its volume with respect to x.

Answer:

Volume of sphere (V) = $\frac{4}{3}\pi r^{3}$
Diameter = $\frac{3}{2}(2x+1)$
So, radius (r) = $\frac{3}{4}(2x+1)$
$\frac{dV}{dx} = \frac{d(\frac{4}{3}\pi r^{3})}{dx} = \frac{d\left(\frac{4}{3}\pi \left(\frac{3}{4}(2x+1)\right)^{3}\right)}{dx} = \frac{4}{3}\pi \times 3 \times \frac{27}{64}(2x+1)^{2} \times 2$
$= \frac{27}{8}\pi (2x+1)^{2}$

Question 14: Sand is pouring from a pipe at the rate of 12 cm 3 /s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?

Answer:

Given = $\frac{dV}{dt} = 12\ \text{cm}^{3}/\text{s}$ and $h = \frac{1}{6}r$
To find = $\frac{dh}{dt}$ at $h = 4\ \text{cm}$
Solution:-

Volume of cone (V) = $\frac{1}{3}\pi r^{2}h$
$\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt} = \frac{d\left(\frac{1}{3}\pi (6h)^{2}h\right)}{dh} \cdot \frac{dh}{dt} = \frac{1}{3}\pi \times 36 \times 3h^{2} \cdot \frac{dh}{dt}$
$= 36\pi \times (4)^{2} \cdot \frac{dh}{dt}$
$\frac{dV}{dt} = 576\pi \cdot \frac{dh}{dt}$

Question 15: The total cost C(x) in Rupees associated with the production of x units of an
item is given by $C ( x) = 0.007 x^3 - 0.003 x^2 + 15 x + 4000$

Find the marginal cost when 17 units are produced.

Answer:

Marginal cost (MC) = $\frac{dC}{dx}$
$C ( x) = 0.007 x^3 - 0.003 x^2 + 15 x + 4000$
$\frac{dC}{dx} = \frac{d(.007x^{3} - .003x^{2}+15x+400)}{dx} = 3\times .007x^{2} - 2\times.003x+15$
$= .021x^{2} - .006x + 15$
Now, at x = 17
MC $= .021(17)^{2} - .006(17) + 15$
$= 6.069 - .102 + 15$
$= 20.967$
Hence, marginal cost when 17 units are produced is 20.967

Question 16: The total revenue in Rupees received from the sale of x units of a product is
given by $R ( x) = 13 x^2 + 26 x + 15$

Find the marginal revenue when x = 7

Answer:

Marginal revenue = $\frac{dR}{dx}$
$R ( x) = 13 x^2 + 26 x + 15$
$\frac{dR}{dx} = \frac{d(13x^{2}+26x+15)}{dx} = 13\times2x+ 26 = 26(x+1)$
at x = 7
$\frac{dR}{dx} = 26(7+1) = 26\times8 = 208$
Hence, marginal revenue when x = 7 is 208

Question 17: The rate of change of the area of a circle with respect to its radius r at r = 6 cm is

(A) $10\pi$

(B) $12\pi$

(C) $8\pi$

(D) $11\pi$

Answer:

Area of circle (A) = $\pi r^{2}$
$\frac{dA}{dr} = \frac{d(\pi r^{2})}{dr} = 2\pi r$
Now, at $r = 6\ \text{cm}$
$\frac{dA}{dr} = 2\pi \times 6 = 12\pi\ \text{cm}^{2}/\text{s}$
Hence, the rate of change of the area of a circle with respect to its radius $r$ at $r = 6\ \text{cm}$ is $12\pi\ \text{cm}^{2}/\text{s}$
Hence, the correct answer is $12\pi$

Question 18: The total revenue in Rupees received from the sale of x units of a product is given by $R(x) = 3x^2 + 36x + 5$ . The marginal revenue, when x = 15 is

(A) $116$

(B) $96$

(C) $90$

(D) $126$

Answer:

Marginal revenue = $\frac{dR}{dx}$
$R ( x) = 3 x^2 + 36 x + 5$
$\frac{dR}{dx} = \frac{d(3x^{2}+36x+5)}{dx} = 3\times2x+ 36 = 6(x+6)$
at x = 15
$\frac{dR}{dx} = 6(15+6) = 6\times21 = 126$
Hence, marginal revenue when x = 15 is 126
Hence, the correct answer is 126

Also Read,

Students can also read,

Topics covered in Chapter 6 Application of Derivatives: Exercise 6.1

  • Introduction
  • Rate of change of quantities: Similarly, whenever one quantity $y$ varies with another quantity $x$, satisfying some rule $y=f(x)$, then $\frac{d y}{d x}$ (or $\left.f^{\prime}(x)\right)$ represents the rate of change of $y$ with respect to $x$ and $\left.\frac{d y}{d x}\right]_{x=x_0}$ (or $\left.f^{\prime}\left(x_0\right)\right)$ represents the rate of change of $y$ with respect to $x$ at $x=x_0$
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NCERT Solutions Subject Wise

Here are links to NCERT textbook solutions for other subjects. Students can explore and review these organised solutions to gain a better understanding.

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Subject-wise NCERT Exemplar solutions

Students may refer to these NCERT exemplar links for additional practice purposes.

CBSE Class 12th Syllabus: Subjects & Chapters
Select your preferred subject to view the chapters

Frequently Asked Questions (FAQs)

Q: How many exercise are in Class 12th chapter Application of Derivative?
A:

Including miscellaneous, there are 6 exercises in the unit Applications of Derivatives

Q: What is the number of multiple-choice questions present in the NCERT Solutions for Class 12 Maths chapter 6 exercise 6.1?
A:

Two multiple-choice questions are discussed in the NCERT exercise 6.1

Q: What number of solved examples are there in the NCERT topic rate of change of quantities?
A:

There are six solved examples

Q: What is the meaning of dy/dx
A:

dy/dx means the rate of change of y with respect to x

Q: When dy/dx is positive?
A:

dy/dx is positive if y increases as x increases

Q: What is the topic discussed in the NCERT Class 12th Maths book after exercise 6.1?
A:

The topic number 6.3 increasing and decreasing function is coming after exercise 6.1. 

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If you want to improve the Class 12 PCM results, you can appear in the improvement exam. This exam will help you to retake one or more subjects to achieve a better score. You should check the official website for details and the deadline of this exam.

I hope it will clear your query!!

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After 12th, if you are interested in computer science, the best courses are:

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I am a CBSE class 12 private candidate will my maksheet of CBSE supplement exam will be delivered on main exam form address for supplementary exam form address ,my both addresses are different please tell

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CBSE generally forwards the marksheet for the supplementary exam to the correspondence address as identified in the supplementary exam application form. It is not sent to the address indicated in the main exam form. Addresses that differ will use the supplementary exam address.

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With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
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