NCERT Solutions for Class 12 Maths Chapter 6 - Exercise

Exercise 6.1

1. Find the rate of change to the area of a circle with respect to its radius when 

a) $r=3cm$ 

b) $r=4cm$ 

Ans: The area of a circle (A) with radius $(r)$ is $A=\pi {{r}^{2}}$.

Now, for a given radius, the change in the area of the circle is given by, 

$\frac{dA}{dr}=\frac{d}{dr}\left( \pi {{r}^{2}} \right)=2\pi r$ 

1. When $r=3cm$

$  \frac{dA}{dr}=2\pi (3) $  

$  =6\pi  $  

As a result, when the radius of the circle is 3cm, the area of the circle changes at a rate of $6\pi $ cm.

2. When $r=4cm$

$  \frac{dA}{dr}=2\pi (4) $  

$  =8\pi  $  

As a result, when the radius of the circle is 4cm, the area of the circle changes at a rate of $8\pi  $cm.

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

Ans:

Let $x$ be the length of a side, $v$ be the volume, and $s$ be the surface area of the cube. 

When $x$ is a function of time $t$, we have

$V={{x}^{3}}$ and $S=6{{x}^{2}}$

It is given that $\frac{dV}{dt}=8c{{m}^{3}}/s$ 

Then, by chain rule we have

$  \therefore 8=\frac{dv}{dt}=\left( {{x}^{3}} \right) $  

$  =\frac{d}{x}\left( x^3 \right).\frac{dx}{dt}=3{{x}^{2}}.\frac{dx}{dt} $  

$\Rightarrow \frac{dx}{dt}=\frac{8}{3{{x}^{2}}}$                                                                                    …… (1)

Now, $\Rightarrow \frac{ds}{dt}=\frac{d}{dt}\left( 6{{x}^{2}} \right)=\frac{d}{dx}\left( 6{{x}^{2}} \right).\frac{dx}{dt}$                                 (By chain rule)

$=12x.\frac{dx}{dt}=\frac{32}{x}$ 

Thus, when $x=12cm$,

$  \frac{dS}{dt}=\frac{32}{12}c{{m}^{2}}/s $

$  =\frac{8}{3}c{{m}^{2}}/s. $  

As a result, if the cube's edge length is $12cm$, the surface area is rising at a rate of $\frac{8}{3}c{{m}^{2}}/s.$

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.

Ans: For the radius $(r)$, the area of a circle (A) is $A=\pi {{r}^{2}}$.

Now, for time ($t$), the rate of change of area (A) is given by,

$  \frac{dA}{dt}=\frac{d}{d}\left( \pi {{r}^{2}} \right) $  

$  =2\pi r\frac{dr}{dt} $                                                          (By chain rule)

It is given that the increase in radius of circle is,

$\frac{dr}{dt}=3cm/s$

$\frac{dA}{dt}=2\pi r\left( 3 \right)=6\pi r$ 

Thus, when $r=10cm$ 

$\frac{dA}{dt}=6\pi \left( 10 \right)=60\pi c{{m}^{2}}/s$ 

As a result, when the radius of the circle is $10cm$, the rate at which the area of the circle increases is $60\pi c{{m}^{2}}/s$.

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?

Ans: Let $x$ be the length of a side and $v$ be the volume of the cube. Then,        $V={{x}^{3}}$ 

$\therefore \frac{dV}{dt}=3{{x}^{2}}.\frac{dx}{dt}$  (By chain rule)

It is given that,

$\frac{dx}{dt}=3cm/s$ 

$\therefore \frac{dV}{dt}=3{{x}^{2}}(3)=9{{x}^{2}}$ 

Then, when $x=10cm$,

$\frac{dV}{dt}=9{{\left( 10 \right)}^{2}}=900c{{m}^{3}}/s$  

As a result, when the edge is $10cm$ long, the volume of the cube increases at the rate of $900c{{m}^{3}}/s$.

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?

Ans: The area of a circle (A) with radius $(r)$ is given by,

$A=\pi {{r}^{2}}$

Now, for time ($t$), the rate of change of area (A) is given by,

$  \therefore \frac{dA}{dt}=\frac{d}{dt}\left( \pi {{r}^{2}} \right) $

$  =\frac{d}{dr}\left( \pi {{r}^{2}} \right)\frac{dr}{dt}=2\pi r\frac{dr}{dt} $ (By chain rule)

It is given that $\frac{dr}{dt}=5cm/s$.

Thus, when $r=8cm$

$\frac{dA}{dt}=80\pi $

As a result, when the circular wave's radius is $8$cm, the enclosed area grows at a pace of $80\pi c{{m}^{2}}/s$.

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?

Ans. The circumference of a circle (C) with the radius $\left( r \right)$ is given by

$C=2\pi r$.

Now, the rate of change of circumference (C) for time ($t$) is given by,

$\frac{dC}{dt}=\frac{dC}{dr}.\frac{dr}{dt}$                                                            (By chain rule)

$  =\frac{d}{dr}\left( 2\pi r \right)\frac{dr}{dt} $  

$  =2\pi .\frac{dr}{dt} $

It is given that $\frac{dr}{dt}=0.7cm/s$.

Hence, the rate of increase of the circumference is $1.4\pi cm/s$.

7. 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 $\left( a \right)$ the perimeter, and $\left( b \right)$ the area of the rectangle.

Ans: It is given that the length $x$ is decreasing at the rate of $5cm/\min $ and the width $y$ is increasing at the rate of $4cm/\min $i.e,

$  \frac{dx}{dt}=-5cm/\min , $  

$  \frac{dy}{dt}=4cm/\min  $  

(a) The perimeter (P) of a rectangle is given by

$  P=2\left( x+y \right) $  

$  \therefore \frac{dP}{dt}=2\left( \frac{dx}{dt}+\frac{dy}{dt} \right) $  

$  =-2cm/\min  $  

Hence, the rate of decrease in the perimeter of rectangle is $2cm/\min $.

(b) The area (A) of a rectangle is given by,

$A=x\times y$

$  \frac{dA}{dt}=\frac{dx}{dt}.y+x.\frac{dy}{dt} $

$  =-5y+4x $  

When $x=8cm$ and $y=6cm$,

$  \frac{dA}{dt}=\left( -5\times 6+4\times 8 \right)c{{m}^{2}}/\min  $  

$  =2c{{m}^{2}}/\min  $  

Hence, the rate of increase in the area of the rectangle is $2c{{m}^{2}}/\min $.

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.

Ans: The volume of a sphere (V) with radius ($r$) is given by,

$V=\frac{4}{3}\pi {{r}^{3}}$ 

For time ($t$), the rate of change of volume (V) is given by,

$\frac{dV}{dt}=\frac{dV}{dr}.\frac{dr}{dt}$                                                   (By chain rule)

$  =\frac{d}{dr}\left( \frac{4}{3}\pi {{r}^{3}} \right).\frac{dr}{dt} $  

$  =4\pi {{r}^{2}}.\frac{dr}{dt} $  

It is given that 

$  \frac{dv}{dt}=900c{{m}^{3}}/s $  

$  \therefore 900=4\pi {{r}^{2}}.\frac{dr}{dt} $  

$  \Rightarrow \frac{dr}{dt}=\frac{900}{4\pi {{r}^{2}}} $  

$  =\frac{225}{\pi {{r}^{2}}} $  

Therefore, when the radius is $15$cm,

$\frac{dr}{dt}=\frac{225}{\pi {{\left( 15 \right)}^{2}}}=\frac{1}{\pi }$ 

As a result, the rate at which the balloon's radius increases when the radius is $15cm$ is $\frac{1}{\pi }cm/\sec $.

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 latter is $10$ cm.

Ans: The volume of a sphere (V) with radius ($r$) is given by,

$V=\frac{4}{3}\pi {{r}^{3}}$ 

Rate of change of volume (V) for time ($t$) is given by,

$\frac{dV}{dr}=\frac{d}{dr}\left( \frac{4}{3}\pi {{r}^{3}} \right) $  

$=4\pi {{r}^{2}} $  

Therefore, when the radius is $10cm$,

$  \frac{dV}{dr}=4\pi {{\left( 10 \right)}^{2}} $  

$  =400\pi  $  

Hence, the rate of increase in the volume of the balloon is $400\pi c{{m}^{3}}/\sec $.

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

Ans: Let the height of the wall be $y$m and the foot of the ladder be $x$m away from the wall.

Then, by Pythagoras Theorem, we have:

${{x}^{2}}+{{y}^{2}}=25$  (Length of the ladder is $5m$)

$\Rightarrow y=\sqrt{25-{{x}^{2}}}$ 

Then, the rate of change of height $y$for time $t$ is given by,

$\frac{dy}{dt}=\frac{-x}{\sqrt{25-{{x}^{2}}}}.\frac{dx}{dt}$ 

It is given that $\frac{dx}{dt}=2cm/s$ 

$\therefore \frac{dy}{dt}=\frac{-2x}{\sqrt{25-{{x}^{2}}}}$ 

Now, when $x=4m$ we have:

$\frac{dy}{dt}=\frac{8}{3}$ 

Hence, the rate of decrease in the height of the ladder on the wall is    $\frac{8}{2}cm/\sec $.

11. A particle moving 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.

Ans: Given the equation of the curve as $6y={{x}^{3}}+2$.

The rate of change of the position of the particle for time

(t)  is given by,

$  6\frac{dy}{dt}=3{{x}^{2}}\frac{dx}{dt}+0 $

$  \Rightarrow 2\frac{dy}{dt}={{x}^{2}}\frac{dx}{dt} $

When the particle's y-coordinate changes $8$ times as fast as its x-coordinate i.e., $\left( \frac{dy}{dt}=8\frac{dx}{dt} \right)$ , we have:

$  2\left( 8\frac{dx}{dt} \right)={{x}^{2}}\frac{dx}{dt} $  

$  \Rightarrow 16\frac{dx}{dt}={{x}^{2}}\frac{dx}{dt} $

$  \Rightarrow \left( {{x}^{2}}-16 \right)\frac{dx}{dt}=0 $  

$  \Rightarrow {{x}^{2}}=16 $  

$  \Rightarrow x=\pm 4 $  

When $x=-4,$ 

$y=\frac{-31}{3}$

and when $x=4,$ 

$y=11$ 

Hence, the points required on the curve are $\left( 4,11 \right)$ and $\left( -4,\frac{-31}{3} \right)$.

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 bubbles of the bubble increasing when the radius is $1$cm?

Ans: The air bubble is in the shape of a sphere.

The volume of an air bubble (V) with radius (r) is given by,

$V=\frac{4}{3}\pi {{r}^{3}}$

Rate of change of volume (V) for time ($t$) is given by,

$\frac{dV}{dt}=\frac{dV}{dr}.\frac{dr}{dt}$                                                   (By chain rule)

$  =\frac{d}{dr}\left( \frac{4}{3}\pi {{r}^{3}} \right).\frac{dr}{dt} $  

$  =4\pi {{r}^{2}}.\frac{dr}{dt} $  

It is given that $\frac{dr}{dt}=\frac{1}{2}cm/s$ 

Therefore, when r is $1$cm,

$  \frac{dV}{dt}=4\pi {{\left( 1 \right)}^{2}}\left( \frac{1}{2} \right) $  

$  =2\pi c{{m}^{3}}/s $  

Hence, the volume of the bubble increases at the rate of $2\pi $cm3/s.

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

Ans: The volume of a sphere $\left( V \right)$ with radius \[\left( r \right)\] is given by,

$V=\frac{4}{3}\pi {{r}^{2}}$.

Diameter is $\frac{3}{2}\left( 2x+1 \right)$

$  r=\frac{3}{4}\left( 2x+1 \right) $  

$  V=\frac{4}{3}\pi {{r}^{3}} $  

$  =\frac{9}{16}\pi {{\left( 2x+1 \right)}^{3}} $  

Hence, the rate of change of volume with respect to $x$ is as

$\frac{dV}{dt}=\frac{9}{16}\pi \frac{d}{dt}{{\left( 2x+1 \right)}^{3}}$  

$=\frac{27}{8}\pi {{\left( 2x+1 \right)}^{3}}$.

14. Sand is pouring from a pipe at the rate of $12c{{m}^{3}}/\sec $. 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?

Ans: The volume of a cone ($V$) with radius and height ($h$) is given by.

$V=\frac{1}{3}\pi {{r}^{2}}h$ 

It is provided that,

$  h=\frac{1}{6}r $  

$  \Rightarrow r=6h $  

$  \therefore V=\frac{1}{3}\pi {{\left( 6h \right)}^{2}}h $  

$  =12\pi {{h}^{3}} $  

The rate of change of volume for time ($t$) is given by

$\frac{dV}{dt}=12\pi \frac{d}{dh}\left( {{h}^{3}} \right).\frac{dh}{dt}$      (by chain rule)

$  =12\pi \left( 3{{h}^{2}} \right)\frac{dh}{dt} $

$  =36\pi {{h}^{2}}\frac{dh}{dt} $  

It is provided that $\frac{dV}{dt}=12c{{m}^{2}}/s$.

Therefore, when $h=4cm$ we have:

$  12=36\pi {{\left( 4 \right)}^{2}}\frac{dh}{dt} $  

$  \Rightarrow \frac{dh}{dt}=\frac{12}{36\pi \left( 16 \right)} $  

$  \frac{dh}{dt}=\frac{1}{48\pi } $  

Hence, when the height of the sand cone is $4$ cm, its height is increasing at the rate of $\frac{1}{48}cm/\sec $.

15. The total cost $C\left( x \right)$ in Rupees associated with the production of $x$ units of an item is given by \[C\left( x \right)=0.007{{x}^{3}}-0.003{{x}^{2}}+15x+4000\].

Find the marginal cost when $17$ units are produced.

Ans: The rate of change in total cost for output is known as marginal cost.

$M C=\frac{d C}{d x}=0.007\left(3 x^{2}\right)-0.003(2 x)+15$

$=0.021 x^{2}-0.006 x+15$

When 

$  x=17,MC=0.021{{\left( 17 \right)}^{2}}-0.006\left( 17 \right)+15 $  

$  =0.21\left( 289 \right)-0.006\left( 17 \right)+15 $  

$  =6.069-0.102+15 $  

$  =20.967 $  

Hence, the marginal cost when $17$ units are produced is $Rs.20.967$.

15. The total revenue in Rupees received from the sale of $x$ units of a product is given by $R\left( x \right)=13{{x}^{2}}+26x+15$. Find the marginal revenue when $x=7$.

Ans: The rate of change in total cost for the number of units sold is known as marginal cost. Let the number of units sold be x.

$  MR=\frac{dR}{dx}=13\left( 2x \right)+26 $  

$  =26x+26 $

When $x=7$,

$MR=208$ 

Hence, the required marginal revenue is $Rs.208$.

16. The rate of change of the area of a circle with respect to its radius $r$ at $r=6cm$ is $\left( A \right)10\pi $ $\left( B \right)12\pi $ $\left( C \right)=8\pi $ $\left( D \right)11\pi $.

Ans: The area of a circle $\left( A \right)$ with radius $\left( r \right)$ is given by, $A=\pi {{r}^{2}}$

As a result, the area's rate of change in relation to its radius r is  $\frac{dA}{dt}=\frac{d}{dr}\left( \pi {{r}^{2}} \right)=2\pi r$ 

Therefore, when $r=6cm$,

$  \frac{dA}{dr}=2\pi \times 6 $  

$  =12\pi c{{m}^{2}}/s $  

Hence, the required rate of change of the area of a circle is $12\pi c{{m}^{2}}/s$.

Therefore, option B is correct.

17. The total revenue in Rupees received from the sale of $x$units of a product is given by $R\left( x \right)=3{{x}^{2}}+36x+5$. The marginal revenue, when $x=15$ is $\left( A \right)116$ $\left( B \right)96$ $\left( C \right)90$ $\left( D \right)126$.

Ans: The rate of change in total cost for the number of units sold is known as marginal cost.

$\therefore MR=\frac{dR}{dx}=6x+36$ 

Therefore, when $x=15$,

$MR=90+36=126$ 

Hence, the required marginal revenue is $Rs.126$.

The correct option is D.

NCERT Solutions for Class 12 Maths PDF Download

• Chapter 1 - Relations and Functions

• Chapter 2 - Inverse Trigonometric Functions

• Chapter 3 - Matrices

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 9 - Differential Equations

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability

NCERT Solution Class 12 Maths of Chapter 6 All Exercises

An Overview of Application of Derivatives Exercise 6.1

Class 12 Maths NCERT Solutions Chapter 6 Exercise 6.1 - Free PDF Download

Chapter 6 in class 12 maths is the last chapter of the book but is also an incredibly important chapter in terms of the number of marks that it comes for as a whole in the final examination for maths. Calculus is the most important part of class 12 maths as it holds the highest number of weightage marks out of any other topic in the two books for maths. In the two books, chapters 5, 6, 7, 8 and 9 are dedicated to calculus. These are 5 chapters out of a total of 12 chapters which are solely dedicated to calculus, and this hopefully helps you understand the importance of the topic.

Because of this, understanding the first part of the first chapter of calculus, which is chapter 6 i.e. Application of Derivatives, holds immense importance. The NCERT solutions for class 12 chapter 6 exercise 6.1 have been prepared for students for this very reason. Please download these free class 12 maths ex 6.1 solutions so that you can get a hang of the first exercise in calculus and then move forward smoothly.

Class 12 Maths Chapter 6 Exercise 6.1

Chapter 6 of class 12 maths is “Application of Derivatives” and is an important chapter in terms of calculus. This chapter consists of the following:

• A brief introduction to derivatives including what was taught in the previous chapter.

• Rate of change of quantities, which relates to applying what we’ve learnt about derivatives to practical examples.

• Increasing and decreasing functions using graphical methods.

• Tangents and normals.

• Approximations.

• Maxima and minima concepts.

• Several miscellaneous examples of the above topics for a clearer understanding.

The Vedantu class 12 maths exercise 6.1 solutions are helpful for the topic ‘rate of change of quantities’ which has been mentioned above. Other solutions for class 12 maths chapter 6 are also available separately.

Important Concepts Covered in Exercise 6.1 of Class 12 Maths NCERT Solutions

Exercise 6.1 of Class 12 Maths NCERT Solutions is mainly based on the below concept.

• Rate of change of quantities

• This also includes applying the chain rule to find out the changes in two variables depending on a third variable

Exercise 6.1 of Class 12 Maths NCERT Solutions Chapter 6 consists of questions related to determining the rate of change of a quantity with respect to the other and most of these questions are in the word problem format. For solving the problems present in this exercise, we have to differentiate the dependent variable with respect to the variable asked in the questions. While solving the questions, students need to keep in mind that the derivative notation dy/dx depicts that y changes when x is changed. 

The weightage marks of class 12 maths are as given below:

Weightage Marks of Class 12 Maths

Chapter 6 of class 12 maths is Applications of Derivatives and it comes under the unit of Calculus for a part of 35 marks.

Benefits of Ex 6.1 Class 12 Maths NCERT Solutions

There are several benefits of exercise 6.1 class 12 maths NCERT solutions, and some of them are listed as follows:

• Vedantu NCERT maths solutions are provided for all the exercises in the book and are good for unloading pressure off students.

• These solutions are downloadable and printable by all students to make them handier.

• The solutions have been prepared by experienced teachers of maths, ensuring accuracy.

• The solutions are entirely free of cost as provided by Vedantu.