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The RD Sharma Class 12 Solution of Derivatives As A Rate Measure Exercise 12.2 is the most recommended book by the CBSE board institutions to their students. Not only do the teachers suggest this book, but many previous batch students also recommend it to their juniors.

Derivative As a Rate Measure exercise 12.2 question 1

Suppose the side of given square sheet be x cm at any instant time.

Now according to the question,

The rate of side of sheet increasing is cm/min

Now the area of square sheet at any time t will be ... (i)

On differentiating (i) with respect to time both side,

[ From equation number (i)]

Therefore,

Let’s Put value of x = 8

The area is increasing at the rate of when the side is 8 cm long.

Derivative As a Rate Measure exercise 12.2 question 2

**Answer**: The volume of the cube increasing at the rate of 900cm^{3}/sec**Hint:** Here, we use the formula of cube’s volume **Given**: An edge of a variable cube is increasing at the rate of 3 cm per second, the volume of the cube increasing when the edge is 10 cm long, So Here x =10**Solution**: Suppose the edge of the given cube be x cm at any instant time.

Now according to the question,

The rate of edge of the cube increasing is ... … (i)

Now, the formula is

By applying derivative,

Lets put the value of x in above equation

Thus the volume of cube increasing at the rate of 900 cm^{3}/sec

This Story also Contains

- Derivative as a Rate Measurer Excercise: 12.2
- Derivative As a Rate Measure exercise 12.2 question 14
- Derivative As a Rate Measure exercise 12.2 question 22
- Derivative As a Rate Measure exercise 12.2 question 27
- RD Sharma Class 12 Solutions Chapter 12 Derivative as a Rate Measure - Other Exercise
- Benefits of picking RD Sharma class 12th exercise 12.2 from Career360 include:

Derivative As a Rate Measure exercise 12.2 question 3

**Answer:** The rate of increasing of the perimeter of the square will be 0.8 cm/sec**Hint:** Here, the perimeter of the square at any time t will be p = 4x cm**Given: **The side of square is increasing at the rate of 0.2 cm/sec.**Solution**: Suppose the edge of the given cube be x cm at any instant time.

Now according to the question,

The rate of side of the square is increasing is … (i)

By applying derivative in perimeter,

(from eqn 1) and p = 4x (given)

So, Thus the rate of increasing of the perimeter of the square will be 0.8 cm/sec.

Derivative As a Rate Measure exercise 12.2 question 4

**Answer:** The rate of increasing of the circle’s circumference will be cm/sec**Hint:** The circumference of circle at any time t will be cm**Given:** The radius of a circle is increasing at the rate of 0.7 cm/sec.**Solution:** Suppose the radius of the given circle be r cm at any instant time.

Now according to the question,

The rate of radius of a circle is increasing is cm/sec ...… (i)

Now the circumference of the circle at any time t will be cm

By applying derivative on circumference,

cm/sec

Thus the rate of increasing of circle circumference will be cm/sec.

Derivative As a Rate Measure exercise 12.2 question 5

**Answer**: The rate of increase of its surface area, when the radius is 7 cm is cm^{2}/sec**Hint**: The surface area of spherical soap bubble at any time t will be cm^{2}**Given**: The radius of a spherical soap bubble is increasing at the rate of 0.2 cm/sec.**Solution:** Suppose the radius of the given spherical soap bubble be cm at any instant time.

Now according to the question,

cm/sec ….... (i)

By applying derivative on surface area

lets put value of r in above formula (given)

cm/sec

Thus the rate of increase of its surface area, when the radius is 7 cm is cm^{2}/sec

Derivative As a Rate Measure exercise 12.2 question 6

**Answer**: **Hint**: We know that the volume of the spherical balloon is .**Given**: The spherical balloon inflated by pumping in 900 cubic cm of gas per second.**Solution**: Suppose the radius of the given spherical balloon be r cm and let V be the volume of the spherical balloon at any instant time.

The balloon is inflated by pumping 900 cubic cm of gas per second hence the rate of volume of spherical balloon is increased by,

lets put value of (given)

Let’s differentiate w.r.t time

So, lets put value of (given)

Derivative As a Rate Measure exercise 12.2 question 7

**Answer**: **Hint:** As we know that the volume of bubble is .**Given:** The radius of air bubble is increasing at the rate of 0.5 cm/sec**Solution**: Suppose the radius of the given air bubble be cm and let be the volume of the air bubble at any instant time.

The rate of increasing in the radius of the air bubble is

By applying derivative on volume

Let’s differentiate

Let’s put value of (given)

Let’s put value of

The rate is increasing is when the radius is 1 cm.

Derivative As a Rate Measure exercise 12.2 question 8

**Answer**: **Hint:** The rate at which the length of the man’s shadow increase will be .**Given**: A man 2 meters high walk at a uniform speed of 5 km/hr away from lamp post 6 meters high.**Solution: **

Suppose AB the lamp post and let MN be of the man of height 2 m.

Suppose AM = l meter and MS be the shadow of the man

Suppose length of the shadow MS=s

Given as man walk at the speed of 5 km/hr

So considering

................(i)

Then considering

...................(ii)

From equation (i) and (ii)

..............(iii)

By applying derivative, With respect to time on both side

Thus, the rate at which the length of his shadow increases by 2.5 km/hr

Derivative As a Rate Measure exercise 12.2 question 9

**Answer:****Hint: **As we know that the area of circle is .**Given:** As a stone is dropped into a quiet lake and waves move in circle at a speed of 4 cm/sec**Solution**: Suppose be the radius of the circle and be the area of the circle.

Whenever stone is dropped into the lake waves in circle at the speed of 4 cm/sec.

That is the radius of the circle increases at the rate of 4 cm/sec.

cm/sec ……………. (i)

Therefore,

So, when the radius of circle is 10 cm,

Thus the enclosed area is increasing at the rate of cm^{2}/sec.

Derivative As a Rate Measure exercise 12.2 question 10

Suppose the lamp post and let be the man of height 160 cm or 1.6 m.

Suppose AM = l meter and be the shadow of the man.

Suppose length of the shadow

So,

Considering

.....…(i)

So considering

....… (ii)

Therefore, from equation (i) and (ii)

.................(iii)

By applying derivative with respect to time on both side

Thus, the rate at which the length of his shadow increases by 0.4 m/hr.

Derivative As a Rate Measure exercise 12.2 question 11

Suppose AM =I meter and be the shadow of the man.

Given as man walk at the speed of 2 m/sec

So, m/sec …(i)

So considering

Then_{},

So by (i) and (ii)

.......(iii)

By applying derivative with respect to time on both side

Thus, the rate at which the length of his shadow increases by 0.57 m/sec.

Derivative As a Rate Measure exercise 12.2 question 12

at the rate of 1.5 m/sec.

Let the bottom of the ladder be at distance of x m from the wall and it’s top be at a height y m from the ground

Therefore, m/sec .......(i)

By Pythagoras theorem,

Considering,

.......(ii)

......(iii)

Put the value of y in equation (ii) , we get

......(iv)

Differentiate the above equation w.r.t ‘t’

........(v)

Now the foot of the ladder is pulled along the ground away from the wall,

So, x=12m

.......from equation (ii)

.........(iv)

Thus equation (v)

Derivative As a Rate Measure exercise 12.2 question 13

**Answer:** and **Hint:** Here we use the equation of curve are .**Given**: Given as particle moves along the curve .**Solution**: As to find the points at which the curve are the x and y coordinates of the particle changing at the same rate

........(i)

When x and y co-ordinate of the particle are changing at the same rate

Substituting value from eq (i)

So,

and

and .

**Answer:****Hint: **Here we use the equation of given curve.**Given:** Equation of the curve and increases at the rate of 4 units per second**Solution:** As the differentiate, the above equation with respect to , we get the slope of the curve

…….(i)

Suppose be the slope of the given curve then the above equation

Given increases at the rate of 4 units per second therefore,

……… eq (ii)

Differentiate the equation of the slope that is equation (ii)

Let’s put value of

Let’s put value of (given)

The slope cannot be negative,

Thus, the slope of the curve is changing at the rate of 48 units/sec when

Derivative As a Rate Measure exercise 12.2 question 15

**Answer:** and **Hint: **Here we use the equation of curve **Given:** As particle moves along the curve **Solution**: Differentiate the above equation with respect to ,

…(i)

When y-coordinate change three times more rapidly than x-coordinate that is

…(ii)

Then equating (i) and (ii)

When

When

and .

Derivative As a Rate Measure exercise 12.2 question 16(i)

**Answer:**

**Hint:** Here we use the formula of angles

**Given:**

**Solution:**

Hence,

So the value of angle which increases twice as fast as its cosine is

Derivative As a Rate Measure exercise 12.2 question 16(ii)

**Answer**: **Hint:** Here we use the given condition, **Given:** Whose rate of increase twice as twice the rate of decrease of its cosine.**Solution:** As per the condition

As the rate of increase twice as twice the rate of decrease hence the minus sign,

Derivative As a Rate Measure exercise 12.2 question 17

be the position of the ladder after being pulled at the rate of 0.5 m/sec.

So …….. (i)

, let’s apply Pythagoras theorem

……… (ii)

Here h = 6 and x=4

Differentiate equation (ii)

Let’s put value of and and ........equation (iii)

From equation (iii) we get,

as we know from question

Substituting in

We get

Derivative As a Rate Measure exercise 12.2 question 18

**Answer**: **Hint**: Use **Given:** A balloon in the form of a right circular cone surrounded by hemisphere having a diameter equal to the height of the cone is being inflated.**Solution**: Let, the total height of the balloon, then

then

...........(i)

So, volume of the cone volume of the hemisphere

Substituting

So, we get

Let’s take derivative w.r.t to time both side

Let’s put value of

Derivative As a Rate Measure exercise 12.2 question 19

Let r be the radius, h be height of the cone and v be the volume of cone at any time of

Let O be the semi vertical angle of the cone CAB whose height CO is 10m and radius is OB = 5m

........(i)

Let v be the volume of water in the cone then,

Differentiate w.r.t 't' then

Now, when the water stands 7.5 m below the base

Derivative As a Rate Measure exercise 12.2 question 20

Since triangle and are similar

Let’s take derivative with respect to time both side

Let’s put value of

Derivative As a Rate Measure exercise 12.2 question 21

Let the radius of the given spherical bubble be cm at any instant time.

It is given that the surface area of a spherical bubble is increasing at the rate of

Now, Volume of sphere

Let’s put value of and

**Answer:****Hint:** we know the volume of the cylinder is **Given:** The radius of a cylinder is increasing at the rate 2cm/sec and its altitude is increasing at rate of 3cm**Solution:** The radius of a cylinder is increasing at the rate 2 cm/sec and its altitude is decreasing at the rate of 3 cm/sec

To find the rate of change of volume when radius is 3 cm and altitude 5 cm

Let V be the volume of the cylinder, r be its radius and h be its altitude at any instant of time ‘t’.

We know volume of the cylinder is

Differentiating this with respect to time we get

Now will apply the product rule of differentiation

So above equation becomes

But given of a cylinder is increasing at the rate 2 cm/sec, i.e., and its altitude is decreasing at the rate of 3 cm/sec, i.e.,

by substituting the above values in equation we get

When radius of the cylinder, cm and its altitude, cm, the above equation becomes

Hence the rate of change of volume when radius is 3 cm and altitude 5cm is 33 cm^{3}/sec

Derivative As a Rate Measure exercise 12.2 question 23

**Answer**: **Hint:** The volume of the hallow sphere is **Given**: radius cm/sec and outer radius are 4 cm and 8 cm**Solution**: Differentiating the volume with respect to time,

This is the rate of the volume of the hollow sphere

Here radius are 4 cm and 8 cm

Derivative As a Rate Measure exercise 12.2 question 24

**Answer: ****Hint**: We know, volume of the cone is **Given:** Sand is being poured into a conical pile at the constant rate of 50 cm^{3}/min**Solution**: Let the volume be V , height be and radius be of the cone at any instant of time.

We know volume of the cone is .

So, the new volume becomes,

Differentiate the above equation with respect to time,

Here,

(height =5 ).

Derivative As a Rate Measure exercise 12.2 question 25

Then from figure by applying Pythagoras theorem,

.......(i)

Differentiating the above equation with respect to time,

So ,

......(ii)

Therefore, _{}m

So, let use equation (i) and (ii)

Derivative As a Rate Measure exercise 12.2 question 26

**Answer:** and **Hint:** Here we use the equation of curve.**Given:** As particle moves along the curve **Solution**: Equation of curve is

Differentiating the above equation,

.......…(i)

When y-coordinate is changing twice as fast as x coordinate

…......(ii)

Equating equation (i) and (ii)

When

When

and

**Answer: ****Hint: **Here we use the concept of equation curve **Given: **A point on the curve .**Solution:** Differentiating the above equation,

..........(i)

..........(ii)

Let use equation (i) and (ii)

When

Hence the point on the curve at which the abscissa and coordinate change at the same rate is .

Derivative As a Rate Measure exercise 12.2 question 28

**Answer**: **Hint:** We know that **Given: **The volume of a cube is increasing at the rate of 9 cm^{3}/sec**Solution:** Differentiating the equation,

........ …(i)

The cube is increasing at the rate of 9 cm^{3}/sec

So the above equation becomes,

........(ii)

So,

Again differentiating the above equation with respect to time is,

Substitute equation (i) in the above equation,

So, length 10 cm,

Derivative As a Rate Measure exercise 12.2 question 29

Then according to the given criteria,

The rate of volume of the spherical balloon is increasing is ....…(i)

But volume of balloon is

Applying derivative

Substituting the value from equation (i)

.......(ii)

Now the surface area of spherical balloon at any time will be

Applying derivative,

Substituting the value from equation (ii) we get,

So, when the radius is 5 cm,

Derivative As a Rate Measure exercise 12.2 question 30

(i)

And width is increasing at the rate of 4 cm/min

Differentiating both sides,

So,

(ii) Let A be the area of the rectangle,

So the above equation becomes,

Substituting the values from equation (i) and (ii)

When cm and cm the above equation becomes,

**Also Read - **RD Sharma Solution for Class 9 to 12 Maths

- Chapter 12 - Derivative as a Rate Measure -Ex 12.1
- Chapter 12 -Derivative as a rate measure -Ex-FBQ
- Chapter 12 -Derivative as a rate measure -Ex-MCQ
- Chapter 12 -Derivative as a rate measure -Ex-CSBQ
- Chapter 12 -Derivative as a rate measure- Ex-VSA

The RD Sharma class 12th exercise 12.2 includes the answers given by various experts. There is no chance for mistakes as all the solutions are rechecked many times.

The Class 12 RD Sharma Chapter 12 Exercise 12.2 solution is available for free which no one can ever imagine. Anybody who visits the Career 360 website gain the access to obtain and download the RD Sharma Class 12 Chapter 12 Exercise 12.2 to do their homework and assignment sums easily.

The RD Sharma class 12 solution chapter 12 exercise 12.2 consists follows the NCERT pattern which the CBSE schools follow. Thus, the students who study in the CBSE board institutions, need not hesitate to use the RD Sharma class 12th exercise 12.2 reference book. using this single reference material, the students can solve all the questions given in the textbook. they need not look for other options having the RD Sharma set of books with them.

There are 18 questions present in this exercise. The RD Sharma Class 12th exercise 12.2 comprises of the solved sums for those questions. Along with the solved sums, there are numerous practice questions asked in the RD Sharma Class 12th Exercise 12.2 reference material. Some of the important concepts that are present in this exercise are concepts of derivate. Rate of increase of perimeter and circumference of the given figure, and other application-based questions on the derivatives topic.

All the sums can be referred to from a single solution material.

The RD Sharma books are accessible by everyone.

Download option is available that lets the students save the reference material into their device to study later.

All the techniques that can be implied on a sum to arrive at the solution are given.

All the formulae are classified separately for ease of reference.

Many mock tests papers are available which helps the students to get exam-ready in the correct method.

The RD Sharma solution books are used by many students to clear their doubts and understand the sums in-depth. These books guide the students as a teacher would be available at all time to solve their doubts. Moreover, the RD Sharma class 12th exercise 12.2 is present for complete free of cost at the Career 360 website.

The same solution material can be used for preparing for entrance exams like JEE Mains and other competitive exams too.

**RD Sharma Chapter-wise Solutions**

- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable

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Download E-book1. What is the most recommended solution book for the class 12 students?

Most of the CBSE schools as well as the previous batch students recommend the RD Sharma solution books to the class 12 students.

2. What is the assurance that all the answers provided in the RD Sharma Class12 Chapter 12 Exercise 12.2 material is correct?

The answers provided in the Class 12 RD Sharma Chapter 12 Exercise 12.2 are given by numerous experts in the exam point of view. All the answers are rechecked to confirm the accuracy.

3. How many sums are solved in exercise 12.2 in the Class 12 RD Sharma Chapter 12, Exercise 12.2 solution book?

In total, there are 18 sums in the second exercise of the 12th chapter. The Class 12 RD Sharma Chapter 12 Exercise 12.2 provides the solved answers for all these questions.

4. Does the Class 12 RD Sharma Chapter 12 Exercise 12.2 consists of solved sums for part – 2 questions too?

The Class 12 RD Sharma Chapter 12, Exercise 12.2 reference material contains answers for all the part 1 and 2 questions. Additionally, it also includes sample question papers and additional practice questions.

5. How much would it cost to own a copy of the RD Sharma Class 12 Chapter 12 Exercise 12.2 maths solutions material?

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