Here we provide RD Sharma Class 12 Ex 14.1 Solutions Chapter 14 Differentials, Errors, and Approximations for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 14.1 Solutions Chapter 14 Differentials, Errors, and Approximations book pdf download. Now you will get step-by-step solutions to each question.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 14 |

Exercise | 14.1 |

Category | RD Sharma Solutions |

Table of Contents

**RD Sharma Class 12 Ex 14.1 Solutions Chapter 14 Differentials, Errors, and Approximations**

### Question 1: If y=sin x and x changes from π/2 to 22/14, what is the approximate change in y?

**Solution:**

According to the given condition,

x = π/2, and

x+△x = 22/14

△x = 22/14-x = 22/14 – π/2

As, y = sin x

= cos x

= cos (π/2) = 0

△y = △x

△y = 0 △x

△y = 0 (22/14 – π/2)

△y = 0

Hence, there will be no change in y.

### Question 2: The radius of a sphere shrinks from 10 to 9.8 cm. Find approximately the decrease in its volume?

**Solution:**

According to the given condition,

Let’s take radius as x

x = 10, and

Let △x be the error in the radius and △y be the error in the volume

x+△x = 9.8

△x = 9.8-x = 9.8-10 = -0.2

As, Volume of sphere =

= 4πx

^{2}= 4π(10)

^{2}= 400 π△y = △x

△y = (400 π) (-0.2)

△y = -80 π

Hence, approximate decrease in its volume will be -80 π cm^{3}

### Question 3: The circular metal plate expands under heating so that its radius increases by k%. Find the approximate increase in the area of the plate, if the radius of the plate before heating is 10 cm.

**Solution:**

According to the given condition,

Let’s take radius as x

x = 10, and

Let △x be the error in the radius and △y be the error in the surface area

△x/x × 100 = k

△x = (k × 10)/100 = k/10

As, Area of circular metal = πx

^{2}= π(2x) = 2πx

= 2π(10) = 20 π

△y = △x

△y = (20 π) (k/10)

△y = 2kπ

Hence, approximate increase in the area of the plate is 2kπ cm^{2}

### Question 4: Find the percentage error in calculating the surface area of a cubical box if an error of 1% is made in measuring the lengths of edges of the cube.

**Solution:**

According to the given condition,

Let △x be the error in the length and △y be the error in the surface area

Let’s take length as x

△x/x × 100 = 1

△x = x/100

x+△x = x+(x/100)

As, surface area of the cube = 6x

^{2}= 6(2x) = 12x

△y = △x

△y = (12x) (x/100)

△y = 0.12 x

^{2}So, △y/y = 0.12 x

^{2}/6 x^{2 }= 0.02Percentage change in y = △y/y × 100 = 0.02 × 100 = 2

Hence, the percentage error in calculating the surface area of a cubical box is 2%

### Question 5: If there is an error of 0.1% in the measurement of the radius of a sphere, find approximately the percentage error in the calculation of the volume of the sphere.

**Solution:**

According to the given condition,

As, Volume of sphere =

Let △x be the error in the radius and △y be the error in the volume

△x/x × 100 = 0.1

△x/x = 1/1000

As, y =

= 4πx

^{2}dy = 4πx

^{2 }dx△y = (4πx

^{2}) △xChange in volume,

△y/y =

△y/y =

△y/y = = 3(0.001) = 0.003

Percentage change in y = △y/y × 100 = 0.003 × 100 = 0.3

Hence, approximately the percentage error in the calculation of the volume of the sphere is 0.3%

### Question 6: The pressure p and the volume v of a gas are connected by the relation pv^{1.4} = constant. Find the percentage error in p corresponding to a decrease of 1/2% in v.

**Solution:**

According to the given condition,

= – 1/2%

pv^{1.4} = constant = k(say)

Taking log on both sides, we get

log(pv^{1.4}) = log (k)

log(p)+log(v^{1.4}) = log k

log(p) + 1.4 log(v) = log k

Differentiating wrt v, we get

Percentage change in p = △p/p × 100 = × 100 = -1.4

= -1.4

= 0.7 %

**Hence, percentage error in p is 0.7%.**

### Question 7: The height of a cone increases by k%, its semi-vertical angle remaining the same. What is the approximate percentage increase?

**Solution:**

According to the given condition,

Let h be the height, y be the surface area. V be the volume, l be the slant height and r be the radius of the cone.

Let △h be the change in the height. △r be the change in the radius of base and △l be the change in slant height.

Semi-vertical angle remaining the same.

△h/h = △r/r = △l/l

and,

△h/h × 100 = k

△h/h × 100 = △r/r × 100 = △l/l × 100 = k

### (i) in total surface area, and

**Solution:**

Total surface area of the cone

y = πrl + πr

^{2}Differentiating both the sides wrt r, we get

= πl + πr + 2πr

= πl + πr + 2πr

= πl + πl + 2πr

= 2πl + 2πr = 2π(r+l)

△y = △r

△y = (2π(r+l))

Percentage change in y = △y/y × 100 = × 100

= 2k %

Hence, percentage increase in total surface area of cone 2k%.

### (ii) in the volume assuming that k is small?

**Solution:**

Volume of cone (y) =

Differentiating both the sides wrt h, we get

(r

^{2}+ h(2r )(r

^{2}+ h(2r )(r

^{2}+ 2r^{2})= πr

^{2}△y = △h

△y = (πr

^{2})Percentage change in y = △y/y × 100 = × 100

= 3k %

Hence, percentage increase in the volume of cone 3k%.

### Question 8: Show that the relative error in computing the volume of a sphere, due to an error in measuring the radius, is approximately equal to the three times the relative error in the radius.

**Solution:**

According to the given condition,

Let △x be the error in the radius and △y be the error in volume.

Volume of cone (y) =

Differentiating both the sides wrt x, we get

(3x

^{2})= 4πx

^{2}△y = △x

△y = (4πx

^{2}) (△x)△y/y =

△y/y =

Hence proved!!

### Question 9: Using differentials, find the approximate values of the following:

### (i)

**Solution:**

Considering the function as

y = f(x) =

Taking x = 25, and

x+△x = 25.02

△x = 25.02-25 = 0.2

△y = dy = dx

△y = △x

△y = (0.02) = 0.002

Hence, = y+△y = 5 + 0.002 = 5.002

### (ii)

**Solution:**

Considering the function as

y = f(x) =

Taking x = 0.008, and

x+△x = 0.009

△x = 0.009-0.008 = 0.001

△y = dy = dx

△y = △x

△y = (0.001) = = 0.008333

Hence, = y+△y = 0.2 + 0.008333 = 0.208333

### (iii)

**Solution:**

Considering the function as

y = f(x) =

Taking x = 0.008, and

x+△x = 0.007

△x = 0.007-0.008 = -0.001

= 0.2

△y = dy = dx

△y = △x

△y = (-0.001) = = -0.008333

Hence, = y+△y = 0.2 + (-0.008333) = 0.191667

### (iv)

**Solution:**

Considering the function as

y = f(x) =

Taking x = 400, and

x+△x = 401

△x = 401-400 = 1

△y = dy = dx

△y = △x

△y = (1) = 0.025

Hence, = y+△y = 20 + 0.025 = 20.025

### (v)

**Solution:**

Considering the function as

y = f(x) =

Taking x = 16, and

x+△x = 15

△x = 15-16 = -1

△y = dy = dx

△y = △x

△y = (-1) = = -0.03125

Hence, = y+△y = 0.2 + (-0.03125) = 1.96875

### (vi)

**Solution:**

Considering the function as

y = f(x) =

Taking x = 256, and

x+△x = 255

△x = 255-256 = -1

= 4

△y = dy = dx

△y = △x

△y = (-1) = = -0.003906

Hence, = y+△y = 0.2 + (-0.003906) = 3.9961

### (vii)

**Solution:**

Considering the function as

y = f(x) =

Taking x = 2, and

x+△x = 2.002

△x = 2.002-2 = 0.002

△y = dy = dx

△y = △x

△y = (0.002) = -0.0005

Hence, = y+△y = + (-0.005) = 0.2495

### (viii) log_{e} 4.04, it being given that log_{10} 4=0.6021 and log_{10} e=0.4343

**Solution:**

Considering the function as

y = f(x) = log_{e} x

Taking x = 4, and

x+△x = 4.04

△x = 4-4.04 = 0.04

y = log_{e} x

= log_{e} 4 = = 1.386368

△y = dy = dx

△y = △x

△y = (0.04) = 0.01

Hence, log_{e} 4.04 = y+△y = 1.386368 + 0.01 = 1.396368

### (ix) log_{e} 10.02, it being given that log_{e} 10=2.3026

**Solution:**

Considering the function as

y = f(x) = log_{e} x

Taking x = 10, and

x+△x = 10.02

△x = 10.02-10 = 0.02

y = log_{e} x

= log_{e} 10 = 2.3026

△y = dy = dx

△y = △x

△y = (0.02) = 0.002

Hence, log_{e} 10.02 = y+△y = 2.3026 + 0.002 = 2.3046

### (x) log_{10} 10.1, it being given that log_{10} e=0.4343

**Solution:**

Considering the function as

y = f(x) = log_{10} x

Taking x = 10, and

x+△x = 10.1

△x = 10.1-10 = 0.1

y = log_{10} x =

= log_{10} 10 = 1

△y = dy = dx

△y = △x

△y = (0.1) = 0.004343

Hence, log_{e} 10.1 = y+△y = 1 + 0.004343 = 1.004343

### (xi) cos 61°, it being given that sin 60°=0.86603 and 1°=0.01745 radian.

**Solution:**

Considering the function as

y = f(x) = cos x

Taking x = 60°, and

x+△x = 61°

△x = 61°-60° = 1° = 0.01745 radian

y = cos x

= cos 60° = 0.5

= – sin x

= – sin 60° = -0.86603

△y = dy = dx

△y = (-0.86603) △x

△y = (-0.86603) (0.01745) = -0.01511

Hence, cos 61° = y+△y = 0.5 + (-0.01511) = 0.48489

### (xii)

**Solution:**

Considering the function as

y = f(x) =

Taking x = 25, and

x+△x = 25.1

△x = 25.1-25 = 0.1

△y = dy = dx

△y = △x

△y = (0.1) = = -0.0004

Hence, = y+△y = + (-0.0004) = 0.1996

### (xiii)

**Solution:**

Considering the function as

y = f(x) = sin x

Taking x = 22/7, and

x+△x = 22/14

△x = 22/14-22/7 = -22/14

sin (-22/14) = -1

y = sin x

= sin (22/7) = 0

= cos x

= cos (22/7)= -1

△y = dy = dx

△y = (-1) △x

△y = (-1) (-1) = 1

Question 9: Using differentials, find the approximate values of the following:

Hence, sin(22/14) = 0+1 = 1

(xiv)Solution:

Considering the function as

y = f(x) = cos x

Taking x = π/3, and

x+△x = 11π/36

△x = 11π/36-π/3 = -π/36

y = cos x

= cos (π/3) = 0.5

= – sin x

= – sin (π/3) = -0.86603

△y = dy = dx

△y = (-0.86603) (-π/36)

△y = 0.0756

Hence, = 0.5+0.0756 = 0.5755

(xv)Solution:

Considering the function as

y = f(x) =

Taking x = 81, and

x+△x = 80

△x = 80-81 = -1

△y = dy = dx

△y = △x

△y = (-1) = = -0.009259

Hence,

= y+△y = 3 + (-0.009259) = 2.99074

(xvi)Solution:

Considering the function as

y = f(x) =

Taking x = 27, and

x+△x = 29

△x = 29-27 = 2

△y = dy = dx

△y = △x

△y = (2) = 0.074

Hence,

= y+△y = 3+0.074 = 3.074

(xvii)Solution:

Considering the function as

y = f(x) =

Taking x = 64, and

x+△x = 66

△x = 66-64 = 2

△y = dy = dx

△y = △x

△y = (2) = 0.042

Hence,

= y+△y = 4+0.042 = 4.042

(xviii)Solution:

Considering the function as

y = f(x) =

Taking x = 25, and

x+△x = 26

△x = 26-25 = 1

△y = dy = dx

△y = △x

△y = (1) = 0.1

Hence,

= y+△y = 5 + 0.1 = 5.1

(xix)Solution:

Considering the function as

y = f(x) =

Taking x = 36, and

x+△x = 37

△x = 37-36 = 1

△y = dy = dx

△y = △x

△y = (1) = 0.0833

Hence,

= y+△y = 6 + 0.0833 = 6.0833

(xx)Solution:

Considering the function as

y = f(x) =

Taking x = 0.49, and

x+△x = 0.48

△x = 0.48-0.49 = -0.01

△y = dy = dx

△y = △x

△y = (-0.01) = -0.007143

Hence,

= y+△y = 0.7 + (-0.007143) = 0.693

(xxi)Solution:

Considering the function as

y = f(x) =

Taking x = 81, and

x+△x = 82

△x = 82-81 = 1

△y = dy = dx

△y = △x

△y = = 0.009259

Hence,

= y+△y = 3 + 0.009259 = 3.009259

(xxii)Solution:

Considering the function as

y = f(x) =

Taking x = 16/81, and

x+△x = 17/81

△x = 17/81-16/81 = 1/81

△y = dy = dx

△y = △x

△y = = 0.01042

Hence,

= y+△y = 2/3 + 0.01042 = 0.6771

(xxiii)Solution:

Considering the function as

y = f(x) =

Taking x = 32, and

x+△x = 33

△x = 33-32 = 1

△y = dy = dx

△y = △x

△y = = 0.0125

Hence,

= y+△y = 2 + 0.0125 = 2.0125

(xxiv)Solution:

Considering the function as

y = f(x) =

Taking x = 36, and

x+△x = 36.6

△x = 36.6-36 = 0.6

△y = dy = dx

△y = △x

△y = (0.6) = 0.05

Hence,

= y+△y = 6 + 0.05 = 6.05

(xxv)Solution:

Considering the function as

y = f(x) =

Taking x = 27, and

x+△x = 25

△x = 25-27 = -2

△y = dy = dx

△y = △x

△y = (-2) = -0.07407

Hence,

= y+△y = 3+(-0.07407) = 2.9259

(xxvi)Solution:

Considering the function as

y = f(x) =

Taking x = 49, and

x+△x = 49.5

△x = 49.5-49 = 0.5

△y = dy = dx

△y = △x

△y = (0.5) = 0.0357

Hence,

= y+△y = 7 + 0.0357 = 7.0357

Question 10: Find the appropriate value of f(2.01), where f(x) = 4x^{2}+5x+2Solution:

Considering the function as

y = f(x) = 4x^{2}+5x+2

Taking x = 2, and

x+△x = 2.01

△x = 2.01-2 = 0.01

y = 4x^{2}+5x+2

= 4(2)^{2}+5(2)+2 = 28

= 8x+5

= 8(2)+5 = 21

△y = dy = dx

△y = (21) △x

△y = (21) (0.01) = 0.21

Hence,f(2.01) = y+△y = 28 + 0.21 = 28.21

Question 11: Find the appropriate value of f(5.001), where f(x) = x^{3}-7x^{2}+15Solution:

Considering the function as

y = f(x) = x^{3}-7x^{2}+15

Taking x = 5, and

x+△x = 5.001

△x =5.001-5 = 0.001

y = x^{3}-7x^{2}+15

= (5)^{3}-7(5)^{2}+15 = -35

= 3x^{2}-14x

= 3(5)^{2}-14(5) = 5

△y = dy = dx

△y = (5) △x

△y = (5) (0.001) = 0.005

Hence,f(5.001) = y+△y = -35 + 0.005 = -34.995

Question 12: Find the appropriate value of log_{10}1005, given that log_{10}e=0.4343Solution:

Considering the function as

y = f(x) = log_{10}x

Taking x = 1000, and

x+△x = 1005

△x =1005-1000 = 5

y = log_{10}x =

= log_{10}1000 = 3

= 0.0004343

△y = dy = dx

△y = (0.0004343) △x

△y = (0.0004343) (5) = 0.0021715

Hence, log_{10}1005 = y+△y = 3 + 0.0021715 = 3.0021715

Question 13: If the radius of a sphere is measured as 9cm with an error of 0.03m, find the approximate error in calculating its surface area.Solution:

According to the given condition,

As, Surface area = 4πx^{2}

Let △x be the change in the radius and △y be the change in the surface area

x = 9

△x = 0.03m = 3cm

x+△x = 9+3 = 12cm

= 4πx^{2}= 4π(9)^{2}= 324 π

= 8πx

= 8π(9) = 72π

△y = dy = dx

△y = (72π) △x

△y = (72π) (3) = 216 πHence, approximate error in surface area of the sphere is 216 π cm^{2}

Question 14: Find the approximate change in the surface area of a cube as side x meters caused by decreasing the side by 1%.Solution:

According to the given condition,

As, Surface area = 6x^{2}

Let △x be the change in the length and △y be the change in the surface area

△x/x × 100 = 1

= 6(2x) = 12x

△y = △x

△y = (12x) (x/100)

△y = 0.12 x^{2}

Hence, the approximate change in the surface area of a cubical box is 0.12 x^{2}m^{2}

Question 15: If the radius of a sphere is measured as 7m with an error of 0.02m, find the approximate error in calculating its volume.Solution:

According to the given condition,

As, Volume of sphere = πx^{3}

Let △x be the error in the radius and △y be the error in the volume

x = 7

△x = 0.02 cm

π(3x^{2}) = 4πx^{2}

= 4π(7)^{2}= 196π

△y = dy = dx

△y = (196π) △x

△y = (196π) (0.02) = 3.92 π

Hence, approximate error in volume of the sphere is 3.92 π cm^{2}

Question 16: Find the approximate change in the volume of a cube as side x meters caused by increasing the side by 1%.Solution:

According to the given condition,

As, Volume of cube = x^{3}

Let △x be the change in the length and △y be the change in the volume

△x/x × 100 = 1

= 3x^{2}

△y = △x

△y = (3x^{2}) (x/100)

△y = 0.03 x^{3}Hence, the approximate change in the volume of a cubical box is 0.03 x^{3}m^{3}

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.