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Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 14 |

Exercise | 14.1 |

Category | RD Sharma Solutions |

**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}

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