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

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## 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,

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 =

= 4πx2

= 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 cm3

### 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,

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 = πx2

= 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 cm2

### 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 = 6x2

= 6(2x) = 12x

△y =  △x

△y = (12x) (x/100)

△y = 0.12 x2

So, △y/y = 0.12 x2/6 x= 0.02

Percentage 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 =

= 4πx2

dy = 4πxdx

△y = (4πx2) △x

Change 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 pv1.4 = constant. Find the percentage error in p corresponding to a decrease of 1/2% in v.

Solution:

According to the given condition,

= – 1/2%

pv1.4 = constant = k(say)

Taking log on both sides, we get

log(pv1.4) = log (k)

log(p)+log(v1.4) = 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 + πr2

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

(r2 + h(2r )

(r2 + h(2r )

(r2 + 2r2)

= πr2

△y =  △h

△y = (πr2

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

(3x2)

= 4πx2

△y =  △x

△y = (4πx2) (△x)

△y/y =

△y/y =

Hence proved!!

### (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) loge 4.04, it being given that log10 4=0.6021 and log10 e=0.4343

Solution:

Considering the function as

y = f(x) = loge x

Taking x = 4, and

x+△x = 4.04

△x = 4-4.04 = 0.04

y = loge x

= loge 4 =  = 1.386368

△y = dy =  dx

△y =  △x

△y =  (0.04) = 0.01

Hence, loge 4.04 = y+△y = 1.386368 + 0.01 = 1.396368

### (ix) loge 10.02, it being given that loge 10=2.3026

Solution:

Considering the function as

y = f(x) = loge x

Taking x = 10, and

x+△x = 10.02

△x = 10.02-10 = 0.02

y = loge x

= loge 10 = 2.3026

△y = dy =  dx

△y =  △x

△y =  (0.02) = 0.002

Hence, loge 10.02 = y+△y = 2.3026 + 0.002 = 2.3046

### (x) log10 10.1, it being given that log10 e=0.4343

Solution:

Considering the function as

y = f(x) = log10 x

Taking x = 10, and

x+△x = 10.1

△x = 10.1-10 = 0.1

y = log10 x =

= log10 10 = 1

△y = dy =  dx

△y =  △x

△y =  (0.1) = 0.004343

Hence, loge 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

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

Question 9: Using differentials, find the approximate values of the following:
(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) = 4x2+5x+2
Solution:
Considering the function as
y = f(x) = 4x2+5x+2
Taking x = 2, and
x+△x = 2.01
△x = 2.01-2 = 0.01
y = 4x2+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) = x3-7x2+15
Solution:
Considering the function as
y = f(x) = x3-7x2+15
Taking x = 5, and
x+△x = 5.001
△x =5.001-5 = 0.001
y = x3-7x2+15
= (5)3-7(5)2+15 = -35
= 3x2-14x
= 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 log10 1005, given that log10 e=0.4343
Solution:
Considering the function as
y = f(x) = log10 x
Taking x = 1000, and
x+△x = 1005
△x =1005-1000 = 5
y = log10 x =
= log10 1000 = 3

= 0.0004343
△y = dy =  dx
△y = (0.0004343) △x
△y = (0.0004343) (5) = 0.0021715
Hence, log10 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 = 4πx2
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
= 4πx2 = 4π(9)2 = 324 π
= 8πx
= 8π(9) = 72π
△y = dy =  dx
△y = (72π) △x
△y = (72π) (3) = 216 π
Hence, approximate error in surface area of the sphere is 216 π cm2
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 = 6x2
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 x2
Hence, the approximate change in the surface area of a cubical box is  0.12 x2 m2
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 = πx3
Let △x be the error in the radius and △y be the error in the volume
x = 7
△x =  0.02 cm
π(3x2) = 4πx2
= 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 cm2
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 = x3
Let △x be the change in the length and △y be the change in the volume
△x/x × 100 =  1
= 3x2
△y =  △x
△y = (3x2) (x/100)
△y = 0.03 x3
Hence, the approximate change in the volume of a cubical box is 0.03 x3 m3

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