# Class 12 RD Sharma Solutions – Chapter 14 Differentials, Errors and Approximations – Exercise 14.1 | Set 1

### 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 klog(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}xTaking 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}xTaking 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}xTaking 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

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