Class 12 RD Sharma Solutions – Chapter 14 Differentials, Errors and Approximations – Exercise 14.1 | Set 2

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