Skip to content
Related Articles

Related Articles

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

Improve Article
Save Article
Like Article
  • Last Updated : 20 May, 2021

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

(xiv) cos (\frac{11\pi}{36})

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

y_{(x=\pi/3)}  = cos (π/3) = 0.5

\frac{dy}{dx}  = – sin x

(\frac{dy}{dx})_{x=\pi/3}  = – sin (π/3) = -0.86603

△y = dy = (\frac{dy}{dx})_{x=\pi/3}  dx

△y = (-0.86603) (-π/36)

△y = 0.0756

Hence, cos (\frac{11\pi}{36})  = 0.5+0.0756 = 0.5755

(xv) (80)^{\frac{1}{4}}

Solution:

Considering the function as

y = f(x) = (x)^{\frac{1}{4}}

Taking x = 81, and

x+△x = 80

△x = 80-81 = -1

y = (x)^{\frac{1}{4}}

y_{(x=81)} = (81)^{\frac{1}{4}}= 3

\frac{dy}{dx} = \frac{1}{4(x)^{\frac{3}{4}}}

(\frac{dy}{dx})_{x=81} = \frac{1}{4(81)^{\frac{3}{4}}} = \frac{1}{108}

△y = dy = (\frac{dy}{dx})_{x=81}  dx

△y = (\frac{1}{108})  △x

△y = (\frac{1}{108}) (-1) = \frac{-1}{108}  = -0.009259

Hence, 

(80)^{\frac{1}{4}}  = y+△y = 3 + (-0.009259) = 2.99074

(xvi) (29)^{\frac{1}{3}}

Solution:

Considering the function as

y = f(x) = (x)^{\frac{1}{3}}

Taking x = 27, and

x+△x = 29

△x = 29-27 = 2

y = (x)^{\frac{1}{3}}

y_{(x=27)} = (27)^{\frac{1}{3}}= 3

\frac{dy}{dx} = \frac{1}{3(x)^{\frac{2}{3}}}

(\frac{dy}{dx})_{x=27} = \frac{1}{3(27)^{\frac{2}{3}}} = \frac{1}{27}

△y = dy = (\frac{dy}{dx})_{x=27}  dx

△y = (\frac{1}{27})  △x

△y = (\frac{1}{27})  (2) = 0.074

Hence, 

(29)^{\frac{1}{3}}  = y+△y = 3+0.074 = 3.074

(xvii) (66)^{\frac{1}{3}}

Solution:

Considering the function as

y = f(x) = (x)^{\frac{1}{3}}

Taking x = 64, and

x+△x = 66

△x = 66-64 = 2

y = (x)^{\frac{1}{3}}

y_{(x=64)} = (64)^{\frac{1}{3}}= 4

\frac{dy}{dx} = \frac{1}{3(x)^{\frac{2}{3}}}

(\frac{dy}{dx})_{x=64} = \frac{1}{3(64)^{\frac{2}{3}}} = \frac{1}{48}

△y = dy = (\frac{dy}{dx})_{x=64}  dx

△y = (\frac{1}{48})  △x

△y = (\frac{1}{48})  (2) = 0.042

Hence,

(66)^{\frac{1}{3}}  = y+△y = 4+0.042 = 4.042

(xviii) \sqrt{26}

Solution:

Considering the function as

y = f(x) = \sqrt{x}

Taking x = 25, and

x+△x = 26

△x = 26-25 = 1

y = \sqrt{x}

y_{(x=25)} = \sqrt{25} = 5

\frac{dy}{dx} = \frac{1}{2\sqrt{x}}

(\frac{dy}{dx})_{x=25} = \frac{1}{2\sqrt{25}} = \frac{1}{10}

△y = dy = (\frac{dy}{dx})_{x=25}  dx

△y = (\frac{1}{10})  △x

△y = (\frac{1}{10})  (1) = 0.1

Hence, 

\sqrt{26}  = y+△y = 5 + 0.1 = 5.1

(xix) \sqrt{37}

Solution:

Considering the function as

y = f(x) = \sqrt{x}

Taking x = 36, and

x+△x = 37

△x = 37-36 = 1

y = \sqrt{x}

y_{(x=36)} = \sqrt{36} = 6

\frac{dy}{dx} = \frac{1}{2\sqrt{x}}

(\frac{dy}{dx})_{x=36} = \frac{1}{2\sqrt{36}} = \frac{1}{12}

△y = dy = (\frac{dy}{dx})_{x=36}  dx

△y = (\frac{1}{12})  △x

△y = (\frac{1}{12})  (1) = 0.0833

Hence,

\sqrt{26}  = y+△y = 6 + 0.0833 = 6.0833

(xx) \sqrt{0.48}

Solution:

Considering the function as

y = f(x) = \sqrt{x}

Taking x = 0.49, and

x+△x = 0.48

△x = 0.48-0.49 = -0.01

y = \sqrt{x}

y_{(x=0.49)} = \sqrt{0.49} = 0.7

\frac{dy}{dx} = \frac{1}{2\sqrt{x}}

(\frac{dy}{dx})_{x=0.49} = \frac{1}{2\sqrt{0.49}} = \frac{1}{1.4}

△y = dy = (\frac{dy}{dx})_{x=0.49}  dx

△y = (\frac{1}{1.4})  △x

△y = (\frac{1}{1.4})  (-0.01) = -0.007143

Hence,

\sqrt{0.48}  = y+△y = 0.7 + (-0.007143) = 0.693

(xxi) (82)^{\frac{1}{4}}

Solution:

Considering the function as

y = f(x) = (x)^{\frac{1}{4}}

Taking x = 81, and

x+△x = 82

△x = 82-81 = 1

y = (x)^{\frac{1}{4}}

y_{(x=81)} = (81)^{\frac{1}{4}}= 3

\frac{dy}{dx} = \frac{1}{4(x)^{\frac{3}{4}}}

(\frac{dy}{dx})_{x=81} = \frac{1}{4(81)^{\frac{3}{4}}} = \frac{1}{108}

△y = dy = (\frac{dy}{dx})_{x=81}  dx

△y = (\frac{1}{108})  △x

△y = (\frac{1}{108})(1) = \frac{1}{108}  = 0.009259

Hence,

(82)^{\frac{1}{4}}  = y+△y = 3 + 0.009259 = 3.009259

(xxii) (\frac{17}{81})^{\frac{1}{4}}

Solution:

Considering the function as

y = f(x) = (x)^{\frac{1}{4}}

Taking x = 16/81, and

x+△x = 17/81

△x = 17/81-16/81 = 1/81

y = (x)^{\frac{1}{4}}

y_{(x=16/81)} = (16/81)^{\frac{1}{4}}= 2/3

\frac{dy}{dx} = \frac{1}{4(x)^{\frac{3}{4}}}

(\frac{dy}{dx})_{x=16/81} = \frac{1}{4(16/81)^{\frac{3}{4}}} = \frac{27}{32}

△y = dy = (\frac{dy}{dx})_{x=16/81}  dx

△y = (\frac{27}{32})  △x

△y = (\frac{27}{32})(\frac{2}{3}) = \frac{1}{96}  = 0.01042

Hence,

(\frac{17}{81})^{\frac{1}{4}}  = y+△y = 2/3 + 0.01042 = 0.6771

(xxiii) (33)^{\frac{1}{5}}

Solution:

Considering the function as

y = f(x) = (x)^{\frac{1}{5}}

Taking x = 32, and

x+△x = 33

△x = 33-32 = 1

y = (x)^{\frac{1}{5}}

y_{(x=32)} = (32)^{\frac{1}{5}}= 2

\frac{dy}{dx} = \frac{1}{5(x)^{\frac{4}{5}}}

(\frac{dy}{dx})_{x=32} = \frac{1}{5(32)^{\frac{4}{5}}} = \frac{1}{80}

△y = dy = (\frac{dy}{dx})_{x=32}  dx

△y = (\frac{1}{80})  △x

△y = (\frac{1}{80})(1) = \frac{1}{80}  = 0.0125

Hence,

(32)^{\frac{1}{5}}  = y+△y = 2 + 0.0125 = 2.0125

(xxiv) \sqrt{36.6}

Solution:

Considering the function as

y = f(x) = \sqrt{x}

Taking x = 36, and

x+△x = 36.6

△x = 36.6-36 = 0.6

y = \sqrt{x}

y_{(x=36)} = \sqrt{36} = 6

\frac{dy}{dx} = \frac{1}{2\sqrt{x}}

(\frac{dy}{dx})_{x=36} = \frac{1}{2\sqrt{36}} = \frac{1}{12}

△y = dy = (\frac{dy}{dx})_{x=36}  dx

△y = (\frac{1}{12})  △x

△y = (\frac{1}{12})  (0.6) = 0.05

Hence,

\sqrt{26}  = y+△y = 6 + 0.05 = 6.05

(xxv) (25)^{\frac{1}{3}}

Solution:

Considering the function as

y = f(x) = (x)^{\frac{1}{3}}

Taking x = 27, and

x+△x = 25

△x = 25-27 = -2

y = (x)^{\frac{1}{3}}

y_{(x=27)} = (27)^{\frac{1}{3}}= 3

\frac{dy}{dx} = \frac{1}{3(x)^{\frac{2}{3}}}

(\frac{dy}{dx})_{x=27} = \frac{1}{3(27)^{\frac{2}{3}}} = \frac{1}{27}

△y = dy = (\frac{dy}{dx})_{x=27}  dx

△y = (\frac{1}{27})  △x

△y = (\frac{1}{27})  (-2) = -0.07407

Hence,

(25)^{\frac{1}{3}}  = y+△y = 3+(-0.07407) = 2.9259

(xxvi) \sqrt{49.5}

Solution:

Considering the function as

y = f(x) = \sqrt{x}

Taking x = 49, and

x+△x = 49.5

△x = 49.5-49 = 0.5

y = \sqrt{x}

y_{(x=49)} = \sqrt{49} = 7

\frac{dy}{dx} = \frac{1}{2\sqrt{x}}

(\frac{dy}{dx})_{x=49} = \frac{1}{2\sqrt{49}} = \frac{1}{14}

△y = dy = (\frac{dy}{dx})_{x=49}  dx

△y = (\frac{1}{14})  △x

△y = (\frac{1}{14})  (0.5) = 0.0357

Hence,

\sqrt{49.5}  = 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

y_{(x=2)}  = 4(2)2+5(2)+2 = 28

\frac{dy}{dx}  = 8x+5

(\frac{dy}{dx})_{x=2}  = 8(2)+5 = 21

△y = dy = (\frac{dy}{dx})_{x=2}  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

y_{(x=5)}  = (5)3-7(5)2+15 = -35

\frac{dy}{dx}  = 3x2-14x

(\frac{dy}{dx})_{x=5}  = 3(5)2-14(5) = 5

△y = dy = (\frac{dy}{dx})_{x=5}  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 = \frac{log_ex}{log_e10}

y_{(x=1000)}  = log10 1000 = 3

\frac{dy}{dx} = \frac{0.4343}{x}

(\frac{dy}{dx})_{x=1000} = \frac{0.4343}{1000}  = 0.0004343

△y = dy = (\frac{dy}{dx})_{x=1000}  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

y_{(x=9)}  = 4πx2 = 4π(9)2 = 324 π

\frac{dy}{dx}  = 8πx

(\frac{dy}{dx})_{x=9}  = 8π(9) = 72π

△y = dy = (\frac{dy}{dx})_{x=9}  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

\frac{dy}{dx}  = 6(2x) = 12x

△y = (\frac{dy}{dx})  △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 = \frac{4}{3} πx3

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

x = 7

△x =  0.02 cm

\frac{dy}{dx} = \frac{4}{3} π(3x2) = 4πx2

(\frac{dy}{dx})_{x=7}  = 4π(7)2 = 196π

△y = dy = (\frac{dy}{dx})_{x=7}  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

\frac{dy}{dx}  = 3x2

△y = (\frac{dy}{dx})  △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


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!