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

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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

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Last Updated : 20 May, 2021

Related Articles

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) $cos (\frac{11\pi}{36})$

(xv) $(80)^{\frac{1}{4}}$

(xvi) $(29)^{\frac{1}{3}}$

(xvii) $(66)^{\frac{1}{3}}$

(xviii) $\sqrt{26}$

(xix) $\sqrt{37}$

(xx) $\sqrt{0.48}$

(xxi) $(82)^{\frac{1}{4}}$

(xxii) $(\frac{17}{81})^{\frac{1}{4}}$

(xxiii) $(33)^{\frac{1}{5}}$

(xxiv) $\sqrt{36.6}$

(xxv) $(25)^{\frac{1}{3}}$

(xxvi) $\sqrt{49.5}$

Question 10: Find the appropriate value of f(2.01), where f(x) = 4x²+5x+2

Question 11: Find the appropriate value of f(5.001), where f(x) = x³-7x²+15

Question 12: Find the appropriate value of log₁₀ 1005, given that log₁₀ e=0.4343

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.

Question 14: Find the approximate change in the surface area of a cube as side x meters caused by decreasing the side by 1%.

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.

Question 16: Find the approximate change in the volume of a cube as side x meters caused by increasing the side by 1%.

Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

Related Articles

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)

(xv)

(xvi)

(xvii)

(xviii)

(xix)

(xx)

(xxi)

(xxii)

(xxiii)

(xxiv)

(xxv)

(xxvi)

Question 10: Find the appropriate value of f(2.01), where f(x) = 4x2+5x+2

Question 11: Find the appropriate value of f(5.001), where f(x) = x3-7x2+15

Question 12: Find the appropriate value of log10 1005, given that log10 e=0.4343

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.

Question 14: Find the approximate change in the surface area of a cube as side x meters caused by decreasing the side by 1%.

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.

Question 16: Find the approximate change in the volume of a cube as side x meters caused by increasing the side by 1%.

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Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

(xiv) $cos (\frac{11\pi}{36})$

(xv) $(80)^{\frac{1}{4}}$

(xvi) $(29)^{\frac{1}{3}}$

(xvii) $(66)^{\frac{1}{3}}$

(xviii) $\sqrt{26}$

(xix) $\sqrt{37}$

(xx) $\sqrt{0.48}$

(xxi) $(82)^{\frac{1}{4}}$

(xxii) $(\frac{17}{81})^{\frac{1}{4}}$

(xxiii) $(33)^{\frac{1}{5}}$

(xxiv) $\sqrt{36.6}$

(xxv) $(25)^{\frac{1}{3}}$

(xxvi) $\sqrt{49.5}$

Question 10: Find the appropriate value of f(2.01), where f(x) = 4x²+5x+2

Question 11: Find the appropriate value of f(5.001), where f(x) = x³-7x²+15

Question 12: Find the appropriate value of log₁₀ 1005, given that log₁₀ e=0.4343