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

\frac{dy}{dx}     = cos x

(\frac{dy}{dx})_{x=\frac{\pi}{2}}     = cos (π/2) = 0

△y = (\frac{dy}{dx})_{x=\frac{\pi}{2}}     △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 = \frac{4}{3}\pi x^3

\frac{dy}{dx} = \frac{4}{3}(3 \pi x^2)     = 4πx²

(\frac{dy}{dx})_{x=10}     = 4π(10)² = 400 π

△y = (\frac{dy}{dx})_{x=10}     △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,

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 = πx²

\frac{dy}{dx}     = π(2x) = 2πx

(\frac{dy}{dx})_{x=10}     = 2π(10) = 20 π

△y = (\frac{dy}{dx})_{x=10}     △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

\frac{dy}{dx}     = 6(2x) = 12x

△y = (\frac{dy}{dx})     △x

△y = (12x) (x/100)

△y = 0.12 x²

So, △y/y = 0.12 x²/6 x² = 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 = \frac{4}{3}\pi x^3

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 = \frac{4}{3}\pi x^3

\frac{dy}{dx} = \frac{4}{3}(3 \pi x^2)     = 4πx²

dy = 4πx² dx

△y = (4πx²) △x

Change in volume,

△y/y = \frac{(4\pi x^2) \triangle x}{\frac{4}{3}\pi x^3}

△y/y = \frac{3}{x} \triangle x

△y/y = 3(\frac{\triangle x}{x} )     = 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,

\frac{\triangle v}{v} \times 100     = - 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 k

log(p) + 1.4 log(v) = log k

Differentiating wrt v, we get

\frac{1}{p} \frac{dp}{dv} + 1.4 (\frac{1}{p}) = 0

\frac{dp}{p} = \frac{-1.4 dv}{v}

Percentage change in p = △p/p × 100 = \frac{-1.4 \triangle v}{v}     × 100 = -1.4 (\frac{\triangle v}{v} \times 100)

= -1.4 (\frac{-1}{2})

= 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

\frac{dy}{dr}     = πl + πr \frac{dl}{dr}     + 2πr

\frac{dy}{dr}     = πl + πr \frac{l}{r}     + 2πr

\frac{dy}{dr}     = πl + πl + 2πr

\frac{dy}{dr}     = 2πl + 2πr = 2π(r+l)

△y = (\frac{dy}{dr})     △r

△y = (2π(r+l)) (\frac{kr}{100}) = \frac{2\pi kr(r+l)}{100}

Percentage change in y = △y/y × 100 = \frac{\frac{2\pi kr(r+l)}{100}}{\pi r(l+r)}     × 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) = \frac{1}{3}\pi r^2h

Differentiating both the sides wrt h, we get

\frac{dy}{dh} = \frac{1}{3}\pi     (r² + h(2r \frac{dr}{dh}    )

\frac{dy}{dh} = \frac{1}{3}\pi     (r² + h(2r \frac{r}{h}    )

\frac{dy}{dh} = \frac{1}{3}\pi     (r² + 2r²)

\frac{dy}{dh}     = πr²

△y = (\frac{dy}{dh})     △h

△y = (πr²) (\frac{kh}{100}) = \frac{kh \pi r^2}{100}

Percentage change in y = △y/y × 100 = \frac{\frac{kh \pi r^2}{100}}{\frac{1}{3}\pi r^2h}     × 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) = \frac{4}{3}\pi x^3

Differentiating both the sides wrt x, we get

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

\frac{dy}{dx}     = 4πx²

△y = (\frac{dy}{dx})     △x

△y = (4πx²) (△x)

△y/y = \frac{(4 \pi x^2) (\triangle x)}{\frac{4}{3}\pi x^3}

△y/y = 3 \frac{\triangle x}{x}

Hence proved!!

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

(i) \sqrt{25.02}

Solution:

Considering the function as

y = f(x) = \sqrt{x}

Taking x = 25, and

x+△x = 25.02

△x = 25.02-25 = 0.2

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})     (0.02) = 0.002

Hence, \sqrt{25.02}     = y+△y = 5 + 0.002 = 5.002

(ii) (0.009)^{\frac{1}{3}}

Solution:

Considering the function as

y = f(x) = (x)^{\frac{1}{3}}

Taking x = 0.008, and

x+△x = 0.009

△x = 0.009-0.008 = 0.001

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

y_{(x=0.008)} = (0.008)^{\frac{1}{3}} = 0.2

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

(\frac{dy}{dx})_{x=0.008} = \frac{1}{3(0.008)^{\frac{2}{3}}} = \frac{1}{0.12}

△y = dy = (\frac{dy}{dx})_{x=0.008}     dx

△y = (\frac{1}{0.12})     △x

△y = (\frac{1}{0.12})     (0.001) = \frac{1}{120}     = 0.008333

Hence, (0.009)^{\frac{1}{3}}     = y+△y = 0.2 + 0.008333 = 0.208333

(iii) (0.007)^{\frac{1}{3}}

Solution:

Considering the function as

y = f(x) = (x)^{\frac{1}{3}}

Taking x = 0.008, and

x+△x = 0.007

△x = 0.007-0.008 = -0.001

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

y_{(x=0.008)} = (0.008)^{\frac{1}{3}}     = 0.2

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

(\frac{dy}{dx})_{x=0.008} = \frac{1}{3(0.008)^{\frac{2}{3}}} = \frac{1}{0.12}

△y = dy = (\frac{dy}{dx})_{x=0.008}     dx

△y = (\frac{1}{0.12})     △x

△y = (\frac{1}{0.12})     (-0.001) = \frac{-1}{120}     = -0.008333

Hence, (0.007)^{\frac{1}{3}}     = y+△y = 0.2 + (-0.008333) = 0.191667

(iv) \sqrt{401}

Solution:

Considering the function as

y = f(x) = \sqrt{x}

Taking x = 400, and

x+△x = 401

△x = 401-400 = 1

y = \sqrt{x}

y_{(x=400)} = \sqrt{400} = 20

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

(\frac{dy}{dx})_{x=400} = \frac{1}{2\sqrt{400}} = \frac{1}{40}

△y = dy = (\frac{dy}{dx})_{x=400}     dx

△y = (\frac{1}{40})     △x

△y = (\frac{1}{40})     (1) = 0.025

Hence, \sqrt{401}     = y+△y = 20 + 0.025 = 20.025

(v) (15)^{\frac{1}{4}}

Solution:

Considering the function as

y = f(x) = (x)^{\frac{1}{4}}

Taking x = 16, and

x+△x = 15

△x = 15-16 = -1

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

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

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

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

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

△y = (\frac{1}{32})     △x

△y = (\frac{1}{32})     (-1) = \frac{-1}{32}     = -0.03125

Hence, (15)^{\frac{1}{4}}     = y+△y = 0.2 + (-0.03125) = 1.96875

(vi) (255)^{\frac{1}{4}}

Solution:

Considering the function as

y = f(x) = (x)^{\frac{1}{4}}

Taking x = 256, and

x+△x = 255

△x = 255-256 = -1

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

y_{(x=256)} = (256)^{\frac{1}{4}}     = 4

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

(\frac{dy}{dx})_{x=256} = \frac{1}{4(256)^{\frac{3}{4}}} = \frac{1}{256}

△y = dy = (\frac{dy}{dx})_{x=256}     dx

△y = (\frac{1}{256})     △x

△y = (\frac{1}{256})     (-1) = \frac{-1}{256}     = -0.003906

Hence, (255)^{\frac{1}{4}}     = y+△y = 0.2 + (-0.003906) = 3.9961

(vii) \frac{1}{(2.002)^2}

Solution:

Considering the function as

y = f(x) = \frac{1}{x^2}

Taking x = 2, and

x+△x = 2.002

△x = 2.002-2 = 0.002

y = \frac{1}{x^2}

y_{(x=2)} = \frac{1}{2^2} = \frac{1}{4}

\frac{dy}{dx} = \frac{-2}{x^3}

(\frac{dy}{dx})_{x=2} = \frac{-2}{2^3} = \frac{-1}{4}

△y = dy = (\frac{dy}{dx})_{x=2}     dx

△y = (\frac{-1}{4})     △x

△y = (\frac{-1}{4})     (0.002) = -0.0005

Hence, \frac{1}{(2.002)^2}     = y+△y = \frac{1}{4}     + (-0.005) = 0.2495

(viii) log_e 4.04, it being given that log₁₀ 4=0.6021 and log₁₀ e=0.4343

Solution:

Considering the function as

y = f(x) = log_e x

Taking x = 4, and

x+△x = 4.04

△x = 4-4.04 = 0.04

y = log_e x 

y_{(x=4)}     = log_e 4 = \frac{log_{10}4}{log_{10}e} = \frac{0.6021}{0.4343}     = 1.386368

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

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

△y = dy = (\frac{dy}{dx})_{x=4}     dx

△y = (\frac{1}{4})     △x

△y = (\frac{1}{4})     (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 x

Taking x = 10, and

x+△x = 10.02

△x = 10.02-10 = 0.02

y = log_e x

y_{(x=10)}     = log_e 10 = 2.3026

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

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

△y = dy = (\frac{dy}{dx})_{x=10}     dx

△y = (\frac{1}{10})     △x

△y = (\frac{1}{10})     (0.02) = 0.002

Hence, log_e 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 = \frac{log_ex}{log_e10}

y_{(x=10)}     = log₁₀ 10 = 1

\frac{dy}{dx} = \frac{1}{2.3025x}

(\frac{dy}{dx})_{x=10} = \frac{1}{23.025}

△y = dy = (\frac{dy}{dx})_{x=10}     dx

△y = (\frac{1}{23.025})     △x

△y = (\frac{1}{23.025})     (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

y_{(x=60 \degree)}     = cos 60° = 0.5

\frac{dy}{dx}     = - sin x

(\frac{dy}{dx})_{x=60\degree}     = - sin 60° = -0.86603

△y = dy = (\frac{dy}{dx})_{x=60\degree}     dx

△y = (-0.86603) △x

△y = (-0.86603) (0.01745) = -0.01511

Hence, cos 61° = y+△y = 0.5 + (-0.01511) = 0.48489

(xii) \frac{1}{\sqrt{25.1}}

Solution:

Considering the function as

y = f(x) = \frac{1}{\sqrt{x}}

Taking x = 25, and

x+△x = 25.1

△x = 25.1-25 = 0.1

y = \frac{1}{\sqrt{x}}

y_{(x=25)} = \frac{1}{\sqrt{25}} = \frac{1}{5}

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

(\frac{dy}{dx})_{x=25} = \frac{-1}{2(25)^{\frac{3}{2}}} = \frac{-1}{250}

△y = dy = (\frac{dy}{dx})_{x=25}     dx

△y = (\frac{-1}{250})     △x

△y = (\frac{-1}{250})     (0.1) = \frac{-1}{2500}     = -0.0004

Hence, \frac{1}{\sqrt{25.1}}     = y+△y = \frac{1}{5}     + (-0.0004) = 0.1996

(xiii) sin (\frac{22}{14})

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

y_{(x=22/7)}     = sin (22/7) = 0

\frac{dy}{dx}     = cos x

(\frac{dy}{dx})_{x=22/7}     = cos (22/7)= -1

△y = dy = (\frac{dy}{dx})_{x=22/7}     dx

△y = (-1) △x

△y = (-1) (-1) = 1

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

Practice Questions on Differentials, Errors and Approximations

1. If y = x³ - 2x + 1, find dy when x changes from 2 to 2.05.

2. Find the approximate value of (1.03)³ using differentials.

3. Use differentials to estimate the percentage error in the volume of a cube if its edge is measured as 10 cm with a possible error of 0.1 cm.

4. If y = sin x, find dy when x changes from π/4 to π/4 + 0.01 radians.

5. Estimate the change in the area of a circle when its radius increases from 5 cm to 5.1 cm.

6. Use differentials to approximate √17.

7. If y = e^x, find the approximate change in y when x changes from 1 to 1.02.

8. Estimate the percentage error in the surface area of a sphere if its radius is measured as 5 cm with a possible error of 0.05 cm.

9. Use differentials to approximate the value of (0.99)⁵.

10. If y = ln x, find dy when x changes from e to e + 0.1.

Summary

Chapter 14 of RD Sharma's Class 12 mathematics textbook focuses on Differentials, Errors and Approximations. Exercise 14.1 | Set 1 introduces the fundamental concepts and techniques for using differentials to estimate changes and errors. Key points covered in this exercise set include:

• Understanding the concept of differentials
• Using differentials to approximate small changes in functions
• Applying differentials to estimate errors in measurements
• Calculating percentage errors using differentials
• Approximating values of complex expressions
• Relating differentials to the derivative of a function
• Applying differentials to various mathematical functions (polynomial, trigonometric, exponential, logarithmic)
• Understanding the limitations of differential approximations