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

Last Updated : 12 Dec, 2021

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 = 4πx²dx
△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