Rd Sharma XII Vol 1 2020 for Class 12 Science Maths Chapter 13 - Differentials, Errors And Approximations

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Rd Sharma XII Vol 1 2020 Solutions for Class 12 Science Maths Chapter 13 Differentials, Errors And Approximations are provided here with simple step-by-step explanations. These solutions for Differentials, Errors And Approximations are extremely popular among Class 12 Science students for Maths Differentials, Errors And Approximations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2020 Book of Class 12 Science Maths Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2020 Solutions. All Rd Sharma XII Vol 1 2020 Solutions for class Class 12 Science Maths are prepared by experts and are 100% accurate.

Page No 13.10:

Question 10:

Find the approximate value of f (2.01), where f (x) = 4x² + 5x + 2.

Answer:

$Let : x = 2 x + ∆ x = 2.01 \Rightarrow ∆ x = 0.01 f (x) = 4 x^{2} + 5 x + 2 \Rightarrow f (x = 2) = 16 + 10 + 2 = 28 Now, y = f (x) \Rightarrow \frac{d y}{d x} = 8 x + 5 ∴ d y = ∆ y = \frac{d y}{d x} d x = (8 x + 5) \times 0.01 = (16 + 5) \times 0.01 = 0.21 ∴ f (2.01) = y + ∆ y = 28.21$

Page No 13.10:

Question 11:

Find the approximate value of f (5.001), where f (x) = x³ − 7x² + 15.

Answer:

$Let : x = 5 x + ∆ x = 5.001 \Rightarrow ∆ x = 0.001 f (x) = x^{3} - 7 x^{2} + 15 \Rightarrow y = f (x = 3) = 125 - 175 + 15 = - 35 Now, y = f (x) \Rightarrow \frac{d y}{d x} = 3 x^{2} - 14 x ∴ d y = ∆ y = \frac{d y}{d x} d x = (3 x^{2} - 14 x) \times 0.001 = (75 - 70) \times 0.001 = 0.005 ∴ f (5.001) = y + ∆ y = - 35 + 0.005 = - 34.995$

Page No 13.10:

Question 12:

Find the approximate value of log₁₀ 1005, given that log₁₀ e = 0.4343.

Answer:

$Let : y = f (x) = \log_{10} x Here, x = 1000, x + ∆ x = 1005 \Rightarrow ∆ x = 5 \Rightarrow d x = ∆ x = 5 For x = 1000, y = \log_{10} 1000 = \log_{10} {(10)}^{3} = 3 Now, y = \log_{10} x = \frac{\log_{e} x}{\log_{e} 10} ∴ \frac{d y}{d x} = \frac{0.4343}{x} \Rightarrow {(\frac{d y}{d x})}_{x = 1000} = \frac{0.4343}{1000} = 0.0004343 ∆ y = d y = \frac{d y}{d x} d x = 0.0004343 \times 5 = 0.0021715 ∴ \log_{10} 1005 = y + ∆ y = 3.0021715$

Page No 13.10:

Question 13:

If the radius of a sphere is measured as 9 cm with an error of 0.03 m, find the approximate error in calculating its surface area.

Answer:

Let x be the radius and y be the surface area of the sphere.

$Then, x = 9 ∆ x = 0.03 m = 3 cm \Rightarrow x + ∆ x = 9 + 3 = 12 cm y = 4 {πx}^{2} For x = 9, y = 4 π \times 9^{2} = 324 π \frac{dy}{dx} = 8 πx \Rightarrow {(\frac{dy}{dx})}_{x = 9} = 72 π ∴ ∆ y = dy = \frac{dy}{dx} dx = 72 π \times 3 = 216 π {cm}^{2} Therefore, the approximate error in the surface area is 216 π c m^{2} . Disclaimer : This solution has been created according to the question given in the book . However, the solution given in the book is incorrect .$

Page No 13.10:

Question 14:

Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%.

Answer:

Let y be the surface area of the cube.

$y = 6 x^{2} We have \frac{△ x}{x} \times 100 = 1 Now, \frac{d y}{d x} = 12 x \Rightarrow △ y = d y = \frac{d y}{d x} d x = 12 x \times \frac{x}{100} = 0.12 x^{2} m^{2} Hence, approximate change in the surface area of the cube is 0.12 x^{2} m^{2} .$

Page No 13.10:

Question 15:

If the radius of a sphere is measured as 7 m with an error of 0.02 m, find the approximate error in calculating its volume.

Answer:

Let x be the radius of the sphere and y be its volume.
$y = \frac{4}{3} π x^{3} Let ∆ x be the error in the radius . x = 7 ∆ x = 0.02 \frac{d y}{d x} = 4 π x^{2} \Rightarrow {(\frac{d y}{d x})}_{x = 7} = 196 π ∴ ∆ y = d y = \frac{d y}{d x} d x = 196 π \times 0.02 = 3.92 π Hence, the approximate error in calculating the volume of the sphere is 3.92 π m^{3} .$

Page No 13.10:

Question 16:

Find the approximate change in the value V of a cube of side x metres caused by increasing the side by 1%.

Answer:

$Volume of the c ube, V = x^{3} We have ∆ x = 0.01 x \frac{d V}{d x} = 3 x^{2} \Rightarrow ∆ V = d V = \frac{d V}{d x} d x = 3 x^{2} \times 0.01 x = 0.03 x^{3} Hence, the approximate change in the value V of the cube is 0.03 x^{3} m^{3} . Disclaimer : This solution has been created according to the question given in the book . However, the solution in the book is incorrect .$

Page No 13.12:

Question 1:

If there is an error of 2% in measuring the length of a simple pendulum, then percentage error in its period is
(a)1%
(b) 2%
(c) 3%
(d) 4%

Answer:

(a) 1%
Let l be the length if the pendulum and T be the period.

$Also, let ∆ l be the error in the length and ∆ T be the error in the period . We have \frac{∆ l}{l} \times 100 = 2 \Rightarrow \frac{d l}{l} \times 100 = 2 Now, T = 2 π \sqrt{\frac{l}{g}} Taking \log on both sides, we get \log T = \log 2 π + \frac{1}{2} \log l - \frac{1}{2} \log g Differentiating both sides w . r . t . x, we get \frac{1}{T} \frac{d T}{d l} = \frac{1}{2 l} \Rightarrow \frac{d T}{d l} = \frac{T}{2 l} \Rightarrow \frac{d l}{l} \times 100 = 2 \frac{d T}{T} \times 100 \Rightarrow \frac{d T}{T} \times 100 = \frac{2}{2} \Rightarrow \frac{∆ T}{T} \times 100 = 1 Hence, there is an error of 1 % in calculating the period of the pendulum .$

Page No 13.12:

Question 2:

If there is an error of a% in measuring the edge of a cube, then percentage error in its surface is
(a) 2a%

(b) $\frac{a}{2} %$

(c) 3a%

(d) none of these

Answer:

(a) 2a%

Let x be the side of the cube and y be its surface area.

$\frac{∆ x}{x} \times 100 = a Also, y = 6 x^{2} \Rightarrow \frac{d y}{d x} = 12 x \Rightarrow \frac{∆ y}{y} = \frac{12 x}{y} \times d x = \frac{2}{x} \times \frac{a x}{100} \Rightarrow \frac{∆ y}{y} \times 100 = 2 a Hence, the error in the surface area is 2 a % .$

Page No 13.12:

Question 3:

If an error of k% is made in measuring the radius of a sphere, then percentage error in its volume is
(a) k%
(b) 3k%
(c) 2k%
(d) k/3%

Answer:

(b) 3k%
Let x be the radius of the sphere and y be its volume.
Then,
$\frac{∆ x}{x} \times 100 = k Also, y = \frac{4}{3} π x^{3} \Rightarrow \frac{d y}{d x} = 4 π x^{2} \Rightarrow \frac{∆ y}{y} = \frac{4 π x^{2}}{y} d x = \frac{4 π x^{2}}{\frac{4}{3} π x^{3}} \times \frac{k x}{100} \Rightarrow \frac{∆ y}{y} \times 100 = 3 k Hence, the error in the volume is 3 k % .$

Page No 13.12:

Question 4:

The height of a cylinder is equal to the radius. If an error of α % is made in the height, then percentage error in its volume is
(a) α %
(b) 2α %
(c) 3α %
(d) none of these

Answer:

(c) 3 $α$ $α$ %
Let x be the radius, which is equal to the height of the cylinder. Let y be its volume.

$\frac{∆ x}{x} \times 100 = α Also, y = π x^{2} x = π x^{3} [Radius = Height of the cylinder] \Rightarrow \frac{d y}{d x} = 3 π x^{2} \Rightarrow \frac{∆ y}{y} = \frac{3 π x^{2}}{y} d x = \frac{3}{x} \times \frac{α x}{100} \Rightarrow \frac{∆ y}{y} \times 100 = 3 α Hence, the error in the volume of the cylinder is 3 α % .$

Page No 13.12:

Question 5:

While measuring the side of an equilateral triangle an error of k % is made, the percentage error in its area is
(a) k %
(b) 2k %
(c) $\frac{k}{2} %$
(d) 3k %

Answer:

(b) 2k%

Let x be the side of the triangle and y be its area.

$\frac{∆ x}{x} \times 100 = k Also, y = \frac{\sqrt{3}}{4} x^{2} \Rightarrow \frac{d y}{d x} = \frac{\sqrt{3}}{2} x \Rightarrow \frac{∆ y}{y} = \frac{\sqrt{3} x}{2 y} d x = \frac{2}{x} \times \frac{k x}{100} \Rightarrow \frac{∆ y}{y} \times 100 = 2 k Hence, the error in the area of the triangle is 2 k % .$

Page No 13.13:

Question 6:

If log_e 4 = 1.3868, then log_e 4.01 =
(a) 1.3968
(b) 1.3898
(c) 1.3893
(d) none of these

Answer:

(c) 1.3893
$Consider the function y = f (x) = \log_{e} x . Let : x = 4 x + ∆ x = 4.01 \Rightarrow ∆ x = 0.01 For x = 4, y = l {og}_{e} 4 = 1.3868 y = \log_{e} x \Rightarrow \frac{d y}{d x} = \frac{1}{x} \Rightarrow {(\frac{d y}{d x})}_{x = 4} = \frac{1}{4} \Rightarrow ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{4} \times 0.01 = 0.0025 ∴ \log_{e} 4.01 = y + ∆ y = 1.3893$

Page No 13.13:

Question 7:

A sphere of radius 100 mm shrinks to radius 98 mm, then the approximate decrease in its volume is
(a) 12000 π mm³
(b) 800 π mm³
(c) 80000 π mm³
(d) 120 π mm³

Answer:

(c) 80000 π mm³

Let x be the radius of the sphere and y be its volume.

$x = 100, x + ∆ x = 98 \Rightarrow ∆ x = - 2 y = \frac{4}{3} π x^{3} \Rightarrow \frac{d y}{d x} = 4 π x^{2} \Rightarrow {(\frac{d y}{d x})}_{x = 100} = 40000 π ∴ ∆ y = d y = \frac{d y}{d x} d x = 40000 π \times (- 2) = - 80000 π Hence, the decrease in the volume of the sphere is 80000 π {mm}^{3} .$

Page No 13.13:

Question 8:

If the ratio of base radius and height of a cone is 1 : 2 and percentage error in radius is λ %, then the error in its volume is
(a) λ %
(b) 2 λ %
(c) 3 λ %
(d) none of these

Answer:

(c) 3 λ %

Let the radius of the cone be x, the height be 2x and the volume be y.
$\frac{∆ x}{x} = λ % \Rightarrow y = \frac{1}{3} π x^{2} \times 2 x = \frac{2}{3} π x^{3} \Rightarrow \frac{d y}{d x} = 2 π x^{2} \Rightarrow \frac{∆ y}{y} = \frac{2 π x^{2}}{y} d x = \frac{3}{x} \times λ x \Rightarrow \frac{∆ y}{y} = 3 λ %$

Page No 13.13:

Question 9:

The pressure P and volume V of a gas are connected by the relation PV¹^/4 = constant. The percentage increase in the pressure corresponding to a deminition of 1/2 % in the volume is

(a) $\frac{1}{2} %$

(b) $\frac{1}{4} %$

(c) $\frac{1}{8} %$

(d) none of these

Answer:

(c) $\frac{1}{8}$ %

We have
$\frac{△ V}{V} = \frac{- 1}{2} % P V^{\frac{1}{4}} = constant = k (say) Taking \log on both sides, we get \log (P V^{\frac{1}{4}}) = \log k \Rightarrow \log P + \frac{1}{4} \log V = \log k Differentiating both sides w . r . t . x, we get \frac{1}{P} \frac{d P}{d V} + \frac{1}{4 V} = 0 \Rightarrow \frac{d P}{P} = - \frac{d V}{4 V} = - \frac{1}{4} \times \frac{- 1}{2} = \frac{1}{8} Hence, the increase in the pressure is \frac{1}{8} % .$

Page No 13.13:

Question 10:

If y = xⁿ, then the ratio of relative errors in y and x is
(a) 1 : 1
(b) 2 : 1
(c) 1 : n
(d) n : 1

Answer:

(d) n:1
$Let \frac{∆ x}{x} be the relative error in x and \frac{∆ y}{y} be the error in y . Now, y = x^{n} \Rightarrow \frac{d y}{d x} = n x^{n - 1} \Rightarrow \frac{∆ y}{y} = \frac{n x^{n - 1}}{y} d x \Rightarrow \frac{∆ y}{y} = \frac{n x^{n - 1}}{x^{n}} d x = n \frac{∆ x}{x} \Rightarrow \frac{∆ y}{y} : \frac{∆ x}{x} = n : 1$

Page No 13.13:

Question 11:

The approximate value of (33)^1/5 is
(a) 2.0125
(b) 2.1
(c) 2.01
(d) none of these

Answer:

(a) 2.0125
Consider the function y= f(x)= $x^{\frac{1}{5}}$ .
^{$Let : x = 32 x + ∆ x = 33 \Rightarrow ∆ x = 1 y = {(x)}^{\frac{1}{5}} For x = 32, y = 2 Also, \frac{d y}{d x} = \frac{1}{5 {(x)}^{\frac{4}{5}}} \Rightarrow {(\frac{d y}{d x})}_{x = 32} = \frac{1}{80} \Rightarrow ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{80} \times 1 = 0.0125 ∴ {(33)}^{\frac{1}{5}} = y + ∆ y = 2.0125$}

Page No 13.13:

Question 12:

The circumference of a circle is measured as 28 cm with an error of 0.01 cm. The percentage error in the area is

(a) $\frac{1}{14}$

(b) 0.01

(c) $\frac{1}{7}$

(d) none of these

Answer:

(a) $\frac{1}{14}$
Let x be the radius of the circle and y be its circumference.
$x = 28 cm ∆ x = 0.01 cm x = 2 π r y = π r^{2} = π \times \frac{x^{2}}{4 π^{2}} = \frac{x^{2}}{4 π} \Rightarrow \frac{d y}{d x} = \frac{x}{2 π} \Rightarrow \frac{∆ y}{y} = \frac{x}{2 π y} d x = \frac{2}{x} \times 0.01 \Rightarrow \frac{∆ y}{y} \times 100 = \frac{2}{x} = \frac{1}{14} Hence, the percentage error in the area is \frac{1}{14} .$

Page No 13.13:

Question 13:

If y = x⁴ - 10 and if x changes from 2 to 1.99, the change in y is
(a) 0.32 (b) 0.032 (c) 5.68 (d) 5.968

Answer:

Let x = 2 and x + ∆x = 1.99.

∴ ∆x = 1.99 − 2 = −0.01

$y = x^{4} - 10$ (Given)

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = 4 x^{3}$

$\Rightarrow {(\frac{d y}{d x})}_{x = 2} = 4 \times {(2)}^{3} = 4 \times 8 = 32$

$∴ ∆ y = (\frac{d y}{d x}) ∆ x$

$\Rightarrow ∆ y = 32 \times (- 0.01) = - 0.32$

Thus, the change in y is 0.32.

Hence, the correct answer is option (a).

Page No 13.13:

Question 1:

If y = x³ + 5 and x changes from 3 to 2.99, then the approximate change is y is _________________.

Answer:

Let x = 3 and x + ∆x = 2.99.

∴ ∆x = 2.99 − 3 = −0.01

y = x³ + 5 (Given)

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = 3 x^{2}$

$\Rightarrow {(\frac{d y}{d x})}_{x = 3} = 3 \times {(3)}^{2} = 27$

∴ $∆ y = (\frac{d y}{d x}) ∆ x$

$\Rightarrow ∆ y = 27 \times (- 0.01) = - 0.27$

Thus, the approximate change in y is −0.27.

If y = x³ + 5 and x changes from 3 to 2.99, then the approximate change is y is ___−0.27___.

Page No 13.13:

Question 2:

The approximate change in the volume of a cube of side x metres caused by increasing the side by 2%, is ______________.

Answer:

Let ∆x be the change in side x and ∆V be the change in the volume of the cube.

It is given that, $\frac{∆ x}{x} \times 100 = 2$ .....(1)

Now,

Volume of the cube of side x, V = x³

$V = x^{3}$

Differentiating both sides with respect to x, we get

$\frac{d V}{d x} = 3 x^{2}$

$∴ ∆ V = (\frac{d V}{d x}) ∆ x$

$\Rightarrow ∆ V = 3 x^{2} ∆ x$

$\Rightarrow ∆ V = 3 x^{2} \times \frac{2 x}{100}$ [Using (1)]

$\Rightarrow ∆ V = \frac{6 x^{3}}{100}$

$\Rightarrow ∆ V = 0.06 x^{3}$

Thus, the approximate change in volume of the cube is 0.06x³ m³.

The approximate change in the volume of a cube of side x metres caused by increasing the side by 2%, is ___0.06x³ m³___.

Page No 13.13:

Question 1:

For the function y = x², if x = 10 and ∆x = 0.1. Find ∆y.

Answer:

$y = x^{2} ∆ x = 0.1 x = 10 \frac{d y}{d x} = 2 x \Rightarrow {(\frac{d y}{d x})}_{x = 10} = 20 \Rightarrow ∆ y = d y = \frac{d y}{d x} d x = 20 \times 0.1 = 2$

Page No 13.13:

Question 2:

If y = log_e x, then find ∆y when x = 3 and ∆x = 0.03.

Answer:

We have

$x = 3 ∆ x = 0.03 y = \log_{e} x For x = 3, y = \log_{e} 3 Also, \frac{d y}{d x} = \frac{1}{x} \Rightarrow {(\frac{d y}{d x})}_{x = 3} = \frac{1}{3} \Rightarrow ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{3} \times 0.03 = 0.01$

Page No 13.13:

Question 3:

If the relative error in measuring the radius of a circular plane is α, find the relative error in measuring its area.

Answer:

Let x be the radius and y be the area of the circular plane.
$We have \frac{△ x}{x} = α and y = x^{2} . \Rightarrow \frac{d y}{d x} = 2 x \Rightarrow \frac{△ y}{y} = \frac{2 x}{y} d x = \frac{2 x}{x^{2}} \times α x = 2 α Hence, the relative error in the area of the circular plane is 2 α .$

Page No 13.13:

Question 4:

If the percentage error in the radius of a sphere is α, find the percentage error in its volume.

Answer:

Let V be the volume of the sphere.
$V = \frac{4}{3} π x^{3} We have \frac{∆ x}{x} \times 100 = α \Rightarrow \frac{d V}{d x} = 4 π x^{2} \Rightarrow \frac{d V}{V} = \frac{4 π x^{2}}{V} d x \Rightarrow \frac{∆ V}{V} = \frac{4 π x^{2}}{\frac{4}{3} π x^{3}} \times \frac{x α}{100} \Rightarrow \frac{∆ V}{V} \times 100 = 3 α Hence, the the percentage error in the volume is 3 α .$

Page No 13.13:

Question 5:

A piece of ice is in the form of a cube melts so that the percentage error in the edge of cube is a, then find the percentage error in its volume.

Answer:

Let x be the side and V be the volume of the cube.
$V = x^{3} We have \frac{∆ x}{x} \times 100 = a ∴ \frac{d V}{d x} = 3 x^{2} \Rightarrow \frac{∆ V}{V} = \frac{3 x^{2}}{V} d x = \frac{3 x^{2}}{x^{3}} \times \frac{a x}{100} \Rightarrow \frac{∆ V}{V} \times 100 = 3 a Hence, the percentage e rror in the volume is 3 a .$

Page No 13.9:

Question 1:

If y = sin x and x changes from π/2 to 22/14, what is the approximate change in y?

Answer:

$Let : x = \frac{π}{2} x + △ x = \frac{22}{14} \Rightarrow d x = △ x = \frac{22}{14} - \frac{π}{2} = 0 Now, y = \sin x \Rightarrow \frac{d y}{d x} = \cos x \Rightarrow {(\frac{d y}{d x})}_{x = \frac{π}{2}} = \cos (\frac{π}{2}) = 0 ∴ ∆ y = \frac{d y}{d x} ∆ x = 0 \times 0 = 0 \Rightarrow △ y = 0$

Hence, there is no change in the value of y.

Page No 13.9:

Question 2:

The radius of a sphere shrinks from 10 to 9.8 cm. Find approximately the decrease in its volume.

Answer:

$Let r be the radius of the sphere . r = 10 c m r + ∆ r = 9.8 cm \Rightarrow ∆ r = 10.0 - 9.8 = 0.2 cm Volume of the sphere, V = \frac{4}{3} π r^{3} \Rightarrow \frac{d V}{d r} = \frac{4}{3} π \times 3 r^{2} = 4 π r^{2} \Rightarrow {(\frac{d V}{d r})}_{r = 10 cm} = 4 π {(10)}^{2} = 400 π {cm}^{3} / cm Change in the volume of the sphere, ∆ V = \frac{d V}{d r} \times d r = 400 π \times 0.2 = 80 π {cm}^{3}$

Page No 13.9:

Question 3:

A circular metal plate expends 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.

Answer:

Let at any time, x be the radius and y be the area of the plate.
$Then, y = x^{2} Let ∆ x be the change in the radius and △ y be the change in the area of the plate . We have \frac{∆ x}{x} \times 100 = k When x = 10, w e g e t ∆ x = \frac{10 k}{100} = \frac{k}{10} Now, y = π x^{2} \Rightarrow \frac{d y}{d x} = 2 π x \Rightarrow {(\frac{d y}{d x})}_{x = 10 cm} = 20 π {cm}^{2} / cm ∴ ∆ y = d y = \frac{d y}{d x} d x = 20 π \times \frac{k}{10} = 2 k π {cm}^{2}$

Hence, the approximate change in the area of the plate is 2k $π$ cm².

Page No 13.9:

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.

Answer:

Let x be the edge of the cube and y be the surface area.

$y = x^{2} Let ∆ x be the error in x and ∆ y be the corresponding error in y . We have \frac{∆ x}{x} \times 100 = 1 \Rightarrow 2 x = \frac{x}{100} [Let d x = ∆ x] Now, y = x^{2} \Rightarrow \frac{d y}{d x} = 2 x ∴ ∆ y = \frac{d y}{d x} \times ∆ x = 2 x \times \frac{x}{100} \Rightarrow ∆ y = 2 \frac{x^{2}}{100} \Rightarrow ∆ y = 2 \frac{y}{100} \Rightarrow \frac{∆ y}{y} = \frac{2}{100} \Rightarrow \frac{∆ y}{y} \times 100 = 2$

Hence, the percentage error in calculating the surface area is 2.

Page No 13.9:

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.

Answer:

Let x be the radius and y be the volume of the sphere.
$y = \frac{4}{3} π x^{3} Let ∆ x be the error in the radius and ∆ y be the error in the volume . Then, \frac{∆ x}{x} \times 100 = 0.1 \Rightarrow \frac{d x}{x} = \frac{1}{1000} Now, y = \frac{4}{3} π x^{3} \Rightarrow \frac{d y}{d x} = 4 {πx}^{2} \Rightarrow dy = 4 {πx}^{2} dx \Rightarrow \frac{dy}{y} = \frac{4 {πx}^{2} dx}{\frac{4}{3} π x^{3}} = \frac{3}{x} dx \Rightarrow \frac{dy}{y} = \frac{3}{1000} \Rightarrow \frac{∆ y}{y} \times 100 = 0.3$

Hence, the percentage error in the calculation of the volume of the sphere is 0.3.

Page No 13.9:

Question 6:

The pressure p and the volume v of a gas are connected by the relation pv^1.4 = const. Find the percentage error in p corresponding to a decrease of 1/2% in v.

Answer:

$We have p v^{1.4} = constant = k (say) Taking \log on both the sides, we get \log (p v^{1.4}) = \log k \Rightarrow \log p + 1.4 \log v = \log k Differentiating both the sides w . r . t . x, we get \frac{1}{p} \frac{d p}{d v} + \frac{1.4}{v} = 0 \Rightarrow \frac{d p}{p} = \frac{- 1.4 d v}{v} Now, d p = \frac{d p}{d v} d v = \frac{- 1.4 p}{v} d v \Rightarrow \frac{d p}{p} \times 100 = - 1.4 (\frac{d v}{v} \times 100) = - 1.4 \times (\frac{- 1}{2}) = 0.7 [Since we are given \frac{1}{2} % decrease in v] Hence, the error in p is 0.7 % .$

Page No 13.9:

Question 7:

The height of a cone increases by k%, its semi-vertical angle remaining the same. What is the approximate percentage increase (i) in total surface area, and (ii) in the volume, assuming that k is small?

Answer:

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.

$Let ∆ h be the change in the height, ∆ r be the change in the radius of base and ∆ l be the change in the slant height . Semi - vertical angle ramaining the same . ∴ \frac{∆ h}{h} = \frac{∆ r}{r} = \frac{∆ l}{l} Also, \frac{∆ h}{h} \times 100 = k Then, \frac{∆ h}{h} \times 100 = \frac{∆ r}{r} \times 100 = \frac{∆ l}{l} \times 100 = k . . . (1) (i) Total surface area of the cone, T = π r l + π r^{2} Differentiating both sides w . r . t . r, we get \frac{d T}{d r} = π l + π r \frac{d l}{d r} + 2 π r \Rightarrow \frac{d T}{d r} = π l + π r \frac{l}{r} + 2 π r [From (1), \frac{d l}{d r} = \frac{∆ l}{∆ r} = \frac{l}{r}] \Rightarrow \frac{d T}{d r} = π l + π l + 2 π r \Rightarrow \frac{d T}{d r} = 2 π (l + r) ∴ ∆ T = \frac{d T}{d r} ∆ r = 2 π (l + r) \times \frac{k r}{100} = \frac{2 k r π (l + r)}{100} ∴ \frac{∆ T}{T} \times 100 = \frac{(\frac{2 k r π (l + r)}{100})}{2 π r (l + r)} \times 100 = 2 k % Hence, the percentage increase in total surface area of cone is 2 k . (ii) Volume of cone, V = \frac{1}{3} π r^{2} h Differentiating both sides w . r . t . h, we get \frac{d V}{d h} = \frac{1}{3} π r^{2} + \frac{1}{3} π h 2 r \frac{d r}{d h} \Rightarrow \frac{d V}{d h} = \frac{1}{3} π r^{2} + \frac{1}{3} π h 2 r \frac{r}{h} [From (1), \frac{d r}{d h} = \frac{∆ r}{∆ h} = \frac{r}{h}] \Rightarrow \frac{d V}{d h} = \frac{1}{3} π r^{2} + \frac{2}{3} π r^{2} \Rightarrow \frac{d V}{d h} = π r^{2} ∴ ∆ V = \frac{d V}{d h} d h = π r^{2} \times \frac{k h}{100} = \frac{k π r^{2} h}{100} ∴ \frac{∆ V}{V} \times 100 = \frac{(\frac{k π r^{2} h}{100})}{\frac{1}{3} π r^{2} h} \times 100 = 3 k % Hence, the percentage increase in the volume of the cone is 3 k .$

Page No 13.9:

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 three times the relative error in the radius.

Answer:

Let x be the radius of the sphere and y be its volume.
$Let ∆ x be the error in the radius and ∆ V be the approximate error in the volume . y = \frac{4}{3} π x^{3} \Rightarrow \frac{d y}{d x} = 4 π x^{2} \Rightarrow ∆ y = d y = \frac{d y}{d x} d x = 4 π x^{2} \times ∆ x \Rightarrow ∆ y = 3 \times \frac{4}{3} π x^{3} \times \frac{∆ x}{x} \Rightarrow ∆ y = 3 \times y \times \frac{∆ x}{x} \Rightarrow \frac{∆ y}{y} = 3 \frac{∆ x}{x}$
Hence proved.

Page No 13.9:

Question 9:

Using differentials, find the approximate values of the following:

(i) $\sqrt{25.02}$

(ii) ${(0.009)}^{\frac{1}{3}}$

(iii) ${(0.007)}^{\frac{1}{3}}$

(iv) $\sqrt{401}$

(v) ${(15)}^{\frac{1}{4}}$

(vi) ${(255)}^{\frac{1}{4}}$

(vii) $\frac{1}{(2.002)^{2}}$

(viii) log_e 4.04, it being given that log₁₀4 = 0.6021 and log₁₀e = 0.4343

(ix) log_e 10.02, it being given that log_e10 = 2.3026

(x) log₁₀ 10.1, it being given that log₁₀e = 0.4343

(xi) cos 61°, it being given that sin60° = 0.86603 and 1° = 0.01745 radian

(xii) $\frac{1}{\sqrt{25.1}}$

(xiii) $\sin (\frac{22}{14})$

(xiv) $\cos (\frac{11 π}{36})$

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

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

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

(xviii) $\sqrt{26}$ [CBSE 2000]

(xix) $\sqrt{37}$ [CBSE 2000]

(xx) $\sqrt{0.48}$ [CBSE 2002C]

(xxi) ${(82)}^{\frac{1}{4}}$ [CBSE 2005]

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

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

(xxiv) $\sqrt{36.6}$

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

(xxvi) $\sqrt{49.5}$ [CBSE 2012]

(xxvii) ${(3.968)}^{\frac{3}{2}}$ [CBSE 2014]

(xxviii) ${(1.999)}^{5}$ [NCERT EXEMPLAR]

(xxix) $\sqrt{0.082}$ [NCERT EXEMPLAR]

Answer:

(i)
$Consider the function y = f (x) = \sqrt{x} . Let : x = 25 x + ∆ x = 25.02 Then, ∆ x = 0.02 For x = 25, y = \sqrt{25} = 5 Let : d x = ∆ x = 0.02 Now, y = \sqrt{x} \Rightarrow \frac{d y}{d x} = \frac{1}{2 \sqrt{x}} \Rightarrow {(\frac{d y}{d x})}_{x = 25} = \frac{1}{10} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{10} \times 0.02 = 0.002 \Rightarrow ∆ y = 0.002 ∴ \sqrt{25.02} = y + ∆ y = 5.002$

(ii)
$Consider the function y = f (x) = \sqrt[3]{x} . Let : x = 0.008 x + ∆ x = 0.009 Then, ∆ x = 0.001 For x = 0.008, y = \sqrt{0.008} = 0.2 Let : d x = ∆ x = 0.001 Now, y = \sqrt[3]{x} \Rightarrow \frac{d y}{d x} = \frac{1}{3 {(x)}^{\frac{2}{3}}} \Rightarrow {(\frac{d y}{d x})}_{x = 0.008} = \frac{1}{3 \times 0.04} = \frac{1}{0.12} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{0.12} \times 0.001 = \frac{1}{120} \Rightarrow ∆ y = \frac{1}{120} = 0.008333 ∴ {(0.009)}^{\frac{1}{3}} = y + ∆ y = 0.208333$

(iii)
$Consider the function y = f (x) = \sqrt[3]{x .} Let : x = 0.008 x + ∆ x = 0.007 Then, ∆ x = - 0.001 For x = 0.008, y = \sqrt{0.008} = 0.2 Let : d x = ∆ x = - 0.001 Now, y = \sqrt[3]{x} \Rightarrow \frac{d y}{d x} = \frac{1}{3 {(x)}^{\frac{2}{3}}} \Rightarrow {(\frac{d y}{d x})}_{x = 0.008} = \frac{1}{3 \times 0.04} = \frac{1}{0.12} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{0.12} \times 0.001 = \frac{1}{120} \Rightarrow ∆ y = \frac{1}{120} = 0.008333 ∴ {(0.007)}^{\frac{1}{3}} = y + ∆ y = 0.191667$

(iv).
$Consider the function y = f (x) = \sqrt{x} . Let : x = 400 x + ∆ x = 401 Then, ∆ x = 1 For x = 400, y = \sqrt{400} = 20 Let : d x = ∆ x = 1 Now, y = \sqrt{x} \Rightarrow \frac{d y}{d x} = \frac{1}{2 \sqrt{x}} \Rightarrow {(\frac{d y}{d x})}_{x = 400} = \frac{1}{40} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{40} \times 1 = \frac{1}{40} \Rightarrow ∆ y = \frac{1}{40} = 0.025 ∴ \sqrt{401} = y + ∆ y = 20.025$

(v)
$Consider the function y = f (x) = x^{\frac{1}{4}} . Let : x = 16 x + ∆ x = 15 Then, ∆ x = - 1 For x = 16, y = {(16)}^{\frac{1}{4}} = 2 Let : d x = ∆ x = - 1 Now, y = {(x)}^{\frac{1}{4}} \Rightarrow \frac{d y}{d x} = \frac{1}{4 {(x)}^{\frac{3}{4}}} \Rightarrow {(\frac{d y}{d x})}_{x = 16} = \frac{1}{32} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{32} \times (- 1) = \frac{- 1}{32} \Rightarrow ∆ y = \frac{- 1}{32} = - 0.03125 ∴ {(15)}^{\frac{1}{4}} = y + ∆ y = 1.96875$

(vi)
$Consider the function y = f (x) = {(x)}^{\frac{1}{4}} . Let : x = 256 x + ∆ x = 255 Then, ∆ x = - 1 For x = 256, y = {(256)}^{\frac{1}{4}} = 4 Let : d x = ∆ x = - 1 Now, y = {(x)}^{\frac{1}{4}} \Rightarrow \frac{d y}{d x} = \frac{1}{4 {(x)}^{\frac{3}{4}}} \Rightarrow {(\frac{d y}{d x})}_{x = 256} = \frac{1}{256} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{256} \times - 1 = \frac{- 1}{256} \Rightarrow ∆ y = \frac{- 1}{256} = - 0.003906 ∴ {(255)}^{\frac{1}{4}} = y + ∆ y = 3.99609 \approx 3.9961$

(vii)
$Consider the function y = f (x) = \frac{1}{x^{2}} . Let : x = 2 x + ∆ x = 2.002 Then, ∆ x = - 0.002 For x = 2, y = \frac{1}{2^{2}} = \frac{1}{4} Let : d x = ∆ x = 0.002 Now, y = \frac{1}{x^{2}} \Rightarrow \frac{d y}{d x} = \frac{2}{x^{3}} \Rightarrow {(\frac{d y}{d x})}_{x = 2} = \frac{1}{4} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{4} \times - 0.002 = - 0.0005 \Rightarrow ∆ y = - 0.0005 ∴ \frac{1}{{(2.002)}^{2}} = y + ∆ y = 0.2495$

(viii)
$Consider the function y = f (x) = \log_{e} x . Let : x = 4 x + ∆ x = 4.04 Then, ∆ x = 0.04 For x = 4, y = \log_{e} 4 = \frac{\log_{10} 4}{\log_{10} e} = \frac{0.6021}{0.4343} = 1.386368 Let : d x = ∆ x = 0.04 Now, y = \log_{e} x \Rightarrow \frac{d y}{d x} = \frac{1}{x} \Rightarrow {(\frac{d y}{d x})}_{x = 4} = \frac{1}{4} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{4} \times 0.04 = 0.01 \Rightarrow ∆ y = 0.01 ∴ \log_{e} 4.04 = y + ∆ y = 1.396368$

(ix)
$Consider the function y = f (x) = \log_{e} x . Let : x = 10 x + ∆ x = 10.02 Then, ∆ x = 0.02 For x =, y = \log_{e} 10 = 2.3026 Let : d x = ∆ x = 0.02 Now, y = \log_{e} x \Rightarrow \frac{d y}{d x} = \frac{1}{x} \Rightarrow {(\frac{d y}{d x})}_{x = 10} = \frac{1}{10} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{10} \times 0.02 = 0.002 \Rightarrow ∆ y = 0.002 ∴ \log_{e} 10.02 = y + ∆ y = 2.3046$

(x)
$Consider the function y = f (x) = \log_{10} x . Let : x = 10 x + ∆ x = 10.1 Then, ∆ x = 0.1 For x =, y = \log_{10} 10 = 1 Let : d x = ∆ x = 0.1 Now, y = \log_{10} x = \frac{\log_{e} x}{\log_{e} 10} \Rightarrow \frac{d y}{d x} = \frac{1}{2.3025 x} \Rightarrow {(\frac{d y}{d x})}_{x = 10} = 0.04343 ∴ ∆ y = d y = \frac{d y}{d x} d x = 0.04343 \times 0.1 = 0.004343 \Rightarrow ∆ y = 0.004343 ∴ \log_{10} 10.1 = y + ∆ y = 1.004343$

(xi)
$Consider the function y = f (x) = \cos x ° . Let : x = 60 ° x + ∆ x = 61 ° Then, ∆ x = 1 ° = 0.01745 For x = 60 °, y = \cos 60 ° = 0.5 Let : d x = ∆ x = 0.01745 Now, y = \cos x \Rightarrow \frac{d y}{d x} = - \sin x \Rightarrow {(\frac{d y}{d x})}_{x = 60} = - 0.86603 ∴ ∆ y = d y = \frac{d y}{d x} d x = - 0.86603 \times 0.01745 = - 0.01511 \Rightarrow ∆ y = - 0.01511 ∴ \cos 61 ° = y + ∆ y = 0.48488 \approx 0.48489$

(xii)
$Consider the function y = f (x) = \frac{1}{\sqrt{x}} . Let : x = 25 x + ∆ x = 25.1 Then, ∆ x = 0.1 For x =, y = \frac{1}{\sqrt{25}} = 0.2 Let : d x = ∆ x = 0.1 Now, y = \frac{1}{\sqrt{x}} \Rightarrow \frac{d y}{d x} = \frac{- 1}{2 {(x)}^{\frac{3}{2}}} \Rightarrow {(\frac{d y}{d x})}_{x = 25} = - 0.004 ∴ ∆ y = d y = \frac{d y}{d x} d x = - 0.004 \times 0.1 = - 0.0004 \Rightarrow ∆ y = - 0.0004 ∴ \frac{1}{\sqrt{25.1}} = y + ∆ y = 0.1996$

(xiii)
$Consider the function y = f (x) = \sin x . Let : x = \frac{22}{7} x + ∆ x = \frac{22}{14} Then, ∆ x = \frac{- 22}{14} For x = π, y = \sin (\frac{22}{7}) = 0 Let : d x = ∆ x = \sin \frac{- 22}{14} = - \sin (\frac{π}{2}) = - 1 Now, y = \sin x \Rightarrow \frac{d y}{d x} = \cos x \Rightarrow {(\frac{d y}{d x})}_{x = \frac{22}{7}} = - 1 ∴ ∆ y = d y = \frac{d y}{d x} d x = - 1 \times (- 1) = 1 \Rightarrow ∆ y = 1 ∴ \sin \frac{22}{14} = y + ∆ y = 1$

(xiv)
$Consider the function y = f (x) = \cos x . Let : x = \frac{π}{3} x + ∆ x = \frac{11 π}{36} Then, ∆ x = \frac{- π}{36} = - 5 ° For x = \frac{π}{3}, y = \cos (\frac{π}{3}) = 0.5 Let : d x = ∆ x = - \sin 5 ° = - 0.08716 Now, y = \cos x \Rightarrow \frac{d y}{d x} = - \sin x \Rightarrow {(\frac{d y}{d x})}_{x = \frac{π}{3}} = - 0.86603 ∴ ∆ y = d y = \frac{d y}{d x} d x = - 0.86603 \times (- 0.08716) = 0.075575 \Rightarrow ∆ y = 0.075575 ∴ \cos \frac{11 π}{36} = y + ∆ y = 0.5 + 0.075575 = 0.575575$

(xv)
$Consider the function y = f (x) = {(x)}^{\frac{1}{4}} . Let : x = 81 x + ∆ x = 80 Then, ∆ x = - 1 For x = 81, y = {(81)}^{\frac{1}{4}} = 3 Let : d x = ∆ x = - 1 Now, y = {(x)}^{\frac{1}{4}} \Rightarrow \frac{d y}{d x} = \frac{1}{4 {(x)}^{\frac{3}{4}}} \Rightarrow {(\frac{d y}{d x})}_{x = 81} = \frac{1}{108} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{108} \times (- 1) = - 0.009259 \Rightarrow ∆ y = - 0.009259 ∴ {(80)}^{\frac{1}{4}} = y + ∆ y = 2.99074$

(xvi)
$Consider the function y = f (x) = {(x)}^{\frac{1}{3}} . Let : x = 27 x + ∆ x = 29 Then, ∆ x = 2 For x = 27, y = {(27)}^{\frac{1}{3}} = 3 Let : d x = ∆ x = 2 Now, y = {(x)}^{\frac{1}{3}} \Rightarrow \frac{d y}{d x} = \frac{1}{3 {(x)}^{\frac{2}{3}}} \Rightarrow {(\frac{d y}{d x})}_{x = 27} = \frac{1}{27} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{27} \times 2 = 0.074 \Rightarrow ∆ y = 0.074 ∴ {(29)}^{\frac{1}{3}} = y + ∆ y = 3.074$

(xvii)
$Consider the function y = f (x) = {(x)}^{\frac{1}{3}} . Let : x = 64 x + ∆ x = 66 Then, ∆ x = 2 For x = 64, y = {(64)}^{\frac{1}{3}} = 4 Let : d x = ∆ x = 2 Now, y = {(x)}^{\frac{1}{3}} \Rightarrow \frac{d y}{d x} = \frac{1}{3 {(x)}^{\frac{2}{3}}} \Rightarrow {(\frac{d y}{d x})}_{x = 64} = \frac{1}{48} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{48} \times 2 = 0.042 \Rightarrow ∆ y = 0.042 ∴ {(66)}^{\frac{1}{3}} = y + ∆ y = 4.042$

(xviii)
$Consider the function y = f (x) = \sqrt{x} . Let : x = 25 x + ∆ x = 26 Then, ∆ x = 1 For x = 25, y = \sqrt{25} = 5 Let : d x = ∆ x = 1 Now, y = {(x)}^{1 / 2} \Rightarrow \frac{d y}{d x} = \frac{1}{2 \sqrt{x}} \Rightarrow {(\frac{d y}{d x})}_{x = 25} = \frac{1}{10} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{10} \times 1 = 0.1 \Rightarrow ∆ y = 0.1 ∴ \sqrt{26} = y + ∆ y = 5.1$

(xix)
$Consider the function y = f (x) = \sqrt{x} . Let : x = 36 x + ∆ x = 37 Then, ∆ x = 1 For x = 36, y = \sqrt{36} = 6 Let : d x = ∆ x = 1 Now, y = {(x)}^{\frac{1}{2}} \Rightarrow \frac{d y}{d x} = \frac{1}{2 \sqrt{x}} \Rightarrow {(\frac{d y}{d x})}_{x = 36} = \frac{1}{12} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{12} \times 1 = 0.0833 \Rightarrow ∆ y = 0.0833 ∴ \sqrt{37} = y + ∆ y = 6.0833$

(xx)
$Consider the function y = f (x) = \sqrt{x .} Let : x = 0.49 x + ∆ x = 0.48 Then, ∆ x = - 0.01 For x = 0.49, y = \sqrt{0.49} = 0.7 Let : d x = ∆ x = 0.01 Now, y = {(x)}^{\frac{1}{2}} \Rightarrow \frac{d y}{d x} = \frac{1}{2 \sqrt{x}} \Rightarrow {(\frac{d y}{d x})}_{x = 0.49} = \frac{1}{1.4} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{1.4} \times (- 0.01) = - 0.007143 \Rightarrow ∆ y = - 0.007143 ∴ \sqrt{0.48} = y + ∆ y = 0.693$

(xxi)
$Consider the function y = f (x) = {(x)}^{\frac{1}{4}} . Let : x = 81 x + ∆ x = 82 Then, ∆ x = 1 For x = 81, y = {(81)}^{\frac{1}{4}} = 3 L e t : d x = ∆ x = 1 Now, y = {(x)}^{\frac{1}{4}} \Rightarrow \frac{d y}{d x} = \frac{1}{4 {(x)}^{\frac{3}{4}}} \Rightarrow {(\frac{d y}{d x})}_{x = 81} = \frac{1}{108} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{108} \times 1 = 0.009259 \Rightarrow ∆ y = 0.009259 ∴ {(82)}^{\frac{1}{4}} = y + ∆ y = 3.009259$

(xxii)
$Consider the function y = f (x) = {(x)}^{\frac{1}{4}} . Let : x = \frac{16}{81} x + ∆ x = \frac{17}{81} Then, ∆ x = \frac{1}{81} For x = \frac{16}{81}, y = {(\frac{16}{81})}^{\frac{1}{4}} = \frac{2}{3} Let : d x = ∆ x = \frac{1}{81} Now, y = {(x)}^{\frac{1}{4}} \Rightarrow \frac{d y}{d x} = \frac{1}{4 {(x)}^{\frac{3}{4}}} \Rightarrow {(\frac{d y}{d x})}_{x = \frac{16}{81}} = \frac{27}{32} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{27}{32} \times \frac{1}{81} = \frac{1}{96} = 0.01042 \Rightarrow ∆ y = 0.01042 ∴ {(\frac{17}{81})}^{\frac{1}{4}} = y + ∆ y = 0.6771$

(xxiii)
$Consider the function y = f (x) = {(x)}^{\frac{1}{5}} . Let : x = 32 x + ∆ x = 33 Then, ∆ x = 1 For x = 33, y = {(32)}^{\frac{1}{5}} = 2 Let : d x = ∆ x = 1 Now, y = {(x)}^{\frac{1}{5}} \Rightarrow \frac{d y}{d x} = \frac{1}{5 {(x)}^{\frac{4}{5}}} \Rightarrow {(\frac{d y}{d x})}_{x = 32} = \frac{1}{80} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{80} \times 1 = 0.0125 \Rightarrow ∆ y = 0.0125 ∴ {(33)}^{\frac{1}{5}} = y + ∆ y = 2.0125$

(xxiv)
$Consider the function y = f (x) = \sqrt{x} . Let : x = 36 x + ∆ x = 36.6 Then, ∆ x = 0.6 For x = 36, y = \sqrt{36} = 6 Let : d x = ∆ x = 0.6 Now, y = {(x)}^{\frac{1}{2}} \Rightarrow \frac{d y}{d x} = \frac{1}{2 \sqrt{x}} \Rightarrow {(\frac{d y}{d x})}_{x = 36} = \frac{1}{12} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{12} \times 0.6 = 0.05 \Rightarrow ∆ y = 0.05 ∴ \sqrt{36.6} = y + ∆ y = 6.05$

(xv)
$Consider the function y = f (x) = {(x)}^{\frac{1}{3}} . Let : x = 27 x + ∆ x = 25 Then, △ x = - 2 For x = 27, y = {(27)}^{\frac{1}{3}} = 3 Let : d x = ∆ x = - 2 Now, y = {(x)}^{\frac{1}{3}} \Rightarrow \frac{d y}{d x} = \frac{1}{3 {(x)}^{\frac{2}{3}}} \Rightarrow {(\frac{d y}{d x})}_{x = 27} = \frac{1}{27} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{27} \times (- 2) = - 0.07407 \Rightarrow ∆ y = - 0.07407 ∴ {(25)}^{\frac{1}{3}} = y + ∆ y = 2.9259$

(xxvi)
$Consider the function y = f (x) = \sqrt{x} . Let : x = 49 x + ∆ x = 49.5 Then, ∆ x = 0.5 For x = 49, y = \sqrt{49} = 7 Let : d x = ∆ x = 0.5 Now, y = {(x)}^{\frac{1}{2}} \Rightarrow \frac{d y}{d x} = \frac{1}{2 \sqrt{x}} \Rightarrow {(\frac{d y}{d x})}_{x = 49} = \frac{1}{14} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{1}{14} \times 0.5 = 0.0357 \Rightarrow ∆ y = 0.0357 ∴ \sqrt{49.5} = y + ∆ y = 7.0357$

(xxvii)
$Consider the function y = f (x) = {(x)}^{\frac{3}{2}} . Let : x = 4 x + ∆ x = 3.968 Then, ∆ x = - 0.032 For x = 4, y = {(4)}^{\frac{3}{2}} = 8 Let : d x = ∆ x = - 0.032 Now, y = {(x)}^{\frac{3}{2}} \Rightarrow \frac{d y}{d x} = \frac{3 \sqrt{x}}{2} \Rightarrow {(\frac{d y}{d x})}_{x = 4} = 3 ∴ ∆ y = d y = \frac{d y}{d x} d x = 3 \times (- 0.032) = - 0.096 \Rightarrow ∆ y = - 0.096 ∴ {(3.968)}^{\frac{3}{2}} = y + ∆ y = 7.904$

(xxviii)
$Consider the function y = f (x) = x^{5} . Let : x = 2 x + ∆ x = 1.999 Then, ∆ x = - 0.001 For x = 2, y = 2^{5} = 32 Let : d x = ∆ x = - 0.001 Now, y = x^{5} \Rightarrow \frac{d y}{d x} = 5 x^{4} \Rightarrow {(\frac{d y}{d x})}_{x = 2} = 80 ∴ ∆ y = d y = \frac{d y}{d x} d x = 80 \times (- 0.001) = - 0.08 \Rightarrow ∆ y = - 0.08 ∴ 1 . 999^{5} = y + ∆ y = 31.92$

(xxix)
$Consider the function y = f (x) = \sqrt{x} . Let : x = 0.0841 x + ∆ x = 0.082 Then, ∆ x = - 0.0021 For x = 0.0841, y = \sqrt{0.0841} = 0.29 Let : d x = ∆ x = - 0.0021 Now, y = {(x)}^{\frac{1}{2}} \Rightarrow \frac{d y}{d x} = \frac{1}{2 \sqrt{x}} \Rightarrow {(\frac{d y}{d x})}_{x = 0.0841} = \frac{1}{0.58} = \frac{50}{29} ∴ ∆ y = d y = \frac{d y}{d x} d x = \frac{50}{29} \times (- 0.0021) = - 0.0036 \Rightarrow ∆ y = - 0.0036 ∴ \sqrt{0.082} = y + ∆ y = 0.2864$

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