Skip to content
Related Articles

Related Articles

Class 12 NCERT Solutions- Mathematics Part I – Chapter 6 Application of Derivatives – Exercise 6.4
  • Last Updated : 05 Mar, 2021

Question 1. Using differentials, find the approximate value of each of the following up to 3 places of decimal.

(i)√25.3

(ii)√49.5

(iii) √0.6

(iv) (0.009)1/3

(v) (0.999)1/10



(vi) (15)1/4

(vii) (26)1/3

(viii) (255)1/4

(ix) (82)1/4

(x) (401)1/2

(xi) (0.0037)1/2

(xii) (26.57)1/3

(xiii) (81.5)1/4

(xiv) (3.968)3/2

(xv) (32.15)1/5

Solution:

(i) √25.3

We take y = √x 

Let x = 25 and △x = 0.3

△y = √(x +△x) – √x

= √25.3 – √25

= √25.3 – 5

or √25.3 = 5 + △y 



Here, dy is approximately equal to △y 

dy = (dy/dx)△x

\frac{1}{2√x}(0.3)

\frac{1}{2√25}(0.3)

= 0.3/10

= 0.03

Hence, the approximate value of√25.3 = 5 + 0.03 = 5.03

(ii)√49.5

We take y = √x 

Let x = 49 and △x = 0.5

△y = √(x +△x) – √x

= √49.5 – √49

= √49.5 – 7

or √49.5 = 7 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

\frac{1}{2√x}(0.5)

\frac{1}{2√49}(0.5)

= 0.5/14

= 0.0357

Hence, the approximate value of √49.5 = 7 + 0.0357 = 7.0357

(iii) √0.6

We take y = √x 

Let x = 0.64 and △x = -0.04

△y = √(x +△x) – √x

= √0.6 – √0.64

= √0.6 – 0.8

or √0.6 = 0.8 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

\frac{1}{2√x}(-0.04)

\frac{1}{2√0.64}(-0.04)

= -0.04/1.6

= -0.025

Hence, the approximate value of √0.6 = 0.8 – 0.025 = 0.775

(iv) (0.009)1/3

We take y = x1/3 

Let x = 0.008 and △x = 0.001

△y = (x +△x)1/3 – x1/3

= (0.009)1/3 – (0.008)1/3

= (0.009)1/3 – 0.2

or (0.009)1/3 = 0.2 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

\frac{0.001}{3x^{\frac{2}{3}}}

\frac{0.001}{3(0.008)^{\frac{2}{3}}}

= 0.0083

Hence, the approximate value of (0.009)1/3 = 0.2 + 0.0083 = 0.2083

(v) (0.999)1/10

We take y = x1/10 

Let x = 1 and △x = -0.001

△y = (x +△x)1/10 – x1/10

= (1)1/10 – (0.001)1/10

= (1)1/10 – 1

or (1)1/10 = 1 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

\frac{-0.001}{10x^{\frac{9}{10}}}

\frac{-0.001}{10(1)^{\frac{9}{10}}}

= -0.0001

Hence, the approximate value of (0.999)1/10 = 1 – 0.0001 = 0.9999

(vi) (15)1/4

We take y = x1/4

Let x = 16 and △x = -1

△y = (x +△x)1/4 – x1/4

= (15)1/4 – (16)1/4

= (15)1/4 – 2

or (15)1/4 = 2 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

\frac{-1}{4x^{\frac{3}{4}}}

\frac{-1}{4(16)^{\frac{3}{4}}}

= -1/32

Hence, the approximate value of (15)1/4 = 2 – 1/32 = 1.9687

(vii) (26)1/3

We take y = x1/3

Let x = 27 and △x = -1

△y = (x +△x)1/3 – x1/3

= (26)1/3 – (27)1/3

= (26)1/3 – 3

or (26)1/3 = 3 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

\frac{-1}{3x^{\frac{2}{3}}}

\frac{-1}{3(27)^{\frac{2}{3}}}

= -1/27

Hence, the approximate value of (26)1/3 = 3 – 1/27 = 2.962

(viii) (255)1/4

We take y = x1/4

Let x = 256 and △x = -1

△y = (x +△x)1/4 – x1/4

= (255)1/4 – (256)1/4

= (255)1/4 – 4

or (255)1/4 = 4 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

\frac{-1}{4x^{\frac{3}{4}}}

\frac{-1}{4(256)^{\frac{3}{4}}}

= -1/256

Hence, the approximate value of (255)1/4 = 4 – 1/256 = 3.996

(ix) (82)1/4

We take y = x1/4

Let x = 81 and △x = 1

△y = (x +△x)1/4 – x1/4

= (82)1/4 – (81)1/4

= (82)1/4 – 3

or (82)1/4 = 3 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

\frac{1}{4x^{\frac{3}{4}}}

\frac{1}{4(81)^{\frac{3}{4}}}

= 1/108

Hence, the approximate value of (82)1/4 = 3 + 1/108 = 3.009

(x) (401)1/2

We take y = x1/2

Let x = 400 and △x = 1

△y = (x +△x)1/2 – x1/2

= (401)1/2 – (400)1/2

= (401)1/2 – 20

or (401)1/2 = 20 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

\frac{1}{2\sqrt{x}}

\frac{1}{2\sqrt{400}}

= 1/40

Hence, the approximate value of (401)1/2 = 20 + 1/40 = 20.025

(xi) (0.0037)1/2

We take y = x1/2

Let x = 0.0036 and △x = 0.0001

△y = (x +△x)1/2 – x1/2

= (0.0037)1/2 – (0.0036)1/2

= (0.0037)1/2 – 0.06

or (0.0037)1/2 = 0.06 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

\frac{0.0001}{2\sqrt{x}}

\frac{0.0001}{2\sqrt{0.0036}}

= 0.0001/0.12

Hence, the approximate value of (0.0037)1/2 = 0.06 + 0.0001/0.12 = 0.0608

(xii) (26.57)1/3

We take y = x1/3

Let x = 27 and △x = -0.43

△y = (x +△x)1/3 – x1/3

= (26.57)1/3 – (27)1/3

= (26.57)1/3 – 3

or (26.57)1/3 = 3 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

\frac{-0.43}{3x^{\frac{2}{3}}}

\frac{-0.43}{3(27)^{\frac{2}{3}}}

= -0.43/27

Hence, the approximate value of (26.57)1/3 = 3 – 0.43/27 = 2.984

(xiii) (81.5)1/4

We take y = x1/4

Let x = 81 and △x = 0.5

△y = (x +△x)1/4 – x1/4

= (81.5)1/4 – (81)1/4

= (81.5)1/4 – 3

or (81.5)1/4 = 3 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

\frac{0.5}{4x^{\frac{3}{4}}}

\frac{0.5}{4(81)^{\frac{3}{4}}}

= 0.5/108

Hence, the approximate value of (81.5)1/4 = 3 + 0.5/108 = 3.004

(xiv) (3.968)3/2

We take y = x3/2

Let x = 4 and △x = -0.032

△y = (x +△x)3/2 – x3/2

= (3.968)3/2 – (4)3/2

= (3.968)3/2 – 8

or (3.968)3/2 = 8 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

\frac{(-0.032)(3)\sqrt{x}}{2}

\frac{(-0.032)(3)\sqrt{4}}{2}

= -0.096

Hence, the approximate value of (3.968)3/2 = 8 – 0.096 = 7.904

(xv) (32.15)1/5

We take y = x1/5

Let x = 32 and △x = 0.15

△y = (x +△x)1/5 – x1/5

= (32.15)1/5 – (32)1/5

= (32.15)1/5 – 2

or (32.15)1/5 = 2 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

\frac{0.15}{5x^{\frac{4}{5}}}

\frac{0.15}{5(32)^{\frac{4}{5}}}

= 0.15/80

Hence, the approximate value of (32.15)1/5 = 2 + 0.15/80 = 2.0018

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

Solution:

Find the approximate value of f(2.01)

So, f(2.01) = f(2 + 0.01)

x = 2 and △x = 0.01      

Given that f(x) = y = 4x2 + 5x + 2       -(1)

On differentiating we get    

dy/dx = 8x + 5

dy = (8x + 5)dx = (8x + 5)△x

Now changing x = x + Δx and y = y +Δy in eq(1)

f(x + Δx) = y + Δy

f(2.01) = y + Δy        -(2)

f(2 + .01) = y + Δy

since 

then Δy = dy and Δx = dx

From eq(2), we get

f(2.01) = (4x2 + 5x + 2) + (8x + 5)△x

= (4(2)2 + 5(2) + 2) + (8(2) + 5)(0.01)

= 16 + 10 + 2 + 0.16 + 0.05

= 28.21

Question 3.  Find the approximate value of f(5.001), where f(x) = x3 – 7x2 + 15.

Solution:

Find the approximate value of f(5.001)

So, f(5.001) = f(5 + .001)

x = 5 and △x = 0.01, 

Given that f(x) = y = x3 – 7x2 + 15        -(1)

On differentiating we get    

dy/dx = 3x2 – 14x 

dy = (3x2 – 14x) dx = (3x2 – 14x) △x

Now changing x = x + Δx and y = y +Δy in eq(1)

f(x + Δx) = y + Δy

f(5.001) = y + Δy        -(2)

f(5 + .001) = y + Δy

since 

then Δy = dy and Δx = dx

From eq(2), we get

f(5.001) = (x3 – 7x2 + 15) + (3x2 – 14x) △x

= ((5)3 – 7(5)2 + 15) + (3(5)2 – 14(5)) (0.001)

= 125 – 175 + 15 + 0.075 + 0.07

= -34.995

Question 4. Find the approximate change in the volume of a cube of side x meters caused by increasing the side by 1%.

Solution:

Volume of cube = x3

V = x3

On differentiating we get    

dV/dx = 3x2

Given that Δx = 0.0

dV = (\frac{dV}{dx}).Δx=3x.Δx

dV = (3x2)(-0.01x)

= 0.03x3m3 

Thus, the approximate change in volume is 0.03x3m3 

Question 5.Find the approximate change in the surface area of a cube of side x meters caused by decreasing the side by 1%.

Solution:

Surface area of cube = 6x2

S = 6x2

On differentiating we get    

dS/dx = 12x

Given that Δx = -0.01[-ve sign because of decrease]

dS = (\frac{dS}{dx}).Δx=12x.Δx

dS = (12x)(-0.01x)

= -0.12x2m2    

Thus, the approximate change in volume is -0.12x2m2

Question 6. If the radius of a sphere is measured as 7m with an error of 0.02m, then find the approximate error in calculating its value.

Solution:

Let r be the radius of the sphere and △r be the error in measuring the radius, 

then r = 7m and △r = 0.02m. 

Now the volume V of the sphere is given by,

v = 4/3πr

On differentiating we get    

\frac{dv}{dr}=4πr^2

Therefore, dv = (4πr2)Δr = 4π(7)2.(0.02)

= 12.30m3

Thus, the approximate error in calculating the volume is 12.30m3

Question 7. If the radius of a sphere is measured as 9m with an error of 0.03m, then find the approximate error in calculating its surface area.

Solution:

Let r be the radius of the sphere and Δr be the error in measuring the radius. 

Then r = 9m and Δr = 0.03m.

Now the surface area S of the sphere is given by,

S = 4πr2

On differentiating we get    

dS/dr = 8πr

Therefore, dS = (8πr)Δr

= 8π(9).(0.03)

= 2.16πm2                     

Thus, the approximate error in measuring the surface area is 2.16πm2.

Question 8. If f(x) = 3x2 + 15x + 5, then the approximate value of f(3.02) is 

(A) 47.66          (B) 57.66          (C)67.66          (D)77.66

Solution:

Given: f(x) = y = 3x2 + 15x + 5        -(1)

f(3.02) = f(3 + 0.2)

So, x = 3 and Δx = 0.02   

On differentiating eq(1) we get    

f'(x) = y = 6x + 15

dy = (6x + 15) dx = (6x + 15)Δx

Now changing x = x + Δx and y = y +Δy in eq(1)

f(x + Δx) = y + Δy

f(3.02) = y + Δy        -(2)

f(3 + 0.2) = y + Δy

since 

then Δy = dy and Δx = dx

From eq(2), we get

f(3.02) = (3x2 + 15x + 5) + (6x + 15)Δx

f(3.02) = (3x2 + 15x + 5) + (6x + 15)(0.02)

= (3(3)2 + 15(3) + 5) + (6(3)+ 15)(0.02)

= 27 + 45 + 5 + 0.36 + 0.3

= 77.66

Hence the correct option is 77.66

Question 9. The approximate change in the volume of a cube of side x meters caused by increasing the side by 3% is 

(A) 0.06x3 m3          (B) 0.6x3 m3          (C)0.09x3 m3           (D)0.9x3 m3

Solution:

We have

Volume of cube = v = x3

On differentiating we get

\frac{dv}{dr}=3x^2

Given that the side increasing 3%, so Δx = 0.03x

dv=(\frac{dv}{dr}).Δx=3x^2.(0.03x)

dv=x^2(0.03x)=0.09x^3m^3

Thus, the approximate change is 0.09x3m

The correct option is (C)

My Personal Notes arrow_drop_up
Recommended Articles
Page :