# Class 12 NCERT Solutions- Mathematics Part I – Chapter 6 Application of Derivatives – Exercise 6.4

### 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)√25.3We 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

=

=

= 0.3/10

= 0.03

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

(ii)√49.5We 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

=

=

= 0.5/14

= 0.0357

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

(iii)√0.6We 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

=

=

= -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 = x

^{1/3}Let x = 0.008 and △x = 0.001

△y = (x +△x)

^{1/3}– x^{1/3}= (0.009)

^{1/3}– (0.008)^{1/3}= (0.009)

^{1/3}– 0.2or (0.009)

^{1/3 }= 0.2 + △yHere, dy is approximately equal to △y

dy = (dy/dx)△x

=

=

= 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 = x

^{1/10}Let x = 1 and △x = -0.001

△y = (x +△x)

^{1/10}– x^{1/10}= (1)

^{1/10}– (0.001)^{1/10}= (1)

^{1/10}– 1or (1)

^{1/10 }= 1 + △yHere, dy is approximately equal to △y

dy = (dy/dx)△x

=

=

= -0.0001

Hence, the approximate value of (0.999)

^{1/10}= 1 – 0.0001 = 0.9999

(vi)(15)^{1/4}We take y = x

^{1/4}Let x = 16 and △x = -1

△y = (x +△x)

^{1/4}– x^{1/4}= (15)

^{1/4}– (16)^{1/4}= (15)

^{1/4}– 2or (15)

^{1/4 }= 2 + △yHere, dy is approximately equal to △y

dy = (dy/dx)△x

=

=

= -1/32

Hence, the approximate value of (15)

^{1/4}= 2 – 1/32 = 1.9687

(vii)(26)^{1/3}We take y = x

^{1/3}Let x = 27 and △x = -1

△y = (x +△x)

^{1/3}– x^{1/3}= (26)

^{1/3}– (27)^{1/3}= (26)

^{1/3}– 3or (26)

^{1/3 }= 3 + △yHere, dy is approximately equal to △y

dy = (dy/dx)△x

=

=

= -1/27

Hence, the approximate value of (26)

^{1/3 }= 3 – 1/27 = 2.962

(viii)(255)^{1/4}We take y = x

^{1/4}Let x = 256 and △x = -1

△y = (x +△x)

^{1/4}– x^{1/4}= (255)

^{1/4}– (256)^{1/4}= (255)

^{1/4}– 4or (255)

^{1/4 }= 4 + △yHere, dy is approximately equal to △y

dy = (dy/dx)△x

=

=

= -1/256

Hence, the approximate value of (255)

^{1/4 }= 4 – 1/256 = 3.996

(ix)(82)^{1/4}We take y = x

^{1/4}Let x = 81 and △x = 1

△y = (x +△x)

^{1/4}– x^{1/4}= (82)

^{1/4}– (81)^{1/4}= (82)

^{1/4}– 3or (82)

^{1/4 }= 3 + △yHere, dy is approximately equal to △y

dy = (dy/dx)△x

=

=

= 1/108

Hence, the approximate value of (82)

^{1/4 }= 3 + 1/108 = 3.009

(x)(401)^{1/2}We take y = x

^{1/2}Let x = 400 and △x = 1

△y = (x +△x)

^{1/2}– x^{1/2}= (401)

^{1/2}– (400)^{1/2}= (401)

^{1/2}– 20or (401)

^{1/2 }= 20 + △yHere, dy is approximately equal to △y

dy = (dy/dx)△x

=

=

= 1/40

Hence, the approximate value of (401)

^{1/2 }= 20 + 1/40 = 20.025

(xi)(0.0037)^{1/2}We take y = x

^{1/2}Let x = 0.0036 and △x = 0.0001

△y = (x +△x)

^{1/2}– x^{1/2}= (0.0037)

^{1/2}– (0.0036)^{1/2}= (0.0037)

^{1/2}– 0.06or (0.0037)

^{1/2 }= 0.06 + △yHere, dy is approximately equal to △y

dy = (dy/dx)△x

=

=

= 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 = x

^{1/3}Let x = 27 and △x = -0.43

△y = (x +△x)

^{1/3}– x^{1/3}= (26.57)

^{1/3}– (27)^{1/3}= (26.57)

^{1/3}– 3or (26.57)

^{1/3 }= 3 + △yHere, dy is approximately equal to △y

dy = (dy/dx)△x

=

=

= -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 = x

^{1/4}Let x = 81 and △x = 0.5

△y = (x +△x)

^{1/4}– x^{1/4}= (81.5)

^{1/4}– (81)^{1/4}= (81.5)

^{1/4}– 3or (81.5)

^{1/4 }= 3 + △yHere, dy is approximately equal to △y

dy = (dy/dx)△x

=

=

= 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 = x

^{3/2}Let x = 4 and △x = -0.032

△y = (x +△x)

^{3/2}– x^{3/2}= (3.968)

^{3/2}– (4)^{3/2}= (3.968)

^{3/2}– 8or (3.968)

^{3/2 }= 8 + △yHere, dy is approximately equal to △y

dy = (dy/dx)△x

=

=

= -0.096

Hence, the approximate value of (3.968)

^{3/2 }= 8 – 0.096 = 7.904

(xv)(32.15)^{1/5}We take y = x

^{1/5}Let x = 32 and △x = 0.15

△y = (x +△x)

^{1/5}– x^{1/5}= (32.15)

^{1/5}– (32)^{1/5}= (32.15)

^{1/5}– 2or (32.15)

^{1/5 }= 2 + △yHere, dy is approximately equal to △y

dy = (dy/dx)△x

=

=

= 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) = 4x^{2 }+ 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 = 4x

^{2 }+ 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) = (4x

^{2 }+ 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) = x^{3 }– 7x^{2 }+ 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 = x

^{3 }– 7x^{2 }+ 15 -(1)On differentiating we get

dy/dx = 3x

^{2 }– 14x^{ }dy = (3x

^{2 }– 14x) dx = (3x^{2 }– 14x) △xNow 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) = (x

^{3 }– 7x^{2 }+ 15) + (3x^{2 }– 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 = x

^{3}V = x

^{3}On differentiating we get

dV/dx = 3x

^{2}Given that Δx = 0.0

dV =

dV = (3x

^{2})(-0.01x)= 0.03x

^{3}m^{3}Thus, the approximate change in volume is 0.03x

^{3}m^{3}

### 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 = 6x

^{2}S = 6x

^{2}On differentiating we get

dS/dx = 12x

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

dS =

dS = (12x)(-0.01x)

= -0.12x

^{2}m^{2}Thus, the approximate change in volume is -0.12x

^{2}m^{2}

### 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

^{2 }On differentiating we get

Therefore, dv = (4πr

^{2})Δr = 4π(7)^{2}.(0.02)= 12.30m

^{3}Thus, the approximate error in calculating the volume is 12.30m

^{3}

### 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 = 4πr

^{2}On differentiating we get

dS/dr = 8πr

Therefore, dS = (8πr)Δr

= 8π(9).(0.03)

= 2.16πm

^{2}Thus, the approximate error in measuring the surface area is 2.16πm

^{2}.

### Question 8. If f(x) = 3x^{2 }+ 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 = 3x

^{2 }+ 15x + 5 -(1)f(3.02) = f(3 + 0.2)

So, x = 3 and

Δx = 0.02On differentiating eq(1) we get

f'(x) = y = 6x

^{ }+ 15dy = (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) = (3x

^{2 }+ 15x + 5) + (6x + 15)Δxf(3.02) = (3x

^{2 }+ 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) 0.06x^{3} m^{3} (B) 0.6x^{3} m^{3} (C)0.09x^{3} m^{3} (D)0.9x^{3} m^{3}

**Solution:**

We have

Volume of cube = v = x

^{3}On differentiating we get

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

Thus, the approximate change is 0.09x

^{3}m^{3 }The correct option is (C)