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## Class - 12 : Mathmatics

### Mathmatics - Chapter No. : 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) (0.009)

^{1/10}(vi) (15)

^{1/4}(vii) (26)

^{1/3}(viii) (255)

^{1/4}(ix) (82)

^{1/4}(x) (410)

^{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}

**Answer :**

(i) Given √(25.3)

Consider y = √x

Let x = 25 and ∆x = 0.3

Then, ∆y = √(x + ∆x) - √x

= √(25.3) - √25

= √(25.3) – 5

⇒ √(25.3) = ∆y + 5

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

= 1/2√x * (0.3) [Since y = √x]

= 1/2√25 * (0.3)

= 0.3/10

= 0.03

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

(ii) Given √(49.5)

Consider y = √x

Let x = 49 and ∆x = 0.5

Then, ∆y = √(x + ∆x) - √x

= √(49.5) - √49

= √(49.5) – 7

⇒ √(49.5) = ∆y + 7

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

= 1/2√x * (0.5) [Since y = √x]

= 1/2√49 * (0.5)

= 0.5/14

= 0.035

Hence, the approximate value of √(49.5) is 7 + 0.035 = 7.035

(iii) Given √(0.6)

Consider y = √x

Let x = 1 and ∆x = -0.4

Then, ∆y = √(x + ∆x) - √x

= √(0.6) - √1

= √(0.6) – 1

⇒ √(0.6) = ∆y + 1

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

= 1/2√x * (-0.4) [Since y = √x]

= 1/2√1 * (-0.4)

= -0.2

Hence, the approximate value of √(0.6) is 1 + (−0.2) = 1 − 0.2 = 0.8

(iv) Given (0.009)^{1/3}

Consider y = x^{1/3}

Let x = 0.008 and ∆x = 0.001

Then, ∆y = (x + ∆x)^{1/3} – x^{1/3}

= (0.009)^{1/3} – (0.008)^{1/3}

= (0.009)^{1/3} – 0.2

⇒ (0.009)^{1/3} = ∆y + 0.2

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

= 1/3x^{2/3} * (0.001) [Since y = x^{1/3}]

= 1/(3 * 0.04) * (0.001)

= 0.001/0.12

= 0.008

Hence, the approximate value of (0.009)^{1/3} is 0.2 + 0.008 = 0.208

(v) Given (0.999)^{1/10}

Consider y = x^{1/10}

Let x = 1 and ∆x = -0.001

Then, ∆y = (x + ∆x)^{1/10} – x^{1/10}

= (0.999)^{1/10} – 1

⇒ (0.999)^{1/10} = ∆y + 1

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

= 1/10x^{9/10} * (-0.001) [Since y = x^{1/10}]

= (-0.001)/10

= -0.0001

Hence, the approximate value of (0.999)^{1/10} is 1 + (−0.0001) = 1 – 0.0001 = 0.9999

(vi) Given (15)^{1/4}

Consider y = x^{1/4}

Let x = 16 and ∆x = -1

Then, ∆y = (x + ∆x)^{1/4} – x^{1/4}

= (15)^{1/4} – (16)^{1/4}

= (15)^{1/4} – 2

⇒ (15)^{1/4} = ∆y + 2

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

= 1/4x^{3/4} * (-1) [Since y = x^{1/4}]

= {1/(4(16)^{3/4} }* (-1)

= -1/(4 * 8)

= -1/32

= -0.03125

Hence, the approximate value of (15)^{1/4} is 2 + (−0.03125) = 2 - 0.03125 = 1.96875

(vii) Given (26)^{1/3}

Consider y = x^{1/3}

Let x = 27 and ∆x = -1

Then, ∆y = (x + ∆x)^{1/3} – x^{1/3}

= (26)^{1/3} – (27)^{1/3}

= (26)^{1/4} – 3

⇒ (26)^{1/3} = ∆y + 3

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

= 1/3x^{2/3} * (-1) [Since y = x^{1/3}]

= {1/(3(27)^{2/3}}* (-1)

= -1/(3 * 7)

= -1/27

= -0.0370

Hence, the approximate value of (26)^{1/3} is 3 + (−0.0370) = 3 - 0.0370 = 2.9629

(viii) Given (255)^{1/4}

Consider y = x^{1/4}

Let x = 256 and ∆x = -1

Then, ∆y = (x + ∆x)^{1/4} – x^{1/4}

= (255)^{1/4} – (256)^{1/4}

= (255)^{1/4} – 4

⇒ (255)^{1/4} = ∆y + 4

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

= 1/4x^{3/4} * (-1) [Since y = x^{1/4}]

= {1/(4(256)^{3/4} }* (-1)

= -1/(4 * 64)

= -1/256

= -0.0039

Hence, the approximate value of (255)^{1/4} is 4 + (−0.0039) = 4 - 0.0039 = 3.9961

(ix) Given (82)^{1/4}

Consider y = x^{1/4}

Let x = 81 and ∆x = 1

Then, ∆y = (x + ∆x)^{1/4} – x^{1/4}

= (82)^{1/4} – (81)^{1/4}

= (82)^{1/4} – 3

⇒ (82)^{1/4} = ∆y + 3

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

= 1/4x^{3/4} * (1) [Since y = x^{1/4}]

= {1/(4(81)^{3/4} }

= 1/(4 * 27)

= 1/108

= 0.0009

Hence, the approximate value of (82)^{1/4} is 3 + 0.009 = 3.009

(x) Given (401)^{1/2}

Consider y = x^{1/2}

Let x = 400 and ∆x = 1

Then, ∆y = (x + ∆x)^{1/2} – x^{1/2}

= (401)^{1/2} – (400)^{1/2}

= (401)^{1/2} – 20

⇒ (401)^{1/2} = ∆y + 20

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

= 1/2x^{1/2} * (1) [Since y = x^{1/2}]

= 1/(2 * 20)

= 1/40

= 0.025

Hence, the approximate value of (401)^{1/2} is 20 + 0.025 = 20.025

(xi) Given (0.0037)^{1/2}

Consider y = x^{1/2}

Let x = 0.0036 and ∆x = 0.0001

Then, ∆y = (x + ∆x)^{1/2} – x^{1/2}

= (0.0037)^{1/2} – (0.0036)^{1/2}

= (0.0037)^{1/2} – 0.06

⇒ (0.0037)^{1/2} = ∆y + 0.06

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

= 1/2x^{1/2} * (0.0001) [Since y = x^{1/2}]

= 0.0001/(2 * 0.06)

= 0.0001/0.12

= 0.00083

Thus, the approximate value of (0.0037)^{1/2} is 0.06 + 0.00083 = 0.06083

(xii) Given (26.57)^{1/3}

Consider y = x^{1/3}

Let x = 27 and ∆x = -0.43

Then, ∆y = (x + ∆x)^{1/3} – x^{1/3}

= (26.57)^{1/3} – (27)^{1/3}

= (26.57)^{1/3} – 3

⇒ (26.57)^{1/3} = ∆y + 3

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

= 1/3x^{2/3} * (-0.43) [Since y = x^{1/3}]

= {1/(3(27)^{2/3}}* (-0.43)

= -0.43/(3 * 9)

= -0.43/27

= -0.015

Hence, the approximate value of (26.57)^{1/3} is 3 + (−0.015) = 3 - 0.015 = 2.984

(xiii) Given (81.5)^{1/4}

Consider y = x^{1/4}

Let x = 81 and ∆x = 0.5

Then, ∆y = (x + ∆x)^{1/4} – x^{1/4}

= (81.5)^{1/4} – (81)^{1/4}

= (81.5)^{1/4} – 3

⇒ (81.5)^{1/4} = ∆y + 3

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

= 1/4x^{3/4} * (0.5) [Since y = x^{1/4}]

= 0.5/{(4(81)^{3/4} }

= 0.5/(4 * 27)

= 0.5/108

= 0.0046

Hence, the approximate value of (81.5)^{1/4} is 3 + 0.0046 = 3.0046

(xiv) Given (3.968)^{3/2}

Consider y = x^{3/2}

Let x = 4 and ∆x = -0.032

Then, ∆y = (x + ∆x)^{3/2} – x^{3/2}

= (3.968)^{3/2} – (4)^{3/2}

= (3.968)^{3/2} – 8

⇒ (3.968)^{3/2} = ∆y + 8

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

= (3/2)x^{1/2} * (-0.032) [Since y = x^{3/2}]

= (3/2) * 4^{1/2} * (-0.032)

= -(3/2) * 2 * (0.032)

= -3 * 0.032

= -0.096

Hence, the approximate value of (3.968)^{3/2} is 8 + (−0.096) = 8 - 0.096 = 7.904

(xv) Given (32.15)^{1/5}

Consider y = x^{1/5}

Let x = 32 and ∆x = 0.15

Then, ∆y = (x + ∆x)^{1/5} – x^{1/5}

= (32.15)^{1/5} – (32)^{1/5}

= (32.15)^{1/5} – 2

⇒ (32.15)^{1/5} = ∆y + 2

Now, dy is approximately equal to ∆y and is given by,

dy = (dy/dx) * ∆x

= 1/5x^{4/5} * (0.15) [Since y = x^{1/5}]

= 0.15/{(5(32)^{4/5} }

= 0.15/(5 * 16)

= 0.15/80

= 0.00187

Hence, the approximate value of (32.15)^{1/5} is 2 + 0.00187 = 2.00187

**Question :2:
Find the approximate value of f (2.01), where f (x) = 4x ^{2} + 5x + 2**

**Answer :**

Let x = 2 and ∆x = 0.01. Then, we have:

f(2.01) = f(x + ∆x) = 4(x + ∆x)^{2} + 5(x + ∆x) + 2

Now, ∆y = f(x + ∆x) − f(x)

f(x + ∆x) = f(x) + ∆y

≈ f(x) + f’(x) * ∆x [Since dx = ∆x]

Now, f(2.01) ≈ (4x^{2} + 5x + 2) + (8x + 5) ∆x

= (4 * 2^{2} + 5 * 2 + 2) + (8 * 2 + 5)(0.01) [as x = 2 and ∆x = 0.01]

= (16 + 10 + 2) + (16 + 5)(0.01)

= 28 + 21 * 0.01

= 28 + 0.21

= 28.21

Hence, the approximate value of f (2.01) is 28.21

**Question :3:
Find the approximate value of f (5.001), where f (x) = x ^{3} - 7x^{2} + 15**

**Answer :**

Let x = 5 and ∆x = 0.001. Then, we have:

f(5.001) = f(x + ∆x) = (x + ∆x)^{3} - 7(x + ∆x)^{2} + 15

Now, ∆y = f(x + ∆x) − f(x)

So, f(x + ∆x) = f(x) + ∆y

≈ f(x) + f’(x) * ∆x [Since dx = ∆x]

Now, 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) [as x = 5 and ∆x = 0.001]

= (125 - 175 + 15) + (75 - 70)(0.001)

= -35 + 5 * 0.01

= -35 + 0.005

= -34.995

Hence, the approximate value of f (5.001) is -34.995

**Question :4:
Find the approximate change in the volume V of a cube of side x metres caused by increasing side by 1%.**

**Answer :**

The volume of a cube (V) of side x is given by V = x^{3}

So, dV = (dV/dx) * Δx

= (3x^{2}) * Δx

= (3x^{2}) * 0.01x

= 0.03x^{3} [Since 1% of x = 0.01x]

Hence, the approximate change in the volume of the cube is 0.03x^{3} m^{3}

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

**Answer :**

The surface area of a cube (V) of side x is given by S = 6x^{2}

So, dS = (dS/dx) * Δx

= (12x) * Δx

= (12x) * 0.01x

= 0.12x^{2} [Since 1% of x = 0.01x]

Hence, the approximate change in the surface area of the cube is 0.12x^{2} m^{3}

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

**Answer :**

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

Then, r = 7 m and ∆r = 0.02 m

Now, the volume V of the sphere is given by,

V = 4πr^{3}/3

dV/dr = 4πr^{2}

So, dV = (dV/dr) * ∆r

= (4πr^{2}) * ∆r

= (4π * 7^{2}) * 0.02

= 3.92π

Hence, the approximate error in calculating the volume is 3.92 π m^{3}

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

**Answer :**

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

Then, r = 9 m and ∆r = 0.03 m

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

S = 4πr^{2}

So, dS/dr = 8πr

Now, dS = (dS/dr) * ∆r

= (8πr) * ∆r

= (8π * 9) * 0.03

= 2.16 π

Hence, the approximate error in calculating the surface area is 2.16π m^{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**

**Answer :**

Let x = 3 and ∆x = 0.02. Then, we have:

f(5.001) = f(x + ∆x) = 3(x + ∆x)^{2} + 15(x + ∆x) + 5

Now, ∆y = f(x + ∆x) − f(x)

So, f(x + ∆x) = f(x) + ∆y

≈ f(x) + f’(x) * ∆x [Since dx = ∆x]

Now, f(3.02) ≈ (3x^{2} + 15x + 5) + (6x + 15) ∆x

= (3 * 3^{2} + 15 * 3 + 5) + (6 * 3 + 15)(0.002) [as x = 3 and ∆x = 0.02]

= (27 + 45 + 5) + (18 + 15)(0.02)

= 77 + 33 * 0.01

= 77 + 0.66

= 77.66

Hence, the approximate value of f (3.02) is 77.66

So, the correct answer is option D.

**Question :9:
The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is
A. 0.06 x**

^{3}m

^{3}B. 0.6 x

^{3}m

^{3}C. 0.09 x

^{3}m

^{3}D. 0.9 x

^{3}m

^{3}

**Answer :**

The volume of a cube (V) of side x is given by V = x^{3}

So, dV = (dV/dx) * ∆x

= (3x^{2}) * ∆x

= (3x^{2}) * 0.03x [Since 3% of x is 0.03x]

= 0.09x^{2}

Hence, the approximate change in the volume of the cube is 0.09x^{3} m^{3}

So, the correct answer is option C.

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