NCERT Solutions for Chapter 6 Application of Derivatives Class 12 Maths Exercise 6.4

Class 12 Maths NCERT Solutions for Chapter 6 Application of Derivatives Exercise 6.4

Application of Derivatives Exercise 6.4 Solutions

1. Using differentials, find the approximate value 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  
Consider y = √x . Let x = 25 and Δx = 0.3
Then,
Δy = √(x + Δx) - √x = √(25.3) - √(25) =  √25.3 - 5
⇒ √25.3 = Δy + 5 
Now, dy is approximately equal to Δy and is given by, 

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

(ii) √49.5
Consider y = √x . Let x = 49 and Δx = 0.5
Then,  
Δy = √(x + Δx) - √x = √(49.5) - √(49) =  √49.5 - 7
⇒ √49.5 = 7 + Δy 
Now, dy is approximately equal to Δy and is given by,  

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

(iii) √0.6
Consider y = √x.
Let x  = 1 and Δx = -0.4
Then,
Δy = √(x + Δx) - √x = √0.6 - 1 
⇒ √0.6 = 1 + Δy 
Now, dy is approximately equal to Δy and is given by, 

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

(iv) (0.009)^1/3   
Consider y = x^1/3.
Let x = 0.008 and ∆x = 0.001
Then,
Δy = (x + Δx)^1/3 - (x)^1/3 = (0.009)^1/3 - (0.008)^1/3 = (0.009)^1/3 - 0.2
⇒ (0.009)^1/3 = 0.2 + Δy 
Now, dy is approximately equal to Δy and is given by, 

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

(v) (0.999)^1/10 

Consider y = (x)^1/10.

Let x = 1 and Δx = -0.001.

Then, 
∆y = (x + ∆x)^1/10 - (x)^1/10 = (0.999)^1/10 - 1
⇒ (0.999)^1/10 = 1 + Δy 
Now, dy is approximately equal to ∆y and is given by, 

Hence, the approximate value of (0.999)^1/10 is 1 + (-0.0001) = 0.9999.

(vi) (15)^1/4 

Consider y = x^1/4.

Let x = 16 and Δx = -1.
Then,
∆y = (x + ∆x)^1/4 - (x)^1/4 = (15)^1/4 - (16)^1/4 = (15)^1/4 - 2
⇒ (15)^1/4 = 2 + Δy 
Now, dy is approximately equal to Δy and is given by, 

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

(vii) (26)^1/3

Consider y = (x)^1/3.

Let x = 27 and Δx = -1.

Then,
∆y = (x + ∆x)^1/3 - (x)^1/3 = (26)^1/3 - (27)^1/3 = (26)^1/3 - 3
⇒ (26)^1/3 = 3 + Δy 
Now, dy is approximately equal to ∆y and is given by,  

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

(viii) (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 = 4 + Δy 
Now, dy is approximately equal to ∆y and is given by,  

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

(ix) (82)^1/4 

Consider y = x^1/4.

Let x = 81 and Δx = 1.
Then,
∆y = (x + ∆x)^1/4 - (x)^1/4 = (82)^1/4 - (81)^1/4 = (82)^1/4 - 3
⇒ (82)^1/4 =  Δy + 3

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

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

(x) (401)^1/2

Consider y = x^1/2

Let x = 400 and ∆x = 1.
Then,
∆y = √(x + ∆x) - √x = √401 - √400 = √401 - 20 

⇒ √401 = 20 + ∆y
Now, dy is approximately equal to ∆y and is given by,  

Hence, the approximate value of √401 is  20 +0.025 = 20.025 

(xi) (0.0037)^1/2 

Consider y = x^1/2

Let x = 0.0036 and ∆x = 0.0001. 
Then,
∆y = (x + Δx)^1/2 - (x)^1/2 = (0.0037)^1/2 - (0.0036)^1/2 = (0.0037)^1/2 - 0.06

⇒ (0.0037)^1/2 = 0.06 + Δy 
Now, dy is approximately equal to Δy and is given by,  

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

(xii) (26.57)^1/3 

Consider y = x^1/3.

Let x = 27 and Δx = -0.43 

Then,

∆y = (x + Δx)^1/3 - (x)^1/3 = (26.57)^1/3 - (27)^1/3 = (26.57)^1/3 - 3

⇒ (26.57)^1/3 = 3 + Δy 

Now, dy is approximately equal to Δy and is given by,  

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

(xiii) (81.5)^1/4 
Consider y = x^1/4.

Let x = 81 and Δx = 0.5 

Then,

∆y = (x + Δx)^1/4 - (x)^1/4 = (81.5)^1/4 - (81)^1/4 = (81.5)^1/4 - 3

⇒ (81.5)^1/4 = 3 + Δy 

Now, dy is approximately equal to Δy  and is given by, 

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

(xiv) (3.968)^3/2 
Consider y = x^3/2,

Let x = 4 and Δx = - 0.032.

Then,

∆y = (x + Δx)^3/2 - (x)^3/2 = (3.968)^3/2  - (4)^3/2  = (3.968)^3/2  - 8

⇒ (3.968)^3/2  = 8 + Δy 

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

= -0.096

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

(xv) (32.15)^1/5 

Consider y = x^1/5.

Let x = 32 and Δx = 0.15 
Then,

∆y = (x + Δx)^1/5 - (x)^1/5 = (32.15)^1/5 - (32)^1/5 = (32.15)^1/5 - 2

⇒ (32.15)^1/5 = 2 + Δy 
Now, dy is approximately equal to Δy and is given by,

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

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

Solution

Let x = 2 and ∆x = 0.01 Then, we have : 
f(2.01) = f(x + ∆x) = 4(x + ∆x)² + 5(x + Δx) + 2 
Now, Δy = f(x + Δx) - f(x) 
∴ f(x + Δx) = f(x) + Δx
≈f(x) + f '(x).Δx     (as dx = Δx)
⇒ f(2.01) ≈ (4x² + 5x + 2) + (8x + 5)Δx
= [4(2)² + 5(2) + 2] + [8(2) + 5](0.01)  [as x = 2, Δx = 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. 

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

Solution

Let x = 5 and Δx = 0.001. Then, we have :
f(5.001) = f(x + ∆x) = (x + ∆x)³ - 7(x + Δx)² + 15

Now, Δy = f(x + Δx) - f(x) 
∴ f(x + Δx) = f(x) + Δy
≈ f(x) + f '(x).Δx     (as dx = Δx)
⇒ f(5.001) ≈ (x³ - 7x²  + 15) + (3x² - 14x)Δx

= [(5)³ + 7(5)²  + 15] + [3(5)²  - 14(5)](0.001)  [x = 5, Δx = 0.001]
= (125 - 175 + 15) + (75 - 70) (0.001)
= -35 + (5)(0.001)
= -35 + 0.005
= -34.995 

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

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

Solution

The volume of a cube (V) of side x is given by V = x³ . 
∴ (dv/dx) Δx 
= (3x² )Δx 
= (3x² )(0.01x)  [as 1% of x is 0.01x] 
= 0.03x³ 
Hence, the approximate change in the volume of the cube is 0.03x³ m³ . 

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

Solution

The surface area of a cube (S) of side x is given by S = 6x² . 
∴ dS = (dS/dx) Δx
= (12x) Δx
= (12x) (-0.01x)  [as 1% of x is 0.01 x]
= -0.12x² 
Hence, the approximate change in the surface area of the cube is -0.12x² m². 

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.

Solution

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

Then, 

r = 7m and Δr = 0.02 m 
Now; the volume V of the sphere is given by, 

= (4πr² ) ∆r 
= 4π(7)² (0.02)m³ = 3.92π m³ 
Hence, the approximate error in calculating the volume is 3.92 πm³ . 

7. 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.

Solution

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

Then, 

r = 9 m and Δr = 0.03 m 
Now, the surface area of the sphere (S) is given by, 
S = 4πr² 
∴ dS/dr = 8πr 
∴ dS = (dS/dr) Δr
= (8πr)Δr
= 8π(9)(0.03) m² 
= 2.16π m² 
Hence, the approximate error in calculating the surface area is 2.16π m² . 

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

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

f(3.02) = f(x + Δx) = 3(x + Δx)² + 15(x + ∆x) + 5 
Now, ∆y = f(x + ∆x) - f(x) 
⇒ f(x + ∆x) = f(x) + ∆y 
≈ f(x) + f '(x) ∆x      (As dx = ∆x) 
⇒ f(3.02) ≈ (3x² + 15x + 5) + (6x + 15)∆x 
= [3(3)² + 15(3) + 5] + [6(3) + 15](0.02)    [As x = 3, Δx = 0.02] 

= (27 + 45 + 5) + (18 + 15)(0.02)
= 77+ 33× 0.02
= 77 + 0.66

= 77.66

Hence, the approximate value of f(3.02) is 77.66
The correct answer is D.

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 x³ m³ 
B. 0.6 x³ m³
C. 0.09 x³ m³
D. 0.9 x³ m³

Solution