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Mathmatics - Chapter No. : 6

NCERTworld ncert textbook mathmatics class 12

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 = x1/3
Let x = 0.008 and ∆x = 0.001
Then, ∆y = (x + ∆x)1/3 – x1/3
                 = (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
     = 1/3x2/3 * (0.001)                            [Since y = x1/3]
     = 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 = x1/10
Let x = 1 and ∆x = -0.001
Then, ∆y = (x + ∆x)1/10 – x1/10
                 = (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
     = 1/10x9/10 * (-0.001)                            [Since y = x1/10]
     = (-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 = x1/4
Let x = 16 and ∆x = -1
Then, ∆y = (x + ∆x)1/4 – x1/4
                 = (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
     = 1/4x3/4 * (-1)                            [Since y = x1/4]
     = {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 = x1/3
Let x = 27 and ∆x = -1
Then, ∆y = (x + ∆x)1/3 – x1/3
                 = (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
     = 1/3x2/3 * (-1)                            [Since y = x1/3]
     = {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 = x1/4
Let x = 256 and ∆x = -1
Then, ∆y = (x + ∆x)1/4 – x1/4
                 = (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
     = 1/4x3/4 * (-1)                            [Since y = x1/4]
     = {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 = x1/4
Let x = 81 and ∆x = 1
Then, ∆y = (x + ∆x)1/4 – x1/4
                 = (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
     = 1/4x3/4 * (1)                            [Since y = x1/4]
     = {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 = x1/2
Let x = 400 and ∆x = 1
Then, ∆y = (x + ∆x)1/2 – x1/2
                 = (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
     = 1/2x1/2 * (1)                            [Since y = x1/2]
     = 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 = x1/2
Let x = 0.0036 and ∆x = 0.0001
Then, ∆y = (x + ∆x)1/2 – x1/2
                 = (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
     = 1/2x1/2 * (0.0001)                            [Since y = x1/2]
     = 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 = x1/3
Let x = 27 and ∆x = -0.43
Then, ∆y = (x + ∆x)1/3 – x1/3
                 = (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
     = 1/3x2/3 * (-0.43)                            [Since y = x1/3]
     = {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 = x1/4
Let x = 81 and ∆x = 0.5
Then, ∆y = (x + ∆x)1/4 – x1/4
                 = (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
     = 1/4x3/4 * (0.5)                            [Since y = x1/4]
     = 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 = x3/2
Let x = 4 and ∆x = -0.032
Then, ∆y = (x + ∆x)3/2 – x3/2
                 = (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)x1/2 * (-0.032)                            [Since y = x3/2]
     = (3/2) * 41/2 * (-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 = x1/5
Let x = 32 and ∆x = 0.15
Then, ∆y = (x + ∆x)1/5 – x1/5
                 = (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
     = 1/5x4/5 * (0.15)                            [Since y = x1/5]
     = 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) = 4x2 + 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) ≈ (4x2 + 5x + 2) + (8x + 5) ∆x
                        = (4 * 22 + 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) = x3 - 7x2 + 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) ≈ (x3 - 7x2 + 15) + (3x2 – 14x) ∆x
                           = (53 – 7 * 52 + 15) + (3 * 52 – 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 = x3
So, dV = (dV/dx) * Δx
            = (3x2) * Δx
            = (3x2) * 0.01x    
            = 0.03x3               [Since 1% of x = 0.01x]
Hence, the approximate change in the volume of the cube is 0.03x3 m3


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 = 6x2
So, dS = (dS/dx) * Δx
            = (12x) * Δx
            = (12x) * 0.01x    
            = 0.12x2               [Since 1% of x = 0.01x]
Hence, the approximate change in the surface area of the cube is 0.12x2 m3


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 = 4πr3/3
dV/dr = 4πr2
So, dV = (dV/dr) * ∆r
             = (4πr2) * ∆r
             = (4π * 72) * 0.02
              = 3.92π
Hence, the approximate error in calculating the volume is 3.92 π m3


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 = 4πr2
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 m2


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

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) ≈ (3x2 + 15x + 5) + (6x + 15) ∆x
                        = (3 * 32 + 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 x3 m3                             B. 0.6 x3 m3                          C. 0.09 x3 m3                     D. 0.9 x3 m3

Answer :
The volume of a cube (V) of side x is given by V = x3
So, dV = (dV/dx) * ∆x
            = (3x2) * ∆x
            = (3x2) * 0.03x                            [Since 3% of x is 0.03x] 
            = 0.09x2
Hence, the approximate change in the volume of the cube is 0.09x3 m3
So, the correct answer is option C.


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