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:
We can use differentials to calculate small changes in the dependent variable of a function corresponding to small changes in the independent variable
(i) √25.3
Consider y = √x
Let x = 25 and Δx = 0.3
Then,
Δy = √x + Δx - √x
= √25.3 - √25
= √25.3 - 5
Δy + 5 = √25.3
Now, dy is approximately equal to Δy and is given by,
dy = (dy/dx) Δx
= 1/2√x (0.3) [∵ y = √x]
= 1/2√25 (0.3)
= 0.03
Hence,
√25.3 = 0.03 + 5
= 5.03
Thus, the approximate value of √25.3 = 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
Δy + 7 = √49.5
Now, dy is approximately equal to Δy and is given by,
dy = (dy/dx) Δx
= 1/2√x (0.5) [∵ y = √x]
= 1/2√49 (0.5)
= 0.035
Hence,
√49.5 = 7 + 0.035
= 7.035
Thus, the approximate value of √49.5 = 7.035.
(iii) √0.6
Consider y = √x
Let x = 1 and Δx = - 0.4
Then,
Δy = √x + Δx - √x
= √0.6 - √1
Δy + 1 = √0.6
Now, dy is approximately equal to Δy and is given by,
dy = (dy/dx) Δx
= 1/2√x (- 0.4) [∵ y = √x]
= 1/2 (- 0.4)
= - 0.2
Hence,
√0.6 = 1 + (- 0.2)
= 1- 0.2
= 0.8
Thus, the approximate value of √0.6 = 0.8.
(iv) (0.009)1/3
Consider y = (x)1/3
Let x = 0.008 and Δx = 0.001
Then,
Δy = (x + Δx)1/3 - (x)1/3
= (0.009)1/3 - (0.008)1/3
Δy + 0.02 = (0.009)1/3
Now, dy is approximately equal to Δy and is given by,
dy = (dy/dx) Δx
= 1/3(x)2/3 Δx [∵ y = (x)1/3]
= [1/(3 x 0.04)] (0.001)
= 0.001/0.12
= 0.008
Hence,
(0.009)1/3 = 0.2 + 0.008
= 0.208
Thus, the approximate value of (0.009)1/3 = 0.208.
(v) (0.999)1/10
Consider y = (x)1/10
Let x = 1 and Δx = - 0.001
Then,
Δy = (x + Δx)1/10 - (x)1/10
= (0.999)1/10 - 1
Δy + 1 = (0.999)1/10
Now, dy is approximately equal to Δy and is given by,
dy = (dy/dx) Δx
= 1/10(x)9/10 Δx [∵ y = (x)1/10]
= 1/10 (- 0.001)
= - 0.0001
Hence,
(0.999)1/10 = 1 + (- 0.0001)
= 0.9999
Thus, the approximate value of (0.999)1/10 = 0.9999.
(vi) (15)1/4
Consider y = (x)1/4
Let x = 16 and Δx = - 1
Then,
Δy = (x + Δx)1/4 - (x)1/4
= (15)1/4 - (16)1/4
= (15)1/4 - 2
Δy + 2 = (15)1/4
Now, dy is approximately equal to Δy and is given by,
dy = (dy/dx) Δx
= 1/4(x)3/4 Δx [∵ y = (x)1/4]
= 1/4(16)1/4 (- 1)
= - 1/(4 x 8)
= - 1/32
= - 0.03125
Hence,
(15)1/4= 2 + (- 0.03125)
= 1.96875
Thus, the approximate value of (15)1/4 = 1.96875.
(vii) (26)1/3
Consider y = (x)1/3
Let x = 27 and Δx = - 1
Then,
Δy = (x + Δx)1/3 - (x)1/3
= (26)1/3 - (27)1/3
= (26)1/3 - 3
Δy + 3 = (26)1/3
Now, dy is approximately equal to Δy and is given by,
dy = (dy/dx) Δx
= 1/3(x)2/3 Δx [∵ y = (x)1/3]
= 1/3(27)2/3 (- 1)
= - 0.0370
Hence,
(26)1/3 = 3 + (- 0.0370)
= 2.9629
Thus, the approximate value of (26)1/3= 2.9629.
(viii) (255)1/4
Consider y = (x)1/4
Let x = 256 and Δx = - 1
Then,
Δy = (x + Δx)1/4 - (x)1/4
= (255)1/4 - (256)1/4
= (255)1/3 - 4
Δy + 4 = (255)1/4
Now, dy is approximately equal to Δy and is given by,
dy = (dy/dx) Δx
= 1/4(x)3/4 Δx [∵ y = (x)1/4]
= 1/4(256)3/4 (- 1)
= - 1/4 x 43
= - 1/32
= - 0.0039
Hence,
(255)1/4 = 4 + (- 0.0039)
= 3.9961
Thus, the approximate value of (255)1/4 = 3.9961.
(ix) (82)1/4
Consider y = (x)1/4
Let x = 81 and Δx = 1
Then,
Δy = (x + Δx)1/4 - (x)1/4
= (82)1/4 - (81)1/4
= (82)1/3 - 3
Δy + 3 = (82)1/4
Now, dy is approximately equal to Δy and is given by,
dy = (dy/dx) Δx
= 1/4(x)3/4 Δx [∵ y = (x)1/4]
= 1/4(81)3/4 (1)
= 1/4 x 33
= 1/108
= 0.009
Hence,
(82)1/4 = 3 + 0.009
= 3.009
Thus, the approximate value of (82)1/4 = 3.009.
(x) (401)1/2
Consider y = (x)1/2
Let x = 400 and Δx = 1
Then,
Δy = (x + Δx)1/2 - (x)1/2
= (401)1/2 - (400)1/2
= (401)1/3 - 20
Δy + 20 = (401)1/2
Now, dy is approximately equal to Δy and is given by,
dy = (dy/dx) Δx
= 1/2√x Δx [∵ y = (x)1/2]
= 1/2 (20) (1)
= 1/40
= 0.025
Hence,
(401)1/2 = 20 + 0.025
= 20.025
Thus, the approximate value of (401)1/2 = 20.025.
(xi) (0.0037)1/2
Consider y = (x)1/2
Let x = 0.0036 and Δx = 0.0001
Then,
Δy = (x + Δx)1/2 - (x)1/2
= (0.0037)1/2 - (0.0036)1/2
= (0.0037)1/2 - 0.06
Δy + 0.06 = (0.0037)1/2
Now, dy is approximately equal to Δy and is given by,
dy = (dy/dx) Δx
= 1/2√x Δx [∵ y = (x)1/2]
= 1/2 (0.006) (1)
= 0.0001/0.12
= 0.0008325
Hence,
(0.0037)1/2 = 0.006 + 0.00083
= 0.06083
Thus, the approximate value of (0.0037)1/2 = 0.06083.
(xii) (26.57)1/3
Consider y = (x)1/3
Let x = 27 and Δx = - 0.43
Then,
Δy = (x + Δx)1/3 - (x)1/3
= (26.57)1/3 - (27)1/3
= (26.57)1/3 - 3
Δy + 3 = (26.57)1/3
Now, dy is approximately equal to Δy and is given by,
dy = (dy/dx) Δx
= 1/3(x)2/3 Δx [∵ y = (x)1/3]
= [1/(3 x 9)] (- 0.43)
= - 0.43/27
= - 0.015
Hence,
(26.57)1/3 = 3 + (- 0.015)
= 2.984
Thus, the approximate value of (26.57)1/3 = 2.984.
(xiii) (81.5)1/4
Consider y = (x)1/4
Let x = 81 and Δx = 0.5
Then,
Δy = (x + Δx)1/4 - (x)1/4
= (81.5)1/4 - (81)1/4
= (81.5)1/4 - 3
Δy + 3 = (81.5)1/4
Now, dy is approximately equal to Δy and is given by,
dy = (dy/dx) Δx
= 1/4(x)3/4 Δx [∵ y = (x)1/4]
= 1/4(81)3/4 (1)
= 1/4(33) (0.5)
= 0.5/108
= 0.0046
Hence,
(81.5)1/4 = 3 + 0.0046
= 3.0046
Thus, the approximate value of (81.5)1/4 = 3.0046.
(xiv) (3.968)3/2
Consider y = (x)3/2
Let x = 4 and Δx = - 0.032
Then,
Δy = (x + Δx)3/2 - (x)3/2
= (3.968)3/2 - (4)3/2
= (3.968)3/2 - 8
Δy + 8 = (3.968)3/2
Now, dy is approximately equal to Δy and is given by,
dy = (dy/dx) Δx
= 3/2(x)1/2 Δx [∵ y = (x)3/2]
= 3/2 (2) (- 0.032)
= - 0.096
Hence,
(3.968)3/2 = 8 + (- 0.096)
= 7.904
Thus, the approximate value of (3.968)3/2 = 7.904.
(xv) (32.15)1/5
Consider y = (x)1/5
Let x = 32 and Δx = 0.15
Then,
Δy = (x + Δx)1/5 - (x)1/5
= (32.15)1/5 - (32)1/5
= (32.15)1/5 - 2
Δy + 2 = (32.15)1/5
Now, dy is approximately equal to Δy and is given by,
dy = (dy/dx) Δx
= 1/5(x)4/5 Δx [∵ y = (x)1/5]
= 1/5(2)4/5 (0.15)
= 1/4(33) (0.5)
= 0.15/80
= 0.00187
Hence,
(32.15)1/5 = 2 + 0.00187
= 2.00187
Thus, the approximate value of (32.15)1/5 = 2.00187
NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.4 Question 1
Using differentials, find the approximate value of each of the following up to 3 places of decimals.
Summary:
Using differentials, the approximate value of each of the following up to 3 places of decimals are 5.03, 7.035, 0.8, 0.208, 0.999, 1.968, 2.962, 3.996, 3.009, 20.025, 0.060,2.984, 3.004,7.904, 2.001 respectively
visual curriculum