Using differentials, find the approximate value of each of the following
(i) (17/81)1/4 (ii) (33)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) (17/81)1/4
Let us consider y = (x)1/4
Assume x = 16/81 and Δx = 1/81
Then,
Δy = (x + Δx)1/4 - (x)1/4
= (17/81)1/4 + (16/81)1/4
= (17/81)1/4 - 2/3
2/3 + Δy = (17/81)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(16/81)3/4 (1/81)
= 27/(4 x 8) x (1/81)
= 1/(32 x 3)
= 1/96
= 0.010
Hence,
(17/81)1/4 = 2/3 + 0.010
= 0.667 + 0.010
= 0.677
Thus, the approximate value of (17/81)1/4 is 0.677.
(ii) (33)1/5
Consider, y = (x)1/5
Let, x = 32 and Δx = 1
Then,
Δy = (x + Δx)1/5 - (x)1/5
= (33)1/5 + (32)1/5
= (33)1/5 - 1/2
1/2 + Δy = (33)1/5
Now, dy is approximately equal to Δy and is given by,
dy = (dy/dx)Δx
= - 1/5(x)6/5 Δx
[since y = (x)1/5]
= - 1/5(2)6/5 (1)
= - 1/320
= - 0.003
Hence,
(33)1/5 = 1/2 + (- 0.003)
= 0.5 - 0.003
= 0.497
Thus, the approximate value of (33)1/5 is 0.497
NCERT Solutions Class 12 Maths - Chapter 6 Exercise ME Question 1
Using differentials, find the approximate value of each of the following (i) (17/81)1/4 (ii) (33)1/5
Summary:
Using differentials, the approximate value of (i) (17/81)1/4 is 0.677 and the approximate value of (ii) (33)1/5 is 0.497
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