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Using differentials, find the sum of digits approximate valu

Question

Using differentials, find the sum of digits approximate value of the following up to $3$ places of decimal.
$(26)^{\frac{1}{3}}$

2.963

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Using differentials, find the sum of digits approximate value of the following up to

3

places of decimal.

(26)^{\frac{1}{3}}

6 mins ago

Discuss this question LIVE

6 mins ago

Text solutionVerified

Consider

y = x^{\frac{1}{3}}

Let

x = 27

and

Δ x = - 1.

Then,

Δ y = x + Δ x^{\frac{1}{3}} - x^{\frac{1}{3}} = (26)^{\frac{1}{3}} - (27)^{\frac{1}{3}} = (26)^{\frac{1}{3}} - 3

\Rightarrow (26)^{\frac{1}{3}} -) = 3 + Δ y

Now,

d y

is approximately equal to

Δ y

and is given by,

d y = (\frac{d y}{d x}) Δ x = \frac{1}{3 ( x ) ^{\frac{2}{3}}} (Δ x)

[as

y = x^{\frac{1}{3}}

]

= \frac{1}{3 ( 27 ) ^{2/3}} (- 1) = - \frac{1}{27} = - .037

Hence, the approximate value of

(26)^{\frac{1}{3}}

3 + (- 0.0370) = 2.9629 \approx 2.963

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Question Text	Using differentials, find the sum of digits approximate value of the following up to $3$ places of decimal. $(26)^{\frac{1}{3}}$
Answer Type	Text solution:1
Upvotes	150