PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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Solution for Using Differential, Find the Approximate Value of the 25 1 3 ? - PUC Karnataka Science Class 12 - Mathematics

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Question

Using differential, find the approximate value of the \[{25}^\frac{1}{3}\] ?

Solution

\[\text { Consider the function  }y = f\left( x \right) = \left( x \right)^\frac{1}{3} . \]

\[\text {Let }: \]

\[ x = 27\]

\[ x + ∆ x = 25\]

\[\text { Then,} \]

\[ \bigtriangleup x = - 2\]

\[\text { For } x = 27, \]

\[ y = \left( 27 \right)^\frac{1}{3} = 3\]

\[\text { Let }: \]

\[ dx = ∆ x = - 2\]

\[\text { Now }, y = \left( x \right)^\frac{1}{3} \]

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{3 \left( x \right)^\frac{2}{3}}\]

\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = 27} = \frac{1}{27}\]

\[ \therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{27} \times \left( - 2 \right) = - 0 . 07407\]

\[ \Rightarrow ∆ y = - 0 . 07407\]

\[ \therefore \left( 25 \right)^\frac{1}{3} = y + ∆ y = 2 . 9259\]

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Solution Using Differential, Find the Approximate Value of the 25 1 3 ? Concept: Approximations.
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