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Find the Approximate Change in the Value V Of a Cube of Side X Metres Caused by Increasing the Side by 1% ?

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Question

Find the approximate change in the value V of a cube of side x metres caused by increasing the side by 1% ?

Sum
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Solution

\[\text { Volume of the cube,} V = x^3 \]

\[\text { We have }\]

\[ ∆ x = 0 . 01x\]

\[\frac{dV}{dx} = 3 x^2 \]

\[ \Rightarrow ∆ V = dV = \frac{dV}{dx}dx = 3 x^2 \times 0 . 01x = 0 . 03 x^3 \]

\[\text { Hence, the approximate change in the value V of the cube is } 0 . 03 x^3 m^3 . \]

\[\text{Disclaimer: This solution has been created according to the question given in the book . However, the solution in the book is incorrect } . \]

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Chapter 13: Differentials, Errors and Approximations - Exercise 14.1 [Page 10]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 13 Differentials, Errors and Approximations
Exercise 14.1 | Q 16 | Page 10
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