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Solution for Find the Approximate Change in the Value V Of a Cube of Side X Metres Caused by Increasing the Side by 1% ? - CBSE (Commerce) Class 12 - Mathematics

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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% ?

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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APPEARS IN

 RD Sharma Mathematics for Class 12 by R D Sharma (Set of 2 Volume) (2018-19 Session) (with solutions)
Chapter 14: Differentials, Errors and Approximations
Q: 16 | Page no. 10

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Solution for question: Find the Approximate Change in the Value V Of a Cube of Side X Metres Caused by Increasing the Side by 1% ? concept: Approximations. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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