Question:
If there is an error of $k\%$ in measuring the edge of a cube, then the percent error in estimating its volume is
If there is an error of $k\%$ in measuring the edge of a cube, then the percent error in estimating its volume is
Updated On: Apr 15, 2024
- $k$
- $3k$
- $\frac{k}{3}$
- None of these
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The Correct Option is B
Solution and Explanation
We know that, the volume V of a cube of side x is given by
$V = x^3$
On differenting w. r. t, we get
$\Rightarrow \frac{dv}{dx} = 3 x^2$
Let the change in $x$ be $\Delta x = K \%$ of $x = \frac{kx}{100}$
Now, the change in volume.
$\Delta V = \left( \frac{dV}{dx}\right) \Delta x = 3x^2 ( \Delta x)$
$= 3x^{2} \left(\frac{kx}{100}\right) = \frac{3x^{2} . k}{100} $
$\therefore$ Approximate change in volume
$=\frac{3kx^{3}}{100} = \frac{3k}{100} . x^{3} $
$= 3k \%$ of original volume
$V = x^3$
On differenting w. r. t, we get
$\Rightarrow \frac{dv}{dx} = 3 x^2$
Let the change in $x$ be $\Delta x = K \%$ of $x = \frac{kx}{100}$
Now, the change in volume.
$\Delta V = \left( \frac{dV}{dx}\right) \Delta x = 3x^2 ( \Delta x)$
$= 3x^{2} \left(\frac{kx}{100}\right) = \frac{3x^{2} . k}{100} $
$\therefore$ Approximate change in volume
$=\frac{3kx^{3}}{100} = \frac{3k}{100} . x^{3} $
$= 3k \%$ of original volume
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Application of Derivatives
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Increasing and Decreasing Function:
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