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If there is an error of k% in measuring the edge of a cube , then the percent error in estimating its volume is 

Answer:

Explanation:

Volume V of a cube of side x is given by

V=$x^{3}$

 $\Rightarrow$    $\frac{dv}{dx}=3x^{2}$

Let the change in x be $\triangle$x=K% of 

$x=\frac{kx}{100}$

Now, the change in volume 

$\triangle V=\left(\frac{dV}{dx}\right)\triangle x=3x^{3}(\triangle x)$

$= 3x^{2}\left(\frac{kx}{100}\right)=\frac{3x^{3}k}{100}$

$\therefore$  Aprroximate change in volume

$=\frac{3kx^{3}}{100}=\frac{3k}{100}x^{3}$

=3K% of original volume