The approximate change in the volume V of a c | vedclass

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Question

(A) Let the side of the cube be $x$ meters. The volume of the cube is given by $V = x^{3}$. Differentiating with respect to $x$,we get $\frac{dV}{dx}

Vedclass · Accepted Answer

$0.09$

Vedclass · Answer

(A) Let the side of the cube be $x$ meters. The volume of the cube is given by $V = x^{3}$. Differentiating with respect to $x$,we get $\frac{dV}{dx} = 3x^{2}$,which implies $dV = 3x^{2} dx$. Given that the side increases by $3\%$,we have $\frac{dx}{x} 	imes 100 = 3$,which means $\frac{dx}{x} = 0.03$,or $dx = 0.03x$. Substituting $dx$ into the expression for $dV$: $dV = 3x^{2} (0.03x) = 0.09x^{3} 	ext{ m}^{3}$. Thus,the approximate change in volume is $0.09x^{3} 	ext{ m}^{3}$.

The approximate change in the volume $V$ of a cube of side $x$ meters caused by increasing the side by $3\%$ is: (in $x^{3} \text{ m}^{3}$)

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