The surface area of a cube increases at a rat | vedclass

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(D) Let $x$ be the length of the edge of the cube and $A$ be its surface area. The surface area of a cube is given by $A = 6x^2$. Given that $\frac{dA

Vedclass · Accepted Answer

$45$

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(D) Let $x$ be the length of the edge of the cube and $A$ be its surface area. The surface area of a cube is given by $A = 6x^2$. Given that $\frac{dA}{dt} = 2 \ cm^2/sec$. Differentiating $A$ with respect to $t$,we get $\frac{dA}{dt} = 12x \frac{dx}{dt}$. Substituting the given values: $2 = 12 	imes 90 	imes \frac{dx}{dt}$. Thus,$\frac{dx}{dt} = \frac{2}{12 	imes 90} = \frac{1}{540} \ cm/sec$. The volume of the cube is $V = x^3$. Differentiating $V$ with respect to $t$,we get $\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$. Substituting $x = 90$ and $\frac{dx}{dt} = \frac{1}{540}$: $\frac{dV}{dt} = 3 	imes (90)^2 	imes \frac{1}{540} = 3 	imes 8100 	imes \frac{1}{540} = \frac{24300}{540} = 45 \ cm^3/sec$.

The surface area of a cube increases at a rate of $2 \ cm^2/sec$. The rate at which its volume increases when the length of its edge is $90 \ cm$ is ..... $cm^3/sec$.

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