The volume of a cube is increasing at a rate | vedclass

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(A) Let $x$ be the length of an edge of the cube. Let $V$ be the volume and $S$ be the surface area of the cube. Then $V = x^3$ and $S = 6x^2$, where

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$3.6$

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(A) Let $x$ be the length of an edge of the cube. Let $V$ be the volume and $S$ be the surface area of the cube. Then $V = x^3$ and $S = 6x^2$, where $x$ is a function of time $t$.Given that the rate of change of volume is $\frac{dV}{dt} = 9 	ext{ cm}^3/	ext{s}$.Using the chain rule, $\frac{dV}{dt} = \frac{d}{dt}(x^3) = 3x^2 \cdot \frac{dx}{dt}$.Substituting the given value, $9 = 3x^2 \cdot \frac{dx}{dt}$, which gives $\frac{dx}{dt} = \frac{3}{x^2}$.Now, we need to find the rate of change of the surface area, $\frac{dS}{dt}$.Using the chain rule, $\frac{dS}{dt} = \frac{d}{dt}(6x^2) = 12x \cdot \frac{dx}{dt}$.Substituting $\frac{dx}{dt} = \frac{3}{x^2}$ into the equation, we get $\frac{dS}{dt} = 12x \cdot \left(\frac{3}{x^2}ight) = \frac{36}{x}$.When the edge length $x = 10 	ext{ cm}$, the rate of change of the surface area is $\frac{dS}{dt} = \frac{36}{10} = 3.6 	ext{ cm}^2/	ext{s}$.

The volume of a cube is increasing at a rate of $9 \text{ cm}^3/\text{s}$. How fast is the surface area increasing when the length of an edge is $10 \text{ cm}$ (in $\text{ cm}^2/\text{s}$)?

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