An edge of a variable cube is increasing at the rate of $3\;{\text{cm}}/{\text{s}}$. How fast is the volume of the cube increasing when the edge is $10\;{\text{cm}}$ long?

Q: An edge of a variable cube is increasing at the rate of $3\;{\text{cm}}/{\text{s}}$. How fast is the volume of the cube increasing when the edge is $10\;{\text{cm}}$ long?

Explanation: Let the side length of the cube be $x$, and its volume be $V$. \[\] Since $V = x^3$, we differentiate with respect to time $t$:\[\] \[\frac{dV}{dt} = \frac{d}{dt}(x^3) = \frac{d}{dx}(x^3) \cdot \frac{dx}{dt} = 3x^2 \cdot \frac{dx}{dt}\] Given $\frac{dx}{dt} = 3\,\text{cm/s}$, we substitute: \[\frac{dV}{dt} = 3x^2 \cdot 3 = 9x^2\] When $x = 10\,\text{cm}$, \[\]we find: $\frac{dV}{dt} = 9(10)^2 = 900\,\text{cm}^3/\text{s}$ \[\] Hence, the volume of the cube is increasing at the rate of $900\,\text{cm}^3/\text{s}$ when the edge is 10 cm long.

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Explanation: Let the side length of the cube be $x$, and its volume be $V$.  
Since $V = x^3$, we differentiate with respect to time $t$:

$$\frac{dV}{dt} = \frac{d}{dt}(x^3) = \frac{d}{dx}(x^3) \cdot \frac{dx}{dt} = 3x^2 \cdot \frac{dx}{dt}$$

Given $\frac{dx}{dt} = 3\,\text{cm/s}$, we substitute:

$$\frac{dV}{dt} = 3x^2 \cdot 3 = 9x^2$$

When $x = 10\,\text{cm}$, we find:

$\frac{dV}{dt} = 9(10)^2 = 900\,\text{cm}^3/\text{s}$

Hence, the volume of the cube is increasing at the rate of $900\,\text{cm}^3/\text{s}$ when the edge is 10
cm long.

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