Pregunta
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The side of a cube increases at a rate of \( \frac{1}{2} \mathrm{~m} / \mathrm{sec} \). Find the rate at which the volume of the cube increases when the side of the cube is 4 m .

Ask by Reeves Craig. in the United States
Mar 17,2025

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The volume of the cube is increasing at a rate of 24 m³/sec.

Solución

Given the volume of a cube as \[ V = s^3, \] where \( s \) is the side length, differentiate both sides with respect to time \( t \) to obtain \[ \frac{dV}{dt} = 3s^2 \frac{ds}{dt}. \] We are given that \[ \frac{ds}{dt} = \frac{1}{2} \text{ m/sec} \] and \( s = 4 \text{ m} \). Substitute these into the differentiated equation: \[ \frac{dV}{dt} = 3(4)^2 \left(\frac{1}{2}\right) = 3 \times 16 \times \frac{1}{2}. \] Simplify the expression: \[ \frac{dV}{dt} = 48 \times \frac{1}{2} = 24. \] Thus, the volume of the cube is increasing at a rate of \[ \boxed{24 \text{ m}^3/\text{sec}}. \]

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The Deep Dive

To find the rate at which the volume of the cube increases, we start by recalling the formula for the volume \( V \) of a cube, which is given by \( V = s^3 \), where \( s \) is the side length. To find the rate of change of the volume with respect to time, we can use the chain rule: \[ \frac{dV}{dt} = \frac{dV}{ds} \cdot \frac{ds}{dt} \] Calculating \( \frac{dV}{ds} \): \[ \frac{dV}{ds} = 3s^2 \] Given that \( \frac{ds}{dt} = \frac{1}{2} \, \text{m/s} \) and \( s = 4 \, \text{m} \): \[ \frac{dV}{dt} = 3(4^2) \cdot \frac{1}{2} = 3 \cdot 16 \cdot \frac{1}{2} = 24 \, \text{m}^3/\text{s} \] Thus, the volume of the cube increases at a rate of \( 24 \, \text{m}^3/\text{s} \) when the side of the cube is 4 m. --- Did you know that cubes have been a staple in architecture and construction for centuries? From the ancient pyramids of Egypt to modern minimalist designs, the cube symbolizes stability and integrity. The geometric properties of cubes are essential not just in art and architecture but also in computer graphics and modeling in virtual environments! In the real world, understanding how shapes change can be critical in fields like engineering and design. For instance, when designing packaging, knowing the rate of volume change helps manufacturers optimize material usage and shipping logistics. This knowledge ensures that products are delivered efficiently while minimizing waste—talk about a win-win!

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