91Ó°ÊÓ

Differentiation is a powerful mathematical tool used to determine how a function changes with respect to one of its variables. In simpler terms, it's like looking at the rate at which something is changing. In our exercise, the length of the edge of the cube is decreasing, and we want to see how this affects the surface area and volume of the cube.

To find these rates of change, we need to use differentiation. We take the formula for surface area and volume and differentiate them with respect to time. This process gives us a derivative function, which provides the rate of change for each of these aspects. For example:

• When differentiating the surface area formula, we found that: \( \frac{dS}{dt} = 12x\frac{dx}{dt} \)


• For volume, differentiating gave us: \( \frac{dV}{dt} = 3x^2\frac{dx}{dt} \)

These derivatives help us understand how quickly the surface area and volume of the cube are changing at the moment we are interested in, specifically when the side is 3 meters long.

Understanding the surface area of a cube is quite straightforward. A cube has six identical square faces, which means that if you know the length of one edge, you can easily find the total surface area by using the formula:

Surface area \( S = 6x^2 \)

Here, \( x \) represents the length of each edge of the cube. For example, if \( x = 3 \) meters, then the surface area can be calculated as \( 6 \times 3^2 = 54 \; \text{m}^2 \).

Now, when the cube shrinks, the surface area changes. Using differentiation, we can determine how rapid is this change. By plugging the values into the derivative \( \frac{dS}{dt} = 12x\frac{dx}{dt} \), we substitute \( x = 3 \text{ m} \) and the rate of change of the edge \( \frac{dx}{dt} = -5 \text{ m/min} \). This calculation tells us the surface area decreases at \(-180\; \text{m}^2/\text{min} \). In essence, it rapidly reduces as the edges shorten.

The volume of a cube gives us a sense of how much space it occupies. Knowing the edge length allows us to calculate this using a simple cubic formula:

Volume \( V = x^3 \)

Here, each edge is multiplied together thrice since a cube is consistent on all sides. Suppose \( x = 3 \) meters, so the volume would be \( 3^3 = 27 \; \text{m}^3 \). It's the space within the boundaries of the cube.

When the cube starts to shrink, we see a change in volume. By differentiating the volume formula with respect to time using \( \frac{dV}{dt} = 3x^2\frac{dx}{dt} \), and then substituting \( x = 3 \text{ m} \) and the rate \( \frac{dx}{dt} = -5 \text{ m/min} \), the result is \(-135\; \text{m}^3/\text{min} \).

This means that the volume is decreasing rapidly as the edges get shorter, highlighting the dynamic nature of shapes in response to changes in their dimensions.