Problem 11 - University Calculus: Early Transcendentals

Question

If the original 24 m edge length \(x\) of a cube decreases at the rate of \(5 \mathrm{m} / \mathrm{min},\) when \(x=3 \mathrm{m}\) at what rate does the cube's a. surface area change? b. volume change?

Step-by-Step Solution

Answer

a. The surface area decreases at a rate of -180 m²/min; b. The volume decreases at a rate of -135 m³/min.

Step 1: Understand the Problem

We are given a cube whose edge length decreases over time. The original edge length is 24 m, but it decreases at a rate of 5 m/min. We need to find the rate of change in the cube's surface area and volume when the edge length is 3 m.

Step 2: Define the Variables and Formulas

Let \( x \) be the edge length of the cube, which changes over time. The rate of change of the edge length is \( \frac{dx}{dt} = -5 \, \text{m/min} \). The surface area \( A \) of a cube is \( A = 6x^2 \), and the volume \( V \) is \( V = x^3 \).

Step 3: Differentiate the Surface Area with Respect to Time

The rate of change of the surface area is given by the derivative of \( A \) with respect to \( t \):\[ \frac{dA}{dt} = \frac{d}{dt}(6x^2) = 12x\frac{dx}{dt} \] Substituting \( x = 3 \, \text{m} \) and \( \frac{dx}{dt} = -5 \, \text{m/min} \), we get:\[ \frac{dA}{dt} = 12(3)(-5) = -180 \, \text{m}^2/\text{min} \]

Step 4: Differentiate the Volume with Respect to Time

The rate of change of the volume is given by the derivative of \( V \) with respect to \( t \):\[ \frac{dV}{dt} = \frac{d}{dt}(x^3) = 3x^2\frac{dx}{dt} \] Substituting \( x = 3 \, \text{m} \) and \( \frac{dx}{dt} = -5 \, \text{m/min} \), we get:\[ \frac{dV}{dt} = 3(3^2)(-5) = 3(9)(-5) = -135 \, \text{m}^3/\text{min} \]

Key Concepts

Derivatives

Derivatives are fundamental in understanding how functions change. They come into play any time we are dealing with quantities that depend on each other. In related rates problems like this one, derivatives help us understand the rate at which one quantity, such as the length of a cube's edge, impacts other quantities, like surface area and volume.

Here's how it works in this context:

• We know the edge length of a cube decreases at a constant rate: \(\frac{dx}{dt} = -5 \, \text{m/min}\).
• To find how this affects the surface area or volume, we find the derivative of those quantities with respect to time.

Differentiating the expressions for surface area and volume with respect to time lets us understand how rapid these measures change as the edge length shrinks. These are known as related rates problems, which closely link derivatives to physical changes.

Surface Area of a Cube

The surface area of a cube is found by calculating the area of all six faces. Since each face of the cube is a square with sides of length \(x\), we can express the surface area \(A\) mathematically as \(A = 6x^2\).

As the edge length of a cube changes, so does its surface area. If the edge length decreases, you would expect the surface area to decrease as well. To find out exactly how quick the surface area changes (its rate of change), we differentiate the surface area formula:

• The derivative \(\frac{dA}{dt} = 12x\frac{dx}{dt}\) shows us how the rate of change in edge length \(\frac{dx}{dt}\) affects the surface area.
• Plugging in values \(x = 3 \, \text{m}\) and \(\frac{dx}{dt} = -5 \, \text{m/min}\), the surface area decreases at \(-180 \, \text{m}^2/\text{min}\).

This tells us that as the cube's edge shrinks, the surface gets smaller quite rapidly.

Volume of a Cube

The volume of a cube is the space contained within its boundaries. This can be calculated using the formula \(V = x^3\), where \(x\) is the side length of the cube.

Just like with surface area, the volume will also decrease if the side length decreases. The change in volume over time can be found by differentiating the volume with respect to time:

• The derivative \(\frac{dV}{dt} = 3x^2\frac{dx}{dt}\) points to how the change in edge length directly impacts the volume.
• By substituting \(x = 3 \, \text{m}\) and \(\frac{dx}{dt} = -5 \, \text{m/min}\), we calculate that the volume decreases at a rate of \(-135 \, \text{m}^3/\text{min}\).

This indicates that as the cube's side shrinks, the loss in volume is substantial. Understanding these rates bridges the gap between abstract mathematics and tangible real-world phenomena.