Problem 26 - Calculus Single Variable

Question

The volume of a cube is increasing at the rate of \(60 \mathrm{~mm}^{3} / \mathrm{s}\). How fast is the surface area of the cube increasing when each edge is \(20 \mathrm{~mm} ?\)

Step-by-Step Solution

Answer

The surface area is increasing at a rate of 12 mm²/s when each edge is 20 mm.

Step 1: Understand the Relationship

The volume of a cube is given by the formula \( V = s^3 \), where \( s \) is the length of a side of the cube. The surface area of the cube is given by \( A = 6s^2 \). We know the rate of change of volume, \( \frac{dV}{dt} = 60 \text{ mm}^3/\text{s} \), and we need to find the rate of change of the surface area, \( \frac{dA}{dt} \), when \( s = 20 \text{ mm} \).

Step 2: Differentiate Volume with respect to Time

Since \( V = s^3 \), we differentiate both sides with respect to time \( t \), using the chain rule:\[ \frac{dV}{dt} = 3s^2 \frac{ds}{dt} \]We will use this equation to find \( \frac{ds}{dt} \).

Step 3: Differentiate Surface Area with respect to Time

Using the surface area formula \( A = 6s^2 \), differentiate both sides with respect to time \( t \):\[ \frac{dA}{dt} = 12s \frac{ds}{dt} \]This equation will be used to find \( \frac{dA}{dt} \) once we know \( \frac{ds}{dt} \).

Step 4: Solve for \( \frac{ds}{dt} \)

We know \( \frac{dV}{dt} = 60 \text{ mm}^3/\text{s} \) and when \( s = 20 \text{ mm} \), substitute into \( \frac{dV}{dt} = 3s^2 \frac{ds}{dt} \):\[ 60 = 3(20)^2 \frac{ds}{dt} \]\[ 60 = 3 \times 400 \frac{ds}{dt} \]\[ 60 = 1200 \frac{ds}{dt} \]\[ \frac{ds}{dt} = \frac{60}{1200} = \frac{1}{20} \text{ mm/s} \]

Step 5: Solve for \( \frac{dA}{dt} \)

Substitute \( \frac{ds}{dt} = \frac{1}{20} \text{ mm/s} \) and \( s = 20 \text{ mm} \) into \( \frac{dA}{dt} = 12s \frac{ds}{dt} \):\[ \frac{dA}{dt} = 12(20) \left(\frac{1}{20}\right) \]\[ \frac{dA}{dt} = 12 \times 1 = 12 \text{ mm}^2/\text{s} \]

Step 6: Conclusion

The surface area of the cube is increasing at a rate of \( 12 \text{ mm}^2/\text{s} \) when each edge is \( 20 \text{ mm} \).

Key Concepts

Differentiation

Differentiation is a powerful concept in calculus used to determine the rate at which one quantity changes with respect to another. In the context of this problem, differentiation helps us understand how the volume and surface area of a cube change as the size of its sides changes over time. When we differentiate a function, we find its derivative, which gives us the rate of change.
For a cube, the volume and surface area depend on the length of its edges, denoted as \( s \). By differentiating the volume and the surface area formulas with respect to time, we can find how fast these properties are changing as the cube's edges grow.

Chain Rule

The chain rule is a fundamental differentiation method used when dealing with composite functions. In problems like this one, where quantities depend on each other through intermediate variables, the chain rule becomes indispensable.
Since the volume and surface area of a cube are both functions of the edge length \( s \), which in turn changes over time, the chain rule allows us to relate the rate of change of the volume or surface area to the rate of change of \( s \) over time, \( \frac{ds}{dt} \). If we have a function dependent on another, the chain rule states:

• If \( z = f(y) \) and \( y = g(x) \), then \( \frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx} \)

By applying the chain rule in this exercise, we successfully linked changes in volume and surface area to changes in edge length over time.

Volume of a Cube

The volume of a cube is a straightforward concept defined by the formula \( V = s^3 \), where \( s \) stands for the length of a side of the cube. This formula stems from the geometric property that the volume of a shape is the product of its three dimensions. For a cube, all three dimensions are equal to its edge length.
In this exercise, since the volume of the cube is increasing at a known rate, \( \frac{dV}{dt} = 60 \text{ mm}^3/\text{s} \), understanding the formula helps us apply differentiation to determine how fast the side length \( s \) is changing, leading us to \( \frac{ds}{dt} \).

Surface Area of a Cube

The surface area of a cube refers to the total area covered by all six faces of the cube. The formula to find the surface area is \( A = 6s^2 \). Each face of the cube is a square with area \( s^2 \), and there are six faces on the cube, hence the product with 6.
When we are given the rate at which the volume changes, it's crucial to also compute how this affects the surface area. Differentiating \( A = 6s^2 \) with respect to time gives us an equation involving \( \frac{dA}{dt} \), which tells us the rate at which the surface area is increasing.

• By knowing both \( s \) and \( \frac{ds}{dt} \), we can substitute these values to find the sought rate \( \frac{dA}{dt} \).

This approach provides a clear path from known information (rate of volume change) to unknown variables (rate of surface area increase).