Problem 13 - Calculus an Applied Approach

Question

Volume All edges of a cube are expanding at a rate of 3 centimeters per second. How fast is the volume changing when each edge is (a) 1 centimeter and (b) 10 centimeters?

Step-by-Step Solution

Answer

When each edge is 1 cm, the volume is changing at 9 cm^3/s and when each edge is 10 cm, the volume is changing at 900 cm^3/s.

Step 1: Define given and required quantities

We are given that \( ds/dt = 3 \) cm/s and we are required to find \( dV/dt \) for two specific situations: when s = 1 cm and when s = 10 cm.

Step 2: Differentiate the volume equation

We start with the equation \( V = s^3 \) and differentiate with respect to time: \( dV/dt = 3s^2 \cdot ds/dt \). This equation relates the rate of change of volume with the rate of change of edge length.

Step 3: Substitute \( ds/dt \) and s values for the case \( s = 1 cm \)

If we put \( ds/dt = 3 \) cm/s and s = 1 cm in the derived equation, we get: \( dV/dt = 3 \times (1 cm)^2 \times 3 cm/s = 9 cm^3/s \).

Step 4: Substitute \( ds/dt \) and s values for the case \( s = 10 cm \)

If we put \( ds/dt = 3 \) cm/s and s = 10 cm in the derived equation, we get: \( dV/dt = 3 \times (10 cm)^2 \times 3 cm/s = 900 cm^3/s \).

Key Concepts

Differentiation

In the world of calculus, differentiation is an essential concept that helps us understand how a function changes at any given point. When we differentiate a function, we're essentially finding its derivative, which tells us the rate at which one variable changes with respect to another.

In the context of related rates problems, like the one we have with the expanding cube, the function we're interested in is the volume of the cube. By differentiating the equation of the cube's volume, we determine how this volume changes as its edges expand.

This process is important because it allows us to link multiple changing quantities—like the edge length and the volume—via differentiation, thus deciphering their relationship.

Volume of a Cube

The volume of a cube is a straightforward concept once you understand the basic geometry involved. A cube is a three-dimensional shape with six equal square faces. To find its volume, we multiply the length of one edge by itself twice:

• This gives us the volume formula: \( V = s^3 \), where \( s \) is the length of one side.
  
• The formula captures the notion of three-dimensionality, as you multiply the side length three times.
  

However, in real-world scenarios, dimensions of objects can change over time. Understanding the basic formula helps us progress further and lead into how these changes affect related quantities, like the volume.

In our problem, as each edge of the cube grows, the volume naturally increases. Our task then becomes finding out precisely how that volume changes with the edges getting longer.

Rate of Change of Volume

In problems involving related rates, determining how fast a quantity is changing often comes down to utilizing the rate of change of volume. For an expanding cube, the rate at which the volume changes is directly tied to how fast its edges expand.

This involves differentiating the volume formula: \( V = s^3 \), giving us the derivative \( dV/dt = 3s^2 \cdot ds/dt \).

• \( dV/dt \) represents the rate of change of volume with respect to time.
  
• \( ds/dt \) stands for the rate at which the edge length changes per unit time.
  

Given that the edge expansion rate is constant (3 cm/s in our case), this derivative helps us calculate how the volume changes at specific instances — when the edge length is 1 cm and when it's 10 cm.

Understanding these rates provides a way to predict how the cube's volume reacts to the growth of its edges, equipping us with a quantitative grasp on the situation.