Question

All edges of a cube are expanding at a rate of 6 centimeters per second. How fast is the volume changing when each edge is (a) 2 centimeters (b) 10 centimeters?

Short answer

(a) The volume is increasing at a rate of 72 cubic centimeters per second when the edge length is 2 cm. (b) The volume is increasing at a rate of 1800 cubic centimeters per second when the edge length is 10 cm.

Step-by-step solution

Identifying Variables

The edge of the cube is given as \(a\), and it is expanding at a rate of 6 cm/sec, so \(da/dt = 6\) cm/sec. The volume of the cube \(V = a^3\). The task is to find \(dV/dt\), the rate at which the volume is changing with respect to time.

Derive the volume with respect to time

Take the derivative of both sides of the volume equation with respect to time \(t\) to obtain: \(dV/dt = 3 \cdot a^2 \cdot da/dt\).

Substitute Known Variables - (a) for edge = 2 cm

Now, substitute \(a = 2\) cm and \(da/dt = 6\) cm/sec into the derived equation to find \(dV/dt\) when the edge is 2 cm. This yields: \(dV/dt = 3 \cdot (2)^2 \cdot 6 = 72\) cubic centimeters per second.

Substitute Known Variables - (b) for edge = 10 cm

Next, substitute \(a = 10\) cm and \(da/dt = 6\) cm/sec into the derived equation to find \(dV/dt\) when the edge is 10 cm. This yields: \(dV/dt = 3 \cdot (10)^2 \cdot 6 = 1800\) cubic centimeters per second.

Key concepts

Related Rates

Related rates are a fascinating application of derivatives in calculus, particularly when dealing with real-world scenarios where multiple variables change over time. Students often encounter problems that require understanding how the rate of change of one quantity is related to the rate of change of another. For example, in our cube problem, the edge length and volume change over time, and these changes are interdependent.

To tackle such problems, here's a step-by-step approach: First, we identify all relevant variable rates of change, such as the rate at which the cube's edges are expanding. Then, we express related quantities—like the volume—in terms of those variables. Finally, we apply the derivative to relate the rates of change of these quantities. This process transforms an abstract mathematical concept into a tool for solving practical problems, ultimately revealing a dynamic connection between varying elements of a system.

Volume Change

Volume change in geometric shapes is a classic study area in calculus, often represented through problems involving cubes, spheres, or cylinders. In the context of a cube, the volume is determined by raising the edge length to the third power.

When considering how the volume changes over time, we step into the world of related rates once again. As the edges of the cube expand, the volume increases at a rate which we can calculate by deriving the volume expression with respect to time. This calculated rate of volume change is not constant—it depends on the cube's edge length at a given moment, demonstrating a non-linear relationship. Understanding this concept is crucial for students, not just for solving textbook problems, but for grasping the underlying principles that govern changes in physical dimensions.

Derivative Applications

Derivatives are the building blocks of calculus and have numerous applications across various fields. In our cube example, we applied a derivative to find out how fast the volume changes as the edges extend. This is one instance where we can see the practical use of derivatives—to model rates of change.

Beyond geometry, derivatives help in understanding motion, growth and decay, optimization problems, and much more. They give us the slope at any point along a curve, which corresponds to the instantaneous rate of change. This concept is essential for students to learn because it empowers them to dissect dynamic processes and predict future outcomes based on current trends. Through exercises that emphasise real-world applications, students can appreciate the pervasiveness of derivatives in everyday problem-solving.