Sphere Volume Increase Rate When Radius Is 3 Cm

In the realm of calculus, understanding rates of change is paramount. One classic problem involves exploring how the volume of a sphere changes as its radius expands. This article delves into a specific scenario: A sphere's radius increases at a rate of 2 cm/sec. Our primary goal is to determine the rate at which the volume increases precisely when the radius reaches 3 cm. This exploration will provide a comprehensive understanding of related rates and their applications in geometry and physics.

Our central problem is: Given that the radius of a sphere is increasing at a rate of 2 cm/sec, how fast does the volume increase when the radius is 3 cm? To tackle this, we'll use the concept of related rates, which is a crucial application of differential calculus. Related rates problems involve finding the rate at which one quantity changes by relating it to other quantities whose rates of change are known. In this case, we relate the rate of change of the sphere's volume to the rate of change of its radius. Understanding the fundamental principles of calculus, such as derivatives and their applications in measuring rates of change, is essential for solving this problem. The formula for the volume of a sphere, V = (4/3)πr³, forms the foundation of our calculations, linking the volume (V) to the radius (r). By differentiating this formula with respect to time, we can relate the rates of change of these two quantities, allowing us to find the specific rate of volume increase when the radius is 3 cm.

To effectively solve this problem, we begin by identifying the given information and what we need to find. We are given that the rate of change of the radius, denoted as dr/dt, is 2 cm/sec. This means the sphere's radius is expanding at a constant rate of 2 centimeters every second. Our goal is to find the rate of change of the volume, denoted as dV/dt, when the radius, r, is 3 cm. This represents how quickly the sphere's volume is increasing at the instant the radius reaches 3 centimeters. The next step is to establish the relationship between the volume and the radius of a sphere. The formula for the volume of a sphere is V = (4/3)πr³. This equation is crucial as it mathematically connects the volume (V) and the radius (r), allowing us to use calculus to relate their rates of change. With this foundation, we can proceed to the next step, which involves differentiating the volume formula with respect to time.

To relate the rates of change of the volume and radius, we differentiate the volume formula with respect to time (t). Starting with V = (4/3)πr³, we apply the chain rule, a fundamental concept in calculus, to differentiate both sides of the equation. The chain rule is essential here because both V and r are functions of time. Differentiating V with respect to t gives us dV/dt, which represents the rate of change of the volume. Differentiating (4/3)πr³ with respect to t requires us to first differentiate with respect to r and then multiply by dr/dt. The derivative of r³ with respect to r is 3r², so the derivative of (4/3)πr³ with respect to r is (4/3)π * 3r² = 4πr². Applying the chain rule, we get: dV/dt = 4πr² (dr/dt). This equation now directly relates the rate of change of the volume (dV/dt) to the radius (r) and the rate of change of the radius (dr/dt). This relationship is the key to solving our problem, as it allows us to plug in the given values and find the unknown rate of change of the volume.

With the relationship dV/dt = 4πr² (dr/dt) established, we can now substitute the given values to find the rate of change of the volume. We know that dr/dt = 2 cm/sec, which is the rate at which the radius is increasing. We also know that we want to find dV/dt when the radius, r, is 3 cm. Plugging these values into the equation, we get: dV/dt = 4π(3 cm)² (2 cm/sec). This substitution allows us to calculate the specific rate of volume increase at the instant when the radius is 3 cm. By performing the calculations, we will arrive at the numerical value of dV/dt, which represents the answer to our problem. This step is crucial as it bridges the theoretical relationship between the rates of change and the specific scenario described in the problem statement.

Having plugged in the given values, we now perform the calculations to find dV/dt. Starting with dV/dt = 4π(3 cm)² (2 cm/sec), we first square the radius: (3 cm)² = 9 cm². Then, we multiply this by 4π: 4π * 9 cm² = 36π cm². Finally, we multiply by the rate of change of the radius, 2 cm/sec: 36π cm² * 2 cm/sec = 72π cm³/sec. Therefore, the rate of change of the volume, dV/dt, is 72π cm³/sec. This result tells us that when the radius of the sphere is 3 cm, the volume is increasing at a rate of 72π cubic centimeters per second. This precise calculation provides a clear understanding of how the volume of the sphere changes as its radius expands at the given rate. Understanding this calculation helps us to visualize the dynamic relationship between the radius and volume of the sphere and how their rates of change are interconnected.

The result, dV/dt = 72π cm³/sec, provides a clear and specific answer to our problem. It tells us that at the instant when the sphere's radius is 3 cm, the volume is increasing at a rate of 72π cubic centimeters per second. This rate of increase is not constant; it is specific to the moment when the radius is exactly 3 cm. The units, cm³/sec, are crucial for understanding the result. They indicate that we are measuring the change in volume (cm³) per unit of time (sec). To better grasp the magnitude of this rate, we can approximate 72π. Since π is approximately 3.14159, 72π is roughly 72 * 3.14159, which is about 226.19. This means that the volume is increasing at approximately 226.19 cubic centimeters per second when the radius is 3 cm. This interpretation highlights the practical significance of the calculation, allowing us to visualize and understand the dynamic changes occurring in the sphere's volume as its radius increases.

In this article, we successfully tackled the problem of finding the rate of change of a sphere's volume given the rate of change of its radius. We determined that when the radius of the sphere is 3 cm and is increasing at a rate of 2 cm/sec, the volume is increasing at a rate of 72π cm³/sec. This problem illustrates the power and applicability of related rates in calculus. By using the formula for the volume of a sphere and applying differentiation, we established a relationship between the rates of change of the volume and the radius. This approach can be applied to a wide range of problems involving changing quantities, making it a fundamental skill in various fields, including physics and engineering. The process involved identifying the given information, setting up the problem mathematically, differentiating the relevant equation, substituting known values, and interpreting the result. This systematic approach is essential for solving related rates problems effectively. Understanding the concepts and techniques demonstrated in this article provides a solid foundation for tackling more complex problems involving rates of change.

To deepen your understanding of related rates, consider exploring similar problems with different geometric shapes, such as cylinders or cones. Varying the given rates of change and the specific moments at which you calculate the rates can also provide valuable insights. For example, you could investigate how the surface area of the sphere changes with respect to time or explore scenarios where the rate of change of the radius is not constant. Additionally, consider the applications of related rates in real-world scenarios, such as fluid dynamics, where understanding the rates of flow and volume change is crucial. Practice applying these techniques to a variety of problems to strengthen your skills and gain a more comprehensive understanding of calculus and its applications.

Rate of Change, Sphere Volume, Differential Calculus, Related Rates, Geometric Applications