Calculating The Rate Of Height Increase In A Conical Gravel Pile A Related Rates Problem

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Introduction

In the realm of calculus and related rates problems, we often encounter scenarios where understanding the relationship between changing quantities is crucial. One classic example involves a conical pile of material, such as gravel, being formed by a conveyor belt. This article delves into such a problem, exploring how the height of the pile increases as gravel is dumped onto it at a constant rate. Specifically, we will investigate the scenario where gravel is being dumped from a conveyor belt at a rate of 40 cubic feet per minute, forming a pile in the shape of a right circular cone whose base diameter and height are always equal. Our goal is to determine how fast the height of the pile is increasing when the pile reaches a height of 16 feet. This exploration will not only solidify your understanding of related rates but also showcase the practical applications of calculus in real-world scenarios.

Problem Statement and Setup

The core of the problem lies in understanding how the volume of the conical pile changes with time and how this change affects the height. We are given that gravel is being dumped at a rate of 40 cubic feet per minute, which means the volume (V) of the cone is increasing at a rate of dV/dt = 40 ft³/min. The cone's geometry is defined by the relationship that its base diameter and height (h) are always equal. This is a crucial piece of information because it allows us to relate the radius (r) of the cone's base to its height. Since the diameter is equal to the height, the radius, which is half the diameter, is r = h/2. This relationship will be instrumental in expressing the volume of the cone in terms of a single variable, the height.

The volume of a cone is given by the formula V = (1/3)πr²h. Substituting r = h/2 into this formula, we get V = (1/3)π(h/2)²h = (1/12)πh³. This equation now expresses the volume of the cone solely in terms of its height, which is what we need to relate the rate of change of volume (dV/dt) to the rate of change of height (dh/dt). We are interested in finding dh/dt, the rate at which the height is increasing, when the height h = 16 feet. The setup involves identifying the given rates, establishing the geometric relationships, and expressing the volume in terms of the variable whose rate of change we want to find. This methodical approach is key to solving related rates problems effectively.

Applying Calculus: Differentiation and the Chain Rule

To determine the relationship between the rates of change of volume and height, we need to employ the power of calculus, specifically differentiation. We have the equation V = (1/12)πh³, which relates the volume of the cone to its height. Now, we differentiate both sides of this equation with respect to time (t). This is where the chain rule comes into play. The chain rule is a fundamental concept in calculus that allows us to differentiate composite functions, which are functions within functions. In this case, we are differentiating a function of h with respect to t, so we need to use the chain rule.

Differentiating V = (1/12)πh³ with respect to t, we get dV/dt = (1/12)π * 3h² * (dh/dt). This step is crucial as it connects the rate of change of volume (dV/dt), which we know, to the rate of change of height (dh/dt), which we want to find. The term 3h² comes from the power rule of differentiation, and the (dh/dt) term is the result of applying the chain rule. Simplifying the equation, we have dV/dt = (1/4)πh² (dh/dt). This equation now provides a direct link between the given rate of gravel dumping (dV/dt) and the rate at which the height of the pile is increasing (dh/dt). By carefully applying differentiation and the chain rule, we have transformed the geometric problem into an algebraic one that we can solve for the desired rate.

Solving for the Rate of Height Increase

With the equation dV/dt = (1/4)πh² (dh/dt) established, we are now in a position to solve for dh/dt, the rate at which the height of the pile is increasing. We are given that dV/dt = 40 ft³/min, and we are interested in finding dh/dt when h = 16 feet. Substituting these values into the equation, we get 40 = (1/4)π(16)² (dh/dt). This equation is now a simple algebraic equation with one unknown, dh/dt.

To isolate dh/dt, we first simplify the equation. We have 40 = (1/4)Ï€(256) (dh/dt), which simplifies to 40 = 64Ï€ (dh/dt). Now, we divide both sides of the equation by 64Ï€ to solve for dh/dt: dh/dt = 40 / (64Ï€). Further simplification gives us dh/dt = 5 / (8Ï€) ft/min. This is the exact value of the rate at which the height is increasing when the pile is 16 feet tall.

To get a numerical approximation, we can use the value of π ≈ 3.14159. Plugging this in, we get dh/dt ≈ 5 / (8 * 3.14159) ≈ 0.199 ft/min. Therefore, when the pile is 16 feet tall, its height is increasing at approximately 0.199 feet per minute. This result gives us a tangible understanding of how the height of the conical pile changes as gravel is continuously dumped onto it. The process of substituting the known values, simplifying the equation, and solving for the unknown rate highlights the power and elegance of calculus in solving real-world problems.

Conclusion

In this exploration of a related rates problem, we successfully determined how fast the height of a conical gravel pile increases as gravel is dumped onto it at a constant rate. By carefully setting up the problem, establishing the geometric relationships, applying differentiation and the chain rule, and solving the resulting equation, we found that the height of the pile increases at a rate of approximately 0.199 feet per minute when the pile is 16 feet tall. This problem not only demonstrates the practical applications of calculus in understanding dynamic systems but also underscores the importance of a methodical approach to problem-solving.

Related rates problems often appear challenging at first glance, but by breaking them down into smaller steps – identifying the given rates, establishing relationships between variables, differentiating with respect to time, and solving for the desired rate – we can tackle them effectively. This example serves as a valuable illustration of how calculus can be used to model and analyze real-world phenomena, providing insights into the dynamic interplay of changing quantities.

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