Calculating Water Surface Rise In A Conical Vessel A Related Rates Problem

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In this article, we will explore a classic calculus problem involving related rates. Specifically, we'll delve into the scenario of water being poured into a conical vessel and determine the rate at which the water surface is rising at a particular depth. This problem elegantly demonstrates the application of differentiation in understanding dynamic situations. The core concept we'll be using is related rates, which allows us to find the rate of change of one quantity in terms of the rate of change of another.

Problem Statement

Imagine a vessel shaped like a right circular cone, with a vertical angle of 60 degrees. This cone is positioned with its axis vertical, and water is being poured into it at a constant rate of 8 cubic centimeters per second (8 cm³/s). The challenge is to determine how fast the water surface is rising when the water depth reaches 4 centimeters. This problem is a fascinating blend of geometry and calculus, requiring us to relate the volume of the water in the cone to its depth and then use calculus to find the rate of change.

Understanding the Geometry

Before diving into the calculus, let's first understand the geometry of the cone. The vertical angle of 60 degrees is crucial information, as it allows us to establish a relationship between the radius (r) and the height (h) of the water in the cone at any given time. Visualizing the cone and the water level inside it helps to see that the water forms a smaller cone similar to the larger vessel. This similarity is key to finding the relationship between r and h. Using trigonometry, specifically the tangent function, we can relate the radius and height. In a right triangle formed by the radius, height, and the slant height of the cone, the tangent of half the vertical angle (30 degrees) is equal to the ratio of the radius to the height (r/h). This geometric insight is a cornerstone of solving this problem. By establishing a clear relationship between the cone's dimensions and the water level, we set the stage for applying calculus techniques to solve the problem.

Setting Up the Related Rates Problem

Now, let's translate the problem into mathematical terms. We are given the rate of change of the volume of water with respect to time (dV/dt), which is 8 cm³/s. Our goal is to find the rate of change of the height of the water with respect to time (dh/dt) when the height h is 4 cm. The volume V of a cone is given by the formula V = (1/3)πr²h. However, to relate the rates, we need to express V as a function of a single variable, either r or h. This is where the geometric relationship we derived earlier comes into play. Since tan(30°) = r/h, we can express r in terms of h (or vice versa). This substitution is a crucial step in simplifying the problem and making it amenable to differentiation. By expressing the volume in terms of a single variable, we can then differentiate with respect to time and relate the rates of change.

Solving for the Rate of Change

With the volume V expressed as a function of h, we can now differentiate both sides of the equation with respect to time t. This differentiation yields an equation that relates dV/dt, dh/dt, and h. We know dV/dt is 8 cm³/s, and we want to find dh/dt when h is 4 cm. Plugging in these values into the differentiated equation, we can solve for dh/dt. The solution will give us the rate at which the water surface is rising at the specified depth. This process demonstrates the power of calculus in solving real-world problems involving changing quantities. The steps involved – establishing geometric relationships, setting up the volume equation, differentiating, and solving for the unknown rate – are fundamental techniques in calculus and related rates problems.

Detailed Solution

Let's walk through the solution step by step. First, we recognize that the volume of a cone is given by the formula:

V=13πr2hV = \frac{1}{3} \pi r^2 h

Where:

  • V is the volume of the water in the cone
  • r is the radius of the water surface
  • h is the height (depth) of the water

The vertical angle of the cone is 60°, which means the semi-vertical angle is 30°. We can relate r and h using the tangent function:

tan⁡(30∘)=rh\tan(30^{\circ}) = \frac{r}{h}

Since $ an(30^{\circ}) = \frac{1}{\sqrt{3}}$, we have:

r=h3r = \frac{h}{\sqrt{3}}

Substitute this expression for r into the volume formula:

V=13π(h3)2hV = \frac{1}{3} \pi \left( \frac{h}{\sqrt{3}} \right)^2 h

V=13πh23hV = \frac{1}{3} \pi \frac{h^2}{3} h

V=π9h3V = \frac{\pi}{9} h^3

Now, differentiate both sides of the equation with respect to time t:

dVdt=ddt(π9h3)\frac{dV}{dt} = \frac{d}{dt} \left( \frac{\pi}{9} h^3 \right)

dVdt=π9⋅3h2dhdt\frac{dV}{dt} = \frac{\pi}{9} \cdot 3h^2 \frac{dh}{dt}

dVdt=π3h2dhdt\frac{dV}{dt} = \frac{\pi}{3} h^2 \frac{dh}{dt}

We are given that dV/dt = 8 cm³/s, and we want to find dh/dt when h = 4 cm. Plug in these values:

8=π3(42)dhdt8 = \frac{\pi}{3} (4^2) \frac{dh}{dt}

8=16π3dhdt8 = \frac{16\pi}{3} \frac{dh}{dt}

Now, solve for dh/dt:

dhdt=8⋅316π\frac{dh}{dt} = \frac{8 \cdot 3}{16\pi}

dhdt=2416π\frac{dh}{dt} = \frac{24}{16\pi}

dhdt=32π\frac{dh}{dt} = \frac{3}{2\pi}

Therefore, the rate at which the water surface is rising when the depth of the water is 4 cm is $\frac{3}{2\pi}$ cm/s.

Final Answer

The water surface is rising at a rate of $\frac{3}{2\pi}$ cm/s when the depth of the water is 4 cm. This result provides a concrete answer to our problem, illustrating how the rate of change of the water level is related to the rate at which water is being poured into the cone. This solution not only answers the specific question but also reinforces the broader concepts of related rates and their applications in various fields.

Key Concepts and Takeaways

This problem highlights several key concepts in calculus and problem-solving:

  • Related Rates: Understanding how the rates of change of different quantities are related to each other. This is a fundamental concept in calculus and is applicable in various fields, including physics, engineering, and economics.
  • Geometric Relationships: Establishing relationships between geometric quantities (like radius and height) is crucial in many calculus problems. This often involves using trigonometric functions and geometric properties.
  • Differentiation: The process of differentiation allows us to find the rate of change of a function. In this case, we differentiated the volume equation with respect to time to relate the rates of change of volume and height.
  • Problem-Solving Strategy: Breaking down a complex problem into smaller, manageable steps is essential. This includes understanding the problem, identifying relevant formulas, establishing relationships between variables, and solving for the unknown quantity.

Further Exploration

To deepen your understanding of related rates problems, consider exploring variations of this problem. For example, you could investigate:

  • Different Cone Angles: How does the rate of water surface rise change with different vertical angles of the cone?
  • Variable Inflow Rates: What if the water is poured into the cone at a non-constant rate?
  • Other Shapes: How would you approach a similar problem with a different shaped vessel, such as a cylinder or a sphere?

By exploring these variations, you can further solidify your understanding of related rates and their applications. This problem serves as a valuable exercise in applying calculus concepts to real-world scenarios. The combination of geometry and calculus makes it a compelling example of how mathematical tools can be used to model and understand dynamic systems.

Conclusion

In conclusion, this problem demonstrates the power of calculus in solving real-world problems involving related rates. By carefully setting up the problem, using geometric relationships, and applying differentiation techniques, we were able to determine the rate at which the water surface is rising in a conical vessel. This problem is a testament to the elegance and utility of calculus in understanding dynamic systems. Through practice and exploration, you can master these techniques and apply them to a wide range of problems in mathematics, science, and engineering. The ability to relate changing quantities and solve for unknown rates is a valuable skill in many fields, and this problem provides a solid foundation for further exploration in calculus and its applications.