Leaking Cone Rate Of Surface Area Increase When Halfway Full

Have you ever wondered how the surface area of water changes in a leaking cone? This article delves into a fascinating mathematical problem involving a right circular cone, a leaky tap, and some calculus magic. We'll explore how to determine the rate at which the water's surface area increases when the water level is halfway up the cone. So, grab your thinking caps, and let's dive in!

Understanding the Problem

Before we jump into the calculations, let's clearly define the problem. Imagine a container shaped like a right circular cone standing tall at 100 meters, with a base radius of 10 meters. Now, picture a pesky tap dripping water into this cone at a rate of $0.01 cm^3/s$. Our mission is to find out how fast the surface area of the water is increasing when the water level reaches the halfway point, which is 50 meters. This problem combines geometry, calculus, and a bit of real-world intuition. We'll need to use related rates, a calculus technique that helps us find the rate of change of one quantity in terms of the rate of change of another.

Key Concepts and Formulas

To tackle this problem effectively, we'll need to arm ourselves with some essential concepts and formulas:

• Volume of a Cone: The volume (V) of a cone is given by the formula $V = (1/3)πr^2h$, where 'r' is the radius of the base and 'h' is the height.
• Surface Area of Water in the Cone: The surface area (A) of the water in the cone is the area of the circular surface at the top of the water, which is $A = πr^2$.
• Related Rates: This calculus technique allows us to relate the rates of change of different variables. If we have an equation relating two variables, say x and y, and both are functions of time t, then we can differentiate both sides of the equation with respect to t to get a related rates equation.
• Similar Triangles: Similar triangles have the same shape but different sizes. The ratios of corresponding sides in similar triangles are equal. This concept will be crucial in relating the radius and height of the water in the cone.

Setting Up the Problem

The first step in solving a related rates problem is to set up the problem correctly. This involves identifying the variables, their rates of change, and any relationships between them. In our case:

• Let 'V' be the volume of water in the cone at time 't'.
• Let 'r' be the radius of the water surface at time 't'.
• Let 'h' be the height of the water at time 't'.
• Let 'A' be the surface area of the water at time 't'.
• We are given that $dV/dt = 0.01 cm^3/s$ (the rate at which water is dripping into the cone). Note that we'll need to convert this to cubic meters per second since our dimensions are in meters. $0.01 cm^3/s = 0.01 * (1 m / 100 cm)^3 s = 10^{-8} m^3/s$.
• We want to find $dA/dt$ when $h = 50 m$ (halfway up the cone).

We also know the dimensions of the cone: the total height is 100 m, and the base radius is 10 m. This gives us a crucial relationship through similar triangles. The triangle formed by the entire cone and the triangle formed by the water inside the cone are similar. Therefore, the ratio of their corresponding sides is equal:

r/h = 10/100$ or $r = h/10

This relationship will allow us to express the volume and surface area in terms of a single variable, which simplifies the calculus.

Solving for the Rate of Increase in Surface Area

Now that we've set up the problem, we can move on to the main event: finding $dA/dt$. Here's the breakdown:

1. Express Surface Area in Terms of Height

We know that $A = πr^2$. We also have the relationship $r = h/10$. Substituting the second equation into the first, we get:

$$A = π(h/10)^2 = πh^2/100$$

Now, the surface area is expressed solely in terms of the height 'h'. This is a significant step because it simplifies the differentiation process.

2. Differentiate with Respect to Time

Next, we differentiate both sides of the equation $A = πh^2/100$ with respect to time 't'. Remember, both A and h are functions of time:

$$dA/dt = (π/100) * 2h * (dh/dt) = (πh/50) * (dh/dt)$$

This equation relates the rate of change of the surface area ($dA/dt$) to the rate of change of the height ($dh/dt$).

3. Find $dh/dt$

To find $dA/dt$, we first need to find $dh/dt$, the rate at which the water level is rising. We'll use the volume equation for this.

We know $V = (1/3)πr^2h$. Substituting $r = h/10$, we get:

$$V = (1/3)π(h/10)^2h = πh^3/300$$

Now, differentiate both sides with respect to time:

$$dV/dt = (π/300) * 3h^2 * (dh/dt) = (πh^2/100) * (dh/dt)$$

We know $dV/dt = 10^{-8} m^3/s$. We want to find $dh/dt$ when $h = 50 m$. Plugging these values into the equation:

$$10^{-8} = (π(50)^2/100) * (dh/dt)$$

Solving for $dh/dt$:

$$dh/dt = 10^{-8} * (100 / (π * 50^2)) = 10^{-8} * (100 / (2500π)) = 10^{-8} * (1 / (25π)) m/s$$

So, $dh/dt = 10^{-8} / (25π) m/s$

4. Calculate $dA/dt$

Now we have all the pieces to find $dA/dt$. We have the equation:

$$dA/dt = (πh/50) * (dh/dt)$$

We know $h = 50 m$ and $dh/dt = 10^{-8} / (25π) m/s$. Plugging these values in:

$$dA/dt = (π * 50 / 50) * (10^{-8} / (25π)) = π * (10^{-8} / (25π)) = 10^{-8} / 25 m^2/s$$

Therefore, the rate of increase of the surface area when the water is halfway up is $10^{-8} / 25 m^2/s$, which is approximately $1.273 * 10^{-10} m^2/s$.

Conclusion

Guys, we've successfully navigated the leaking cone problem! By applying the principles of related rates, similar triangles, and some careful calculus, we were able to determine the rate at which the water's surface area increases when the water level is halfway up the cone. This problem highlights the power of calculus in solving real-world scenarios. The key takeaways are the importance of setting up the problem correctly, identifying relationships between variables, and using differentiation to find the desired rates of change. Remember to always convert units to ensure consistency throughout your calculations.

This exploration not only sharpens our mathematical skills but also gives us a new perspective on everyday phenomena. So next time you see a cone-shaped container, you'll have a deeper appreciation for the mathematics at play! Keep exploring, keep questioning, and keep those calculations coming!