Problem 25 - Calculus Volume 1

Question

Consider a right cone that is leaking water. The dimensions of the conical tank are a height of \(16 \mathrm{ft}\) and a radius of \(5 \mathrm{ft}\). How fast does the depth of the water change when the water is \(10 \mathrm{ft}\) high if the cone leaks water at a rate of 10 \(\mathrm{ft}^{3} / \mathrm{min} ?\)

Step-by-Step Solution

Answer

The depth of the water decreases at 0.032 ft/min.

Step 1: Understand the problem

We have a right cone with a height of 16 ft and a base radius of 5 ft. The cone is leaking water at a rate of 10 ft³/min. We need to find how fast the depth of the water changes when the water is 10 ft high.

Step 2: Set up the relationship

Consider the volume of a cone formula \( V = \frac{1}{3} \pi r^2 h \), where \(r\) is the radius and \(h\) is the height. In our scenario, the relationship between the radius and height of the water level is similar to the full cone, i.e., \( \frac{r}{h} = \frac{5}{16} \). This means \( r = \frac{5}{16} h \) inside cone.

Step 3: Differentiate the volume with respect to time

We have the volume of water as \( V = \frac{1}{3} \pi \left( \frac{5}{16} h \right)^2 h \). This simplifies to \( V = \frac{25}{768} \pi h^3 \). Differentiate with respect to time (t), \( \frac{dV}{dt} = \frac{25}{256} \pi h^2 \frac{dh}{dt} \).

Step 4: Solve for \( \frac{dh}{dt} \) when \( h = 10 \)

Plug in \( \frac{dV}{dt} = -10 \) (since the cone is leaking) and \( h = 10 \) into the equation. \(-10 = \frac{25}{256} \pi (10)^2 \frac{dh}{dt} \). Solve for \( \frac{dh}{dt} \): \( \frac{dh}{dt} = \frac{-10 \times 256}{25 \times 100 \pi} \approx -0.032 \) ft/min.

Step 5: Interpret the negative sign

The negative sign in \( \frac{dh}{dt} \) indicates that the height of the water is decreasing at a rate of 0.032 ft/min when the water is 10 ft high.

Key Concepts

Calculus

Calculus is a branch of mathematics that studies how things change. One of its key applications is solving problems that involve changes over time, such as determining how quickly certain quantities change. In the case of the leaking cone problem, we want to understand how the height of water changes as the volume decreases over time. Calculus provides the tools to analyze and work through such real-world problems.

By using the techniques of calculus, we are able to set up relationships between different quantities and explore how they influence each other. We do this by using derivatives, which allow us to compute rates of change and gain insights into the dynamics of the system under observation. This is crucial for solving the problem of the leaking cone where different parts of the conical geometry are changing in relation to time.

Differentiation

Differentiation is the process of finding a derivative, which measures how a function changes as its input changes. In practical terms, a derivative represents a rate of change. For the leaking cone problem, differentiation helps us determine how fast the height of the water is changing, based on the volume change rate.

• The differentiation process allows us to transition from a straightforward formula of volume to a dynamic formula that considers changes over time.
• In our locked cone problem, we start by taking the basic volume formula and then differentiate it with respect to time, exploring how both the radius and height of the cone contribute to changes in water level.

This step is essential because it gives us the mathematical framework to compute the derivative of volume with respect to height, ultimately leading us to understand the movement and depth change of water in the cone.

Volume of a Cone

The volume of a cone is an important concept to understand in geometry, especially when studying problems dealing with space and capacity. The formula to calculate the volume of a cone is given by:\[ V = \frac{1}{3} \pi r^2 h \]where \( V \) is the volume, \( r \) is the radius of the base, and \( h \) is the height of the cone. This formula is derived by considering the area of the base and the linear dimension of height.

• Understanding the volume of a cone helps us to set up the initial scenario where water is leaking from it.
• Since the volume is related to both the radius and height, any change in these dimensions directly affects the total volume.

Grasping how the volume changes when the shape dimensions change is crucial. It allows us to navigate through related rates problems, connecting changes in one quantity (height) to another (volume).

Rate of Change

A rate of change is a measure of how quickly a quantity is changing with respect to another quantity, often time. In mathematics, and in the real world, understanding rates of change is critical as it describes the dynamics of change over time.

• In the context of the leaking cone, the rate at which the water leaks is a negative rate of change of volume, expressed as \( \frac{dV}{dt} \).
• This rate of change in volume directly affects the rate of change in water depth, \( \frac{dh}{dt} \).

Essentially, by understanding these rates, we observe how the decrease in the volume of water translates into a decrease in the water's height. Solving such situations often requires setting up and solving equations that intertwine these different rates, a typical task when dealing with related rates problems.