Problem 35 - Calculus Volume 1

Question

Consider a pool shaped like the bottom half of a sphere, that is being filled at a rate of 25 \(\mathrm{ft}^{3} / \mathrm{min}\). The radius of the pool is \(10 \mathrm{ft}\). Gravel is being unloaded from a truck and falls into a pile shaped like a cone at a rate of \(10 \mathrm{ft}^{3} / \mathrm{min}\). The radius of the cone base is three times the height of the cone. Find the rate at which the height of the gravel changes when the pile has a height of \(5 \mathrm{ft}\).

Step-by-Step Solution

Answer

The height of the gravel changes at approximately 0.014 ft/min.

Step 1: Learn the Sphere Volume Formula

The pool has the shape of a half-sphere. The volume of a full sphere is given by \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius. Since we deal with half a sphere, its volume is \( V = \frac{1}{2} \cdot \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3 \).

Step 2: Understand the Cone Volume Formula

The volume of a cone is calculated by \( V = \frac{1}{3} \pi r^2 h \), where \( r \) is the radius of the base and \( h \) is the height. Given that \( r = 3h \), we substitute to get \( V = \frac{1}{3} \pi (3h)^2 h = \frac{1}{3} \pi 9h^3 = 3\pi h^3 \).

Step 3: Differentiate the Cone Volume Formula with Respect to Time

Given \( V = 3\pi h^3 \), differentiate both sides with respect to time \( t \) using chain rule to find \( \frac{dV}{dt} = 9\pi h^2 \frac{dh}{dt} \). Here, \( \frac{dh}{dt} \) is the rate at which the height of the pile increases, \( h \) is the height of the cone, and \( \frac{dV}{dt} \) is the rate at which the volume increases.

Step 4: Substitute Known Values and Solve for \( \frac{dh}{dt} \)

Substitute \( \frac{dV}{dt} = 10 \mathrm{ft}^3/\mathrm{min} \) (since gravel is falling at this rate), \( h = 5 \mathrm{ft} \), and \( \pi \) into the differentiated formula: \( 10 = 9\pi (5)^2 \frac{dh}{dt} \). Simplify and solve for \( \frac{dh}{dt} \). \( 10 = 225\pi \frac{dh}{dt} \) \( \frac{dh}{dt} = \frac{10}{225\pi} \) \( \frac{dh}{dt} = \frac{2}{45\pi} \).

Step 5: Evaluate the Rate of Change

The rate at which the height of the gravel changes is \( \frac{2}{45\pi} \) feet per minute. Simplifying further or computing numerically, this gives approximately 0.014 ft/min.

Key Concepts

Sphere Volume

Understanding how to calculate the volume of a sphere is crucial when dealing with problems involving spherical shapes. The formula for the volume of a sphere is expressed as \[ V = \frac{4}{3} \pi r^3 \]where \( V \) denotes the volume and \( r \) is the radius. Since the pool in the exercise is shaped like the bottom half of a sphere, we only need half the entire volume. Thus, the volume of the half-sphere becomes\[ V = \frac{1}{2} \cdot \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3 \]Here are the steps involved:

• Calculate the volume for a complete sphere.
• Divide by two to find the volume for the half-sphere.

The calculation shows how geometric shapes can impact the methods we use for determining volume, especially when considering only portions of those shapes.

Cone Volume

In problems involving cones, knowing how to determine the volume is necessary. 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. In our example, we see that the radius \( r \) is three times the height \( h \), leading to a substituted volume formula:\[ V = \frac{1}{3} \pi (3h)^2 h = 3\pi h^3 \]The steps are:

• Substitute \( r = 3h \) into the formula.
• Simplify to express the cone's volume solely in terms of height.

This simplification transforms the volume formula into one that directly relates volume to height, aiding in finding rate changes with respect to these dimensions.

Chain Rule

The chain rule is an essential calculus tool used for finding derivatives of composite functions. When differentiating a function with respect to time, we apply the chain rule to link different rate changes.Suppose you have a function like the cone's volume with respect to its height, expressed in terms of time:\[ V = 3 \pi h^3 \]To differentiate with respect to time, we apply the chain rule:\[ \frac{dV}{dt} = 9\pi h^2 \frac{dh}{dt} \]Here, \( \frac{dV}{dt} \) is the rate of change of volume, while \( \frac{dh}{dt} \) is the rate of change of height. The steps using the chain rule involve:

• Identifying each part's function dependency.
• Applying differentiation rules appropriately.

The chain rule's power is its ability to connect various rate changes, crucial in solving related rates problems.

Derivatives

Derivatives are fundamental in calculus for analyzing how things change. In the context of related rates, a derivative represents the rate of change of one quantity with respect to another. Here, it helps us understand how the height of a cone changes as the volume of material in the cone increases.Given the formula for a cone's volume in terms of height:\[ V = 3\pi h^3 \]Differentiating with respect to time \( t \) gives us:\[ \frac{dV}{dt} = 9\pi h^2 \frac{dh}{dt} \]This tells us how the rate of volume change \( \frac{dV}{dt} \) relates to the rate of height change \( \frac{dh}{dt} \). Solving for \( \frac{dh}{dt} \) provides insight into how the height of the particle pile in the cone is changing at any given moment.In practice, using derivatives:

• Involves applying the chain and product rules effectively.
• Lets you relate multiple changing quantities in a dynamic system.