Problem 55 - Thomas Calculus

Question

Volume of a bowl a. A hemispherical bowl of radius \(a\) contains water to a depth \(h .\) Find the volume of water in the bowl. b. Related rates Water runs into a sunken concrete hemispherical bowl of radius 5 \(\mathrm{m}\) at the rate of 0.2 \(\mathrm{m}^{3} / \mathrm{sec} .\) How fast is the water level in the bowl rising when the water is 4 \(\mathrm{m}\) deep?

Step-by-Step Solution

Answer

a. Volume formula: \( V = \frac{\pi h^2}{3} (3a - h) \). b. \( \frac{dh}{dt} = \frac{0.2}{\pi (5 \times 4 - \frac{4^2}{3})} \).

Step 1: Formula for the Volume of a Hemispherical Cap

To find the volume of water in a hemispherical bowl filled to depth \( h \), we use the formula for the volume of a spherical cap: \[ V = \frac{\pi h^2}{3} (3a - h), \] where \( a \) is the radius of the hemisphere, and \( h \) is the depth of water.

Step 2: Substitute Values into Volume Formula

Substitute the given values into the formula for the volume. For part a, we substitute \( a \) as the radius and \( h \) as the depth of water. For part b, we will evaluate the rate of change in terms of \( h \).

Step 3: Differentiate to Find Related Rate

In part b, differentiate the volume formula \( V = \frac{\pi h^2}{3} (3a - h) \) with respect to time \( t \) to find the rate of change of the water height (\( \frac{dh}{dt} \)). Apply the chain rule during differentiation, considering \( a = 5 \) m and the given rate \( \frac{dV}{dt} = 0.2 \) m³/s.

Step 4: Solve for dh/dt

Using the differentiated equation from Step 3:\[ \frac{dV}{dt} = \pi (ah - \frac{h^2}{3}) \frac{dh}{dt}, \]substitute the known values \( a = 5 \), \( h = 4 \), and \( \frac{dV}{dt} = 0.2 \) to solve for \( \frac{dh}{dt} \).

Key Concepts

Hemispherical Bowl

A hemispherical bowl is essentially half of a sphere, just like a bowl cut right down the middle. It's a three-dimensional shape, and understanding its properties is key in many practical and theoretical applications. In the context of this exercise, we're particularly interested in how much volume this kind of shape can hold or, more specifically, how much water fits inside it when filled to a certain depth.

When dealing with a hemispherical bowl, it's important to note:

• The complete diameter of the bowl is twice its radius, so if the radius is 'a', then the diameter is '2a'.
• The depth of the water or liquid inside will always be less than or equal to the radius when measuring from the center to the top of the water level.

These properties help in calculating how changes in the depth of water affect the bowl's volume.

Related Rates

Related rates are a method in calculus used to find the rate at which one quantity changes with respect to another over time. In our scenario, we're observing how the height of water in the bowl changes as water pours in at a constant rate.

To work with related rates, keep these steps in mind:

• Identify all related variables in your problem (like water depth, water volume, etc.).
• Express the quantities in terms of these variables using known formulas (like the volume of water in terms of its depth).
• Differentiate these equations with respect to time to find relationships between the rates of change.

This is crucial in environments where one action causes another to change, like filling a bowl with water.

Spherical Cap Volume Formula

The spherical cap is the portion of a sphere that is "cut off" at a certain height. This formula is essential to calculate volumes in cases where the filling substance (like water) doesn't reach the very top of the hemispherical bowl.

The Spherical Cap Volume Formula is:\[ V = \frac{\pi h^2}{3} (3a - h) \]

• Here, \( V \) represents the volume of the cap.
• \( h \) is the height from the sphere's base to the top of the water.
• \( a \) is the radius of the sphere.

This formula adapts to the dimensions of your problem, enabling you to calculate the precise volume of any liquid up to a specific depth within a hemispherical shape.

Differentiation

Differentiation is a fundamental concept in calculus that enables the determination of how a function changes as its input changes. Here, it is used to understand the rate at which the volume of water in the bowl increases as the water level rises.

When you differentiate the volume formula \( V = \frac{\pi h^2}{3} (3a - h) \) with respect to time \( t \), you're primarily looking at how fast certain parts are changing:

• Overall, you're finding the rate of change in water height, represented as \( \frac{dh}{dt} \).
• The challenge is linking how fast the volume changes (given as a constant rate) to how fast the water level changes using calculus.

By applying tools like differentiation, you turn these abstract concepts into tangible solutions that make analyzing dynamic systems manageable.

Chain Rule

The chain rule is a critical tool in calculus used for differentiating compositions of functions, essential for solving problems where multiple variables depend on each other. In this exercise, it's applied to the formula for water volume to determine how the depth of water changes over time.

When using the chain rule:

• Understand that it allows you to differentiate a function of a function by taking the derivative of the outer function and multiplying it by the derivative of the inner function.
• It's especially useful when dealing with related rates, where the rate of change in one variable leads to a change in another.
• In our case, to find \( \frac{dh}{dt} \), we look at the rate of change of volume \( \frac{dV}{dt} \) and how that translates to changes in depth.

Applying the chain rule ensures that each variable's interdependency is considered, critical for solving sophisticated calculus problems like those involving the dynamics of a filling hemispherical bowl.