Problem 53 - Thomas Calculus in SI Units

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 \(0.2 \mathrm{m}^{3} / \mathrm{s}\). 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 is \( \frac{\pi h^2}{3} (3a - h) \). b) Water level rises at \( \frac{1}{120\pi} \text{ m/s} \).

Step 1: Introduction to the Problem

We have a hemispherical bowl with water filled to a depth of \( h \). We need to calculate the volume of water in this bowl regarding the given radius \( a \). Additionally, we're interested in how the water level changes over time as more water is added.

Step 2: Find the Volume of the Water

To find the volume of water in the hemispherical bowl, we must use the formula for the volume of a spherical cap. The volume \( V \) of a spherical cap with radius \( a \) and height \( h \) is given by:\[ V = \frac{\pi h^2}{3} (3a - h) \]This formula helps calculate the volume of water based on the depth \( h \) to which the water is filled in the hemisphere.

Step 3: Substitute Known Values for Part (b)

In part (b), we are given a hemispherical bowl with a radius \( a = 5 \text{ m} \) and the water is filled to a depth of \( h = 4 \text{ m} \). We need to find how fast the water level is rising when water flows at \(0.2 \text{ m}^3/\text{s}\). The rate of change of volume is given by \( \frac{dV}{dt} = 0.2\).

Step 4: Use Implicit Differentiation

Using the volume formula from Step 2, differentiate both sides with respect to time \( t \):\[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{\pi h^2}{3} (3a - h) \right) \]Apply the product rule and chain rule:\[ \frac{dV}{dt} = \pi h (2ah - h^2) \frac{dh}{dt} \]Solve for \( \frac{dh}{dt} \), the rate of change of the water level.

Step 5: Solve the Rate of Water Level Rising

With \( a = 5 \text{ m} \) and \( h = 4 \text{ m} \), substitute these into the differentiated equation:\[ 0.2 = \pi \cdot 4 (2 \times 5 - 4) \cdot \frac{dh}{dt} \]Simplify and solve for \( \frac{dh}{dt} \):\[ 0.2 = \pi \cdot 4 \cdot 6 \cdot \frac{dh}{dt} \]\[ \frac{dh}{dt} = \frac{0.2}{24\pi} \approx \frac{1}{120\pi} \text{ m/s} \]

Step 6: Interpretation of Result

The rate at which the water level is rising when the water is 4 meters deep is approximately \( \frac{1}{120\pi} \text{ m/s} \). This indicates how quickly the depth of water increases at the given flow rate.

Key Concepts

Related Rates

Related rates problems involve finding the rate at which one quantity changes in relation to another quantity over time. In our exercise, this relates to how the height of the water in the hemisphere changes as water flows into the bowl.
To solve such problems, we typically follow these steps:

• Identify all relevant variables and their relationships. Here, the volume of water in the bowl, the water's depth, and the flow rate are interconnected.
• Establish a relationship between the variables using equations. The volume of water, given its depth, is expressed using the formula for the volume of a spherical cap.
• Differentiating equations with respect to time gives us the related rates—such as the rate of change of depth with respect to time.
• Finally, substitute known values to find specific rates at given conditions.

The real challenge with related rates is understanding how to link these rates via differentiation, particularly when one rate is known (in this case, the rate of inflow of water) and others need discovery.

Implicit Differentiation

Implicit differentiation is a technique used when an equation involving multiple variables cannot be easily solved for one variable. Instead of solving for a variable explicitly, we differentiate each term of the equation with respect to the variable we are interested in.
This technique becomes invaluable when dealing with related rates, especially when the relationship between variables isn't simple or direct. For the spherical bowl scenario, the volume equation isn't straightforward in terms of the water height changing over time.
Here's how implicit differentiation is applied:

• Take the derivative of the entire equation with respect to time. Terms not explicitly involving time require the application of the chain rule.
• Use the product rule where necessary, as volume involves both the height and depth of water.
• Solve for the desired rate, such as the rate of change of the water depth, after differentiating.

By using implicit differentiation, we derive a complex calculus problem into a manageable calculation, which allows us to discover unknown rates effectively.

Spherical Cap Volume

The volume of water in a hemispherical bowl is calculated using the formula for a spherical cap. A spherical cap is a portion of a sphere that lies between a plane cutting through the sphere and the curvature above it. It helps in determining volumes in spherical shapes when not filled fully.
To calculate the volume of the water up to a certain depth, the following formula is used: \[ V = \frac{\pi h^2}{3} (3a - h) \] Where:

• \( V \) is the volume of the spherical cap (in our case, the volume of water).
• \( h \) is the height or depth of the water in the bowl.
• \( a \) is the radius of the hemisphere or bowl.

This formula comes from integrating the volume of an incremental disk layer from the bottom of the sphere to the height \( h \). When solving for related rates, knowing how to differentiate this expression gives insight into dynamic changes in the volume in practical scenarios. Overall, the spherical cap volume concept enables us to understand and calculate partial volumes effectively, especially in engineering and physics problems.