Problem 39 - Calculus Early Transcendentals

Question

A volume function The volume of a right circular cone of radius \(r\) and height \(h\) is \(V(r, h)=\pi r^{2} h / 3.\) a. Graph the function in the window \([0,5] \times[0,5] \times[0,150].\) b. What is the domain of the volume function? c. What is the relationship between the values of \(r\) and \(h\) when \(V=100 ?\)

Step-by-Step Solution

Answer

Answer: The relationship between the radius (r) and height (h) of a right circular cone when its volume is 100 can be represented by the equation: \(h = \frac{300}{\pi r^2}\).

Step 1: Graph of the volume function

To analyze the graph of the volume function, we need to plot the given function \(V(r, h) = \frac{\pi r^2 h}{3}\) over the window \([0,5] \times[0,5] \times[0,150]\). In this case, the best way is to use a graphing utility or software that supports 3D graphing, such as 'GeoGebra' or 'WolframAlpha'.

Step 2: Domain of the volume function

To find the domain of the volume function, we need to determine the possible values of \(r\) and \(h\). For a right circular cone, the radius \(r\) and height \(h\) must always be non-negative. Therefore, the domain of the function can be represented as: \(\{(r,h) | r \geq 0, h \geq 0\}\).

Step 3: Relationship between radius and height for a volume of 100

Given that the volume (\(V\)) of the right circular cone is 100, we have: \(V(r, h) = 100\). We are given the formula for the volume: \(V(r, h) = \frac{\pi r^2 h}{3}\). Now, we need to find the relationship between \(r\) and \(h\) for a volume of \(100\). To do this, we can substitute \(V(r, h) = 100\) into the equation, and solve for either \(r\) or \(h\). For example, we can solve for \(h\): \(h = \frac{3V}{\pi r^2} = \frac{3 \times 100}{\pi r^2} = \frac{300}{\pi r^2}\). The relationship between the radius \(r\) and height \(h\) of the right circular cone when the volume is \(100\) can be represented by the equation: \(h = \frac{300}{\pi r^2}\).

Key Concepts

Domain of a Function

The domain of a function refers to the set of possible input values that you can put into the function. When it comes to the volume of a cone, defined by the function \[ V(r, h) = \frac{\pi r^{2} h}{3} \] where \( r \) is the radius and \( h \) is the height, understanding the domain is essential. For a cone, both the radius and the height must be non-negative numbers since negative measurements are not physically meaningful.

Thus, the domain of this volume function can be specified as all pairs \( (r, h) \) such that \( r \geq 0 \) and \( h \geq 0 \). This means any positive or zero value for \( r \) and \( h \) is acceptable in this context.

In practical situations, this translates to ensuring both the radius and height are realistic and feasible measurements, which impacts how we represent and utilize the function in different scenarios.

Graphing Functions

To truly grasp how the volume of a cone is affected by changes in its dimensions, graphing provides a visual representation. The function given is:\[ V(r, h) = \frac{\pi r^{2} h}{3} \].Plotting this function in three dimensions allows students to see how the volume changes with different values of \( r \) and \( h \). You would typically use a 3D graphing tool like GeoGebra or WolframAlpha, which helps in visualizing such functions over a selected range.

To graph within the window specified in the exercise, which is \([0,5] \times [0,5] \times [0,150]\), you can see how both the dimensions of the base and the height contribute to the increasing volume. This effectively demonstrates relationships and dependencies visually, reinforcing the mathematical calculations you perform.

Relationship between Variables

Understanding the relationship between variables \( r \) and \( h \) is key in learning about the volume of cones. If a specific volume is given, like 100 cubic units, you can use the equation \[ V(r, h) = \frac{\pi r^2 h}{3} \]to express one variable in terms of the other.

For instance, by setting \( V = 100 \), you can derive a relationship for \( h \) when \( r \) is known:\[ h = \frac{300}{\pi r^2} \]. This equation shows how changing the radius will require a corresponding adjustment in the height to maintain the same volume.

• An increase in the radius \( r \) necessitates a decrease in height \( h \) to keep the volume constant, and vice versa.
• The equation provides a straightforward method for calculating unknown dimensions when aiming for a specific volume.

This interplay helps students understand practical applications, such as maintaining material constraints in manufacturing or adjusting dimensions to fit design specifications.

3D Plotting

Three-dimensional plotting is a powerful way to visualize mathematical functions involving multiple variables. For the volume of the cone \[ V(r, h) = \frac{\pi r^{2} h}{3} \], 3D plots depict how the volume changes with the radius \( r \) and height \( h \).

By plotting this function in a 3D space, you can view a surface where every point corresponds to a set of values \( (r, h, V) \). Such plots vividly show how the volume increases as both dimensions expand. The plotting window \([0,5] \times [0,5] \times [0,150]\) ensures that the most relevant and feasible dimensions for cones are captured, allowing students to observe the continuous growth in volume with increasing size.

• These graphs are essential tools in teaching students about advanced math concepts effortlessly.
• 3D plots open up a clearer understanding, demonstrating how variables interplay to form the geometric objects and their properties.

Integrating 3D visualization in learning can also help in comprehending complex dependencies that are not easily understood through equations alone.