Problem 14 - University Calculus: Early Transcendentals

Question

The radius \(r\) and height \(h\) of a right circular cone are related to the cone's volume \(V\) by the equation \(V=(1 / 3) \pi r^{2} h\). a. How is \(d V / d t\) related to \(d h / d t\) if \(r\) is constant? b. How is \(d V / d t\) related to \(d r / d t\) if \(h\) is constant? c. How is \(d V / d t\) related to \(d r / d t\) and \(d h / d t\) if neither \(r\) nor \(h\) is constant?

Step-by-Step Solution

Answer

Change in volume is affected by changes in both the radius and height of the cone.

Step 1: Part (a): r is Constant

\(V = \frac{1}{3}\pi r^2 h\). With \(r\) constant:
\(\frac{dV}{dt} = \frac{1}{3}\pi r^2 \frac{dh}{dt}\)

Step 2: Part (b): h is Constant

With \(h\) constant:
\(\frac{dV}{dt} = \frac{1}{3}\pi (2rh) \frac{dr}{dt} = \frac{2}{3}\pi rh \frac{dr}{dt}\)

Step 3: Part (c): Neither r nor h is Constant

Using the product rule:
\(\frac{dV}{dt} = \frac{1}{3}\pi \left(2rh \frac{dr}{dt} + r^2 \frac{dh}{dt}\right) = \frac{\pi}{3}\left(2rh\frac{dr}{dt} + r^2\frac{dh}{dt}\right)\)

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus used to find the rate at which something changes. In this exercise, we're interested in how the volume of a cone changes as its dimensions change over time. Imagine a situation where the size of your cone is changing, perhaps because you're inflating or deflating an object shaped like a cone. You need to know how quickly the volume is changing, which is where differentiation comes into play.

In mathematical terms, when we're given a function, in this case, the volume of the cone as a function of its radius and height, differentiating allows us to calculate how fast this volume is increasing or decreasing over time. The changing factors of this cone -- the radius and the height -- can be functions of time, denoted as \( r(t) \) and \( h(t) \). Hence, the rate of change of volume is expressed as \( \frac{dV}{dt} \), which shows how much and how fast the volume changes over time based on changes in radius and height. This is the essence of applying differentiation in problems involving related rates.

It is noteworthy to understand the chain rule in differentiation, which is crucial for understanding how changes in multiple independent variables (like radius and height) affect the dependent variable (volume).

Volume of a Cone

The volume of a cone is a measure of the three-dimensional space occupied by the cone. In mathematical terms, the volume \( V \) of a cone with radius \( r \) and height \( h \) is given by the formula: \[ V = \frac{1}{3} \pi r^2 h \]

This formula shows that volume is proportional to the square of the radius and linear in height. This means if you increase the height or radius, the volume will increase, but the radius has a greater impact since it's squared. Understanding this relationship is key when solving related rates problems because it dictates how changes in the cone's size (via \( r \) and \( h \)) affect its volume.

In real-world applications, this formula helps in scenarios such as filling a cone-shaped container where determining overflow or capacity is essential.

• When \( r \) is constant, changes in volume depend only on \( h \).
• Conversely, when \( h \) is constant, changes only depend on \( r \).
• If neither is constant, both contribute to changes in \( V \).

Understanding these relationships helps solve part c of the exercise, where both variables fluctuate.

Calculus Problem-Solving

Solving calculus problems, especially those involving related rates, requires a systematic approach. Here’s a generalized procedure to tackle such problems:

1. **Identify what you're looking for.** Start by understanding the problem statement and determining which rate of change you need to find. For instance, whether you need \( \frac{dV}{dt} \), \( \frac{dh}{dt} \), or \( \frac{dr}{dt} \).

2. **Know your variables and constants.** In some cases, such as ours, either the radius \( r \) or height \( h \) might stay constant, simplifying the differentiation process.

3. **Use the chain rule.** This mathematical tool allows you to relate the rates of change of different variables through differentiation. For example, if both \( r \) and \( h \) change, you'll use the chain rule to find \( \frac{dV}{dt} \) as it's related to both \( \frac{dr}{dt} \) and \( \frac{dh}{dt} \).

4. **Set up and solve equations.** Substitute the known values and relationships into your differentiated equation. This step involves a bit of algebra to isolate the rate you need to solve for.

• Take partial derivatives when one variable changes.
• Apply the product rule for terms involving two variables, like \( \pi r^2 h \).

By following these steps, you'll systematically solve calculus problems involving rates of change, ensuring understanding and accuracy.