Problem 10 - Calculus an Applied Approach

Question

Volume The radius \(r\) of a right circular cone is increasing at a rate of 2 inches per minute. The height \(h\) of the cone is related to the radius by \(h=3 r\). Find the rates of change of the volume when (a) \(r=6\) inches and (b) \(r=24\) inches.

Step-by-Step Solution

Answer

The rate of change of the volume when the radius \(r\) is 6 inches and 24 inches are \(216\pi\) cubic inches per minute and \(3456\pi\) cubic inches per minute respectively.

Step 1: Express the height in terms of the radius

From the given exercise, we are told that \(h = 3r\).

Step 2: Plug the height equation into the volume equation

Substitute the value of h into the equation of the volume, we would get \(V = \frac{1}{3}\pi r^2 (3r) = \pi r^3\). This is the equation for the volume in terms of r.

Step 3: Differentiate the volume equation with respect to time

To find the rate of change of volume with respect to time, differentiate V with respect to t. The derivative of V is \(\frac{dV}{dt} = 3\pi r^2 \frac{dr}{dt}\). Here, we know that \(\frac{dr}{dt}\) is 2.

Step 4: Find the rate of change of volume when \(r = 6\)

Substitute \(r = 6\) and \(\frac{dr}{dt} = 2\) into the derivative equation, we get \(\frac{dV}{dt} = 3\pi (6)^2 (2) = 216\pi \) cubic inches per minute.

Step 5: Find the rate of change of volume when \(r = 24\)

Next, substitute \(r = 24\) and \(\frac{dr}{dt} = 2\) into the derivative equation, we get \(\frac{dV}{dt} = 3\pi (24)^2 (2) = 3456\pi \) cubic inches per minute.

Key Concepts

Rate of Change of Volume

Understanding the rate of change of volume is essential when examining how a shape's capacity alters over time. For instance, say we're watching a balloon inflate; we're actually observing the change in volume with respect to time.

In calculus, this concept frequently appears in related rates problems, where variables are connected and change together. The rate of change of volume in the example of the right circular cone, \(\frac{dV}{dt}\), denotes how quickly the volume changes as time progresses. It's crucial because it elucidates how fast or slow the volume is expanding or contracting, which has practical implications like filling water tanks or assessing air conditions.

Differentiating with Respect to Time

Differentiation is a key concept in calculus, and differentiating with respect to time is a powerful tool when dealing with real-world dynamic systems. It allows us to calculate the rate of change of one variable in relation to another variable that represents time.

In related rates problems, after expressing the quantities of interest (like volume, surface area) in terms of one another, we use differentiation with respect to time to find out how one quantity’s rate of change affects another. In our cone example, after obtaining the volume formula in terms of the radius, \(V = \pi r^3\), we differentiate it with respect to time (\

Right Circular Cone Volume

The volume of a right circular cone is a fundamental concept that involves spatial understanding. A cone's volume can be calculated using the formula \(V = \frac{1}{3}\pi r^2 h\), where \(r\) represents the radius of the cone's base and \(h\) is the height.

If a cone’s dimensions are changing, it’s essential to express the changing quantities in relation to one another to use calculus efficiently. As in our problem, the height is directly proportional to the radius; therefore, we can say \(h = 3r\). This relation allows us to write the volume solely in terms of the radius, simplifying the problem. Understanding how volume relates to radius and height is crucial, as it is the basis for determining the rate at which the volume of the cone changes over time.

Applying Chain Rule in Calculus

The Chain Rule is an essential differentiation technique in calculus, particularly in related rates problems. It allows us to differentiate composite functions; that is, a function that relies on another function.

Consider the example of the cone volume \(V\). The volume depends on the radius \(r\), which itself is a function of time \(t\). To find the rate at which volume changes with time, we must take the derivative of volume with respect to radius and then multiply by the derivative of radius with respect to time, applying the Chain Rule. Hence, \(\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}\). Mastering this rule helps to unravel complex problems where variables are interdependent, demonstrating the profound interconnectedness of mathematical concepts with physical phenomena.