Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sand is pouring onto a conical pile in such a way that at a certain instant the height is 100 inches and increasing at 3 inches per minute and the base radius is 40 inches and increasing at 2 inches per minute. How fast is the volume increasing at that instant?

Step-by-Step Solution

Answer

The volume is increasing at approximately 6933.33\pi cubic inches per minute.

Step 1: Understand the Problem

We need to find the rate at which the volume of a conical pile is increasing. At a specific moment, the height (\( h \)) of the cone is 100 inches, increasing at 3 inches per minute, and the base radius (\( r \)) is 40 inches, increasing at 2 inches per minute.

Step 2: Write Down the Volume Formula for a Cone

The volume (\( V \)) of a cone is given by the formula:\[ V = \frac{1}{3}\pi r^2 h \]where \( r \) is the radius and \( h \) is the height.

Step 3: Differentiate the Volume Formula with respect to Time

We need to find \(\frac{dV}{dt}\), the rate at which the volume is increasing. Differentiating both sides of \( V = \frac{1}{3} \pi r^2 h \)with respect to time (\( t \)), using the product rule, gives:\[ \frac{dV}{dt} = \frac{1}{3} \pi \left(2rh \frac{dr}{dt} + r^2 \frac{dh}{dt}\right) \]

Step 4: Plug in the Known Values

Substitute \( r = 40 \), \( h = 100 \), \( \frac{dr}{dt} = 2 \), and \( \frac{dh}{dt} = 3 \) into the differentiated equation:\[ \frac{dV}{dt} = \frac{1}{3} \pi \left(2(40)(100)(2) + (40)^2(3)\right) \]

Step 5: Simplify and Calculate the Result

Simplify the equation:\[ \frac{dV}{dt} = \frac{1}{3} \pi \left(16000 + 4800\right) = \frac{1}{3} \pi (20800) \]Calculate:\[ \frac{dV}{dt} = \frac{20800}{3} \pi \]This simplifies to approximately \( 6933.33\pi \) cubic inches per minute.

Key Concepts

Volume of a Cone

Understanding the volume of a cone is crucial in many real-world applications. The formula \[ V = \frac{1}{3}\pi r^2 h \] allows us to calculate how much space is inside a cone-shaped object, like a party hat or a volcanic mountain. Here, \( r \) stands for the radius of the base, while \( h \) represents the cone's height. These geometric elements are essential because the area of the base and its height directly affect the cone's total volume.

• Base Area: Calculated as \( \pi r^2 \).
• Height Influence: Multiplied to the base area before applying the \( \frac{1}{3} \) constant factor.

This factor comes from the fundamental definition of a cone's volume, as a cone is essentially a third of a cylinder with the same base and height. Remembering this relation to a cylinder can help make these concepts more intuitive.
Understanding these basics lays a solid foundation for tackling related rate problems.

Differentiation

Differentiation is a powerful tool in calculus used to determine how a function changes over time, specifically how fast something is changing. In our exercise, we're interested in how quickly the volume of a cone increases as sand pours onto it. This requires the process of differentiating a formula with respect to time.

• Product Rule: Essential here because the volume formula involves products of different functions of time \( r(t) \) and \( h(t) \).
• Chain Rule: Used implicitly in differentiation when a variable, like time, affects others (height and radius) that affect the volume.

By applying these rules, we can find the rate of change, \( \frac{dV}{dt} \), giving us insights into how the volume increases as both radius and height change.
The steps involve substituting known rates of change into the differentiated formula to find the desired rate of volume change.

Calculus Concepts

Calculus is divided into differential and integral calculus, both crucial for understanding dynamic changes in quantities. In our exercise, differential calculus takes the spotlight, focusing on rates of change, using two essential concepts:

• Instantaneous Change: Captures the rate at which one quantity changes instantly with respect to another, like time.
• Function Interrelationships: Recognizes how different functions (like height and radius) affect others (like volume).

By bringing together these aspects, differential calculus allows us to model real-world scenarios mathematically. Recognizing these elements underpins many applied mathematics problems, making it easier to deal with complex relationships between varying quantities.

Applied Mathematics

Applied mathematics takes abstract concepts from mathematics and uses them to solve real-world issues, like determining the rate of change of volume in our conical sand pile. This exercise demonstrates a practical application of calculus in understanding how quantities dynamically interact:

• Modeling Real Systems: Uses mathematical equations to model physical phenomena, like sand pouring onto a surface.
• Predictive Value: Enables us to foresee changes and plan accordingly, like calculating material volumes for construction.

By combining theoretical knowledge with practical application, mathematics becomes a tool for innovation and problem-solving.
This approach is not limited to academic exercises but extends to engineering, economics, physics, and other fields where understanding change is vital.