Problem 1 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

A cylinder 80 centimeters tall with a 10 -centimeter radius is filled with a compressible substance. The density of this substance is given by \(\rho(h)\) grams per cubic centimeter, where \(h\) is the height (in centimeters) from the bottom of the cylinder. Write an expression for the total mass of the substance in the cylinder.

Step-by-Step Solution

Answer

The expression for the total mass of the substance in the cylinder is given by \[m = \int_{0}^{80} \rho(h) \times [100\pi \times dh]\]

Step 1: Identify the Density Function

The density function is given to be \(\rho(h)\). This is important because the density varies at different heights from the bottom of the cylinder.

Step 2: Determine the Volume Function

The volume of a small layer with thickness \(\delta h\) from the bottom is, \(dV = \pi \times (radius^2) \times \(\delta h\) = \pi \times (10 cm)^2 \times \(\delta h\) = 100\(\pi \times \delta h\) cubic cm.

Step 3: Execute the Mass Calculation

The small mass \(\delta m\) contained in this small layer is, \(\delta m = \rho \times dV = \rho(h) \times [100\pi \times \delta h]\). We now integrate over the height of the cylinder to get the total mass. \[m = \int_{0}^{80} \rho(h) \times [100\pi \times dh] grams\] This is the expression for the total mass in the cylinder. The exact total mass can only be determined if the actual function for \(\rho(h)\) is known.

Key Concepts

Density Function

In calculus, a density function, often denoted by \( \rho(h) \), describes how mass is distributed in a particular object depending on a variable, which can be height, in this context. Here, \( h \) represents the height from the bottom of the cylinder. Notice that this density function might change with height due to the compressible nature of the substance.

Understanding this concept is crucial since, unlike uniform substances, the density here isn't constant and can change at different elevations inside our 3D shape. For example, substances compressed due to pressure might be denser at lower sections.

To calculate the total mass, we need to consider these variations of density at different heights and how they contribute to overall mass distribution. By integrating the density function over the height of the cylinder, we correctly account for these differences, accumulating a sum of all tiny mass layers across the entire height.

Remember, without the exact form of \( \rho(h) \), the total mass can't be determined precisely. However, setting up this integral expression helps frame the problem accurately for any given \( \rho(h) \).

Cylinder Volume

The volume of a cylinder can be calculated easily using calculus concepts. The formula for a cylindrical volume of a slice or "layer" comes from determining the area of the circular base and multiplying it by the thickness of the layer. This is especially useful when dealing with substances whose properties change with height.

In this exercise, the volume of an infinitesimally thin layer at height \( h \) with thickness \( \delta h \) is given by the formula:

• \( dV = \pi \times (radius^2) \times \delta h \),
• Here, \( \text{radius} = 10 \) cm, so \( dV = 100\pi\delta h \).

Each of these tiny volumes corresponds to a circular slice across the cylinder. By slicing the cylinder into many such layers from bottom to top, we can analyze how they differ and accumulate them for various purposes, such as mass calculations.

This foundational geometric concept is useful in a variety of contexts, and knowing how to calculate it allows us to tackle complex situations where properties such as density vary with height.

Mass Calculation

Calculating mass from density and volume is a core application in physics and engineering. When encountering substances with varying densities, calculating the total mass involves integrating the mass contributions from all layers of the object, considering their respective densities.

Here's how you approach it:

• The small mass of an individual layer \( \delta m \) at height \( h \) is a product of its density and volume, which is given by \( \delta m = \rho(h) \times dV = \rho(h) \times [100\pi\delta h] \).
• To find the total mass, we integrate over the entire height of the cylinder: \[ m = \int_{0}^{80} \rho(h) \times [100\pi \, dh]. \]

This integral computation effectively "adds up" the small contributions of every slice from bottom to top of the cylinder. Each part with unique density plays its part in the overall mass calculation, which is why providing the exact function of \( \rho(h) \) is essential for the precise mass result.

This exercise highlights calculus's power in solving real-world problems involving continuous distributions and varying properties. By understanding the relationship between these elements, we unlock the potential to model a range of complex physical situations.