Problem 83 - Calculus Volume 1

Question

Find the change in volume \(d V\) or in surface area \(d A\). \(d V\) if a circular cylinder of height 3 changes from \(r=2\) to \(r=1.9 \mathrm{~cm}\).

Step-by-Step Solution

Answer

The change in volume \( dV \) is \( -1.17\pi \) cubic centimeters.

Step 1: Understand the Formula for Volume of a Cylinder

The volume \( V \) of a cylinder is given by the formula \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height of the cylinder. In this problem, the height \( h = 3 \) cm.

Step 2: Calculate Initial Volume

Using the given radius \( r = 2 \) cm, calculate the initial volume \( V_1 \) as follows: \[ V_1 = \pi (2)^2 (3) = 12\pi \] cubic centimeters.

Step 3: Calculate New Volume with Decreased Radius

Calculate the new volume \( V_2 \) when the radius changes to \( r = 1.9 \) cm:\[ V_2 = \pi (1.9)^2 (3) \]. Perform the calculation: \( V_2 = \pi (3.61)(3) = 10.83\pi \) cubic centimeters.

Step 4: Determine the Change in Volume

The change in volume \( dV \) is found by subtracting the new volume from the initial volume:\[ dV = V_2 - V_1 = 10.83\pi - 12\pi = -1.17\pi \] cubic centimeters.

Key Concepts

Volume of a Cylinder

Understanding the volume of a cylinder is key in solving many real-world problems. The formula to find the volume of a cylinder is \( V = \pi r^2 h \). Simply put, this formula calculates how much space is inside the cylinder. In the formula, \( \pi \) is a constant roughly equal to 3.14159. It represents the ratio of a circle's circumference to its diameter. The variable \( r \) stands for the radius of the cylindrical base, and it is squared in the equation. Squaring the radius accounts for the circular area of the base. Finally, \( h \) represents the height of the cylinder, which gives the formula its three-dimensional aspect. Let's break it down with an example: If you have a cylinder with a radius of 2 cm and height of 3 cm, plug these numbers into the formula to find the volume:

• Square the radius: \( 2^2 = 4 \)
• Multiply by the height: \( 4 \times 3 = 12 \)
• Multiply by \( \pi \): \( 12 \pi \)

Thus, the volume is \( 12 \pi \) cubic centimeters. Understanding this formula helps in calculating how the volume changes when the radius or height changes.

Change in Volume

When we're talking about a change in volume, we're interested in how the volume of the cylinder changes as some of its dimensions change. In the exercise, the cylinder's radius decreases from 2 cm to 1.9 cm, while the height remains constant. To see how this change affects the volume, we calculate the volume with both radii and then find the difference. This difference is known as \( dV \) or the change in volume.Here's how it's done:

• Calculate the initial volume with radius 2 cm: \( V_1 = 12\pi \).
• Calculate the new volume with radius 1.9 cm: Find \( V_2 = \pi (1.9)^2 (3) = 10.83\pi \).
• Determine the change: \( dV = V_2 - V_1 = 10.83\pi - 12\pi = -1.17\pi \).

This calculation shows a decrease in volume, which makes sense since the radius became smaller. The negative sign in \( dV \) indicates a reduction. Such operations are essential in fields like engineering and manufacturing where precision in volume adjustments are critical.

Differentiation Techniques

Differentiation is a powerful tool in calculus that deals with how a function changes as its input changes. In simpler terms, differentiation helps us understand the rate of change or slope of a function at any given point. When applied to scenarios involving cylinders, it can specifically allow us to predict changes in volume when dimensions change.In the given exercise, although we manually calculated the change in volume, differentiation could offer a faster approach. We would use the derivative of the volume formula \( V = \pi r^2 h \) with respect to \( r \). Differentiation provides: \[ \frac{dV}{dr} = \frac{d}{dr}(\pi r^2 h) = 2\pi rh \]This derivative, \( 2\pi rh \), represents the rate of change of volume with respect to the radius. For small changes in \( r \), multiplying this rate by the change in radius \( \Delta r \) gives an approximation of the change in volume:\[ dV \approx 2\pi rh \Delta r \]For this problem, with \( h = 3 \) cm and \( r = 2 \) cm, and the change in \( r \) being -0.1 cm:

• \( \frac{dV}{dr} = 2\pi (2)(3) = 12\pi \)
• Approximate change: \( dV \approx 12\pi \times (-0.1) = -1.2\pi \) cubic centimeters.

Differentiation offers a quick and efficient method for estimating small changes in volume, streamlining calculations in complex scenarios.