Problem 23 - Technical Mathematics with Calculus

Question

Where needed, assume the earth to be a sphere with a radius of 3960 mi. Actually, the distance from pole to pole is about 27 mi less than the diameter at the equator. One circular "track" on a magnetic disk used for computer data storage is located at a radius of 155 mm from the center of the disk. If 1000 "bits" of data can be stored in \(1 \mathrm{mm}\) of this track, how many bits can be stored in the length of this track subtending an angle of \(\pi / 12\) rad?

Step-by-Step Solution

Answer

There can be \( 1000 \times \frac{\pi}{12} \times 155 \) bits stored in that length of the track.

Step 1: Calculate the Circumference of the Circular Track

The circumference (C) of a circle is calculated using the formula: \( C = 2\pi r \) where \( r \) is the radius of the circle. For a circular track with a radius of 155 mm, the circumference is \( C = 2\pi \times 155 \) millimeters.

Step 2: Find the Length of the Track Subtending the Given Angle

The length (L) of the track corresponding to the angle \( \theta \) in radians is given by the formula: \( L = \theta \times r \). With \( \theta = \frac{\pi}{12} \) radians and \( r = 155 \) mm, we find the length by \( L = \frac{\pi}{12} \times 155 \) millimeters.

Step 3: Calculate the Number of Bits that Can Be Stored

Since 1000 bits can be stored per millimeter of the track's length, we simply multiply the length of the track subtending the given angle by 1000 to find the number of bits. Therefore, the number of bits is \( 1000 \times L \).

Step 4: Compute the Final Answer

Substituting L from Step 2 into the multiplication with 1000 bits/mm, we get the total bits that can be stored which is \( 1000 \times \frac{\pi}{12} \times 155 \). Calculate this to find the final answer.

Key Concepts

Circumference Calculation

When we talk about the circumference of a circle, we're referring to the distance around the circle's edge. It's an important concept in many areas of mathematics and science, including data storage on circular tracks, like those found in a computer's magnetic disk.

To calculate the circumference, we use a simple formula: \( C = 2\pi r \), where \( r \) is the radius of the circle, or the distance from the center to the edge. This formula is vital for understanding the total data storage capacity of circular tracks, as the larger the circumference, the more storage space available. For instance, if a magnetic disk has a track with a radius of 155 mm, its circumference can be found by plugging the radius into the formula as follows: \( C = 2\pi \times 155 \).

Arc Length in Radians

An arc is a portion of the circumference of a circle. Arc length becomes especially relevant when dealing with circular tracks and data storage. When we describe an arc in terms of radians, we’re using a measure for angles based on the radius of the circle.

For calculating the length of an arc (L) subtending a given angle (\( \theta \)), in radians, the simple formula is: \( L = \theta \times r \). Assume you have a disk track radius (\( r \)) of 155 mm and an angle of \( \frac{\pi}{12} \) radians; the length of the arc is then \( L = \frac{\pi}{12} \times 155 \) mm. This particular measurement of arc length helps in determining how much data can be stored in that specific section of the track.

Data Storage Capacity

Data storage capacity refers to the total amount of data that can be saved on a storage medium. In the context of a magnetic disk, where data is stored along circular tracks, it's crucial to know how much data can fit on a given length of track.

Continuing from our arc length calculation, if a track can store 1000 bits per millimeter, the data storage capacity of an arc can be determined by multiplying the arc length in millimeters by the storage density (bits per millimeter). For example, using the previously calculated arc length subtending a \( \frac{\pi}{12} \) radian angle on a track of radius 155 mm, the data storage capacity for that arc is \( 1000 \times \frac{\pi}{12} \times 155 \) bits. This computation helps in assessing the efficiency and space utilization of the storage medium.