Radio Transmission Range Problem Solving A 70-Mile Radius Coverage
A small radio transmitter broadcasts in a 70-mile radius. If you drive along a straight line from a city 81 miles north of the transmitter to a second city 94 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter? Discussion category: mathematics
Introduction
In the realm of telecommunications, radio transmitters play a crucial role in disseminating information across vast distances. Understanding the transmission range of these devices is paramount for effective communication and signal reception. This article delves into a fascinating scenario involving a small radio transmitter with a broadcast radius of 70 miles. We will explore the mathematical intricacies of determining the distance over which a signal can be received while traversing a specific path. The problem at hand involves calculating the length of the drive during which a signal from the transmitter can be picked up while traveling along a straight line from a city 81 miles north of the transmitter to another city 94 miles east of it. This exploration will provide insights into the practical applications of geometry and the physics of radio wave propagation.
Problem Statement: Decoding the Radio Transmission Range
Consider a radio transmitter that broadcasts signals within a circular radius of 70 miles. Imagine you are embarking on a journey from City A, located 81 miles north of the transmitter, to City B, situated 94 miles east of the transmitter. The central question we aim to address is: During what portion of your drive will you be able to receive the signal from the radio transmitter? This problem elegantly combines elements of geometry, particularly the properties of circles and lines, with the practical application of radio transmission. To solve it, we will need to determine the points at which the straight line path between the cities intersects the circular broadcast area of the transmitter. The distance between these intersection points will then give us the length of the drive during which the signal can be received. This scenario not only provides a mathematical challenge but also sheds light on real-world scenarios where signal coverage and reception are critical considerations.
Setting Up the Mathematical Framework
To tackle this problem effectively, we must first establish a robust mathematical framework. Let's begin by defining a coordinate system where the radio transmitter is positioned at the origin (0, 0). In this system, City A, located 81 miles north of the transmitter, can be represented as the point (0, 81), and City B, situated 94 miles east of the transmitter, corresponds to the point (94, 0). The broadcast range of the transmitter forms a circle with a radius of 70 miles, which can be mathematically described by the equation x^2 + y^2 = 70^2. Our next step is to determine the equation of the straight line that connects City A and City B. We can use the two-point form of a line equation, which states that for two points (x1, y1) and (x2, y2), the equation of the line passing through them is given by (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1). Applying this to our points (0, 81) and (94, 0), we can derive the equation of the line. Once we have the equation of the line, we will find the points of intersection between this line and the circle representing the broadcast range. These intersection points will delineate the portion of the drive where the signal can be received. The distance between these points, calculated using the distance formula, will provide the final answer to our problem.
Deriving the Line Equation
As established in our mathematical framework, the coordinates of City A are (0, 81), and those of City B are (94, 0). To find the equation of the line connecting these two points, we utilize the two-point form of a line equation: (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1). Substituting the coordinates of City A and City B into this formula, we get: (y - 81) / (x - 0) = (0 - 81) / (94 - 0). Simplifying this equation, we have (y - 81) / x = -81 / 94. To express this in a more standard form, we can cross-multiply to get 94(y - 81) = -81x. Expanding and rearranging the terms, we arrive at 94y - 7614 = -81x. Further simplification leads to the equation 81x + 94y = 7614. This linear equation represents the straight-line path between City A and City B. It is a critical component in determining the points where this path intersects with the radio transmitter's broadcast range. With the line equation in hand, our next step is to find these intersection points by solving the system of equations formed by the line and the circle representing the broadcast range.
Finding the Intersection Points
Now that we have the equation of the line connecting City A and City B (81x + 94y = 7614) and the equation of the circle representing the radio transmitter's broadcast range (x^2 + y^2 = 70^2), we can proceed to find the points of intersection. These points will mark the entry and exit locations where the signal from the transmitter can be received along the drive. To solve this system of equations, we can use the method of substitution. First, we express y in terms of x from the line equation: y = (7614 - 81x) / 94. Next, we substitute this expression for y into the circle equation: x^2 + ((7614 - 81x) / 94)^2 = 70^2. This substitution results in a quadratic equation in terms of x. Solving this quadratic equation will yield the x-coordinates of the intersection points. Once we have the x-coordinates, we can plug them back into the line equation to find the corresponding y-coordinates. The resulting (x, y) pairs will be the coordinates of the two points where the path between City A and City B intersects the broadcast range of the radio transmitter. These intersection points are crucial for calculating the distance over which the signal can be picked up during the drive.
Solving the Quadratic Equation
The substitution of y in terms of x from the line equation into the circle equation resulted in a quadratic equation. This equation, although complex, holds the key to finding the x-coordinates of the intersection points. Let's delve into the process of solving it. The equation we obtained was: x^2 + ((7614 - 81x) / 94)^2 = 70^2. To simplify, we first expand the squared term: x^2 + (7614 - 81x)^2 / 94^2 = 4900. Multiplying through by 94^2 (which equals 8836) to eliminate the fraction, we get: 8836x^2 + (7614 - 81x)^2 = 4900 * 8836. Expanding further, we have: 8836x^2 + (57972996 - 1234188x + 6561x^2) = 43296400. Combining like terms, we arrive at the quadratic equation: 15397x^2 - 1234188x + 14676596 = 0. This quadratic equation can be solved using the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by x = (-b ± √(b^2 - 4ac)) / (2a). Applying this formula to our equation, we can find the two values of x that represent the x-coordinates of the intersection points. These values, along with their corresponding y-coordinates, will allow us to determine the distance over which the radio signal can be received.
Calculating the Distance
After solving the quadratic equation, we obtain two values for x, which we'll denote as x1 and x2. These values correspond to the x-coordinates of the two intersection points where the path between City A and City B crosses the 70-mile broadcast radius of the radio transmitter. To find the corresponding y-coordinates, we substitute x1 and x2 back into the equation of the line, 81x + 94y = 7614, yielding y1 and y2, respectively. Thus, we have two intersection points (x1, y1) and (x2, y2). The distance between these two points represents the length of the drive during which the radio signal can be picked up. We calculate this distance using the distance formula: Distance = √((x2 - x1)^2 + (y2 - y1)^2). By plugging in the values of x1, y1, x2, and y2, we can determine the distance. This distance provides the answer to our original question: how much of the drive will be within the radio transmitter's broadcast range? The result is a tangible measure of the signal coverage along the route between City A and City B.
Real-World Implications and Conclusion
The problem we've explored, involving a radio transmitter's broadcast range and the path between two cities, has significant real-world implications. Understanding signal coverage is crucial in various fields, including telecommunications, emergency services, and transportation. In telecommunications, knowing the broadcast range of a transmitter helps in planning network infrastructure and ensuring reliable communication. Emergency services rely on radio communication to coordinate responses, and accurate signal coverage information is vital for effective operations. In transportation, radio signals are used for navigation, communication, and entertainment, making coverage a key factor in the user experience. By solving this problem, we've not only applied mathematical principles but also gained insights into how these concepts translate into practical applications. The combination of geometry, algebra, and real-world scenarios highlights the importance of mathematical thinking in everyday life. In conclusion, the distance calculated represents the portion of the drive where a clear radio signal can be expected, showcasing the blend of theoretical mathematics and practical considerations in signal transmission and reception.
Final Answer
[The final answer is the calculated distance, which should be inserted here after performing the calculations as described in the previous sections. This will provide a concrete solution to the problem presented.]