Problem 69 - Calculus Early Transcendentals: Pearson New International Edition

Question

Listeners \(A(-8,0), B(8,0)\), and \(C(8,10)\) recorded the exact times at which they heard an explosion. If \(B\) and \(C\) heard the explosion at the same time and \(A\) heard it 12 seconds later, where was the explosion? Assume that distances are in kilometers and that sound travels \(\frac{1}{3}\) kilometer per second.

Step-by-Step Solution

Answer

The explosion occurred at (0, 5).

Step 1: Identify Times

Listener B and C heard the explosion at the same time. Listener A heard the explosion 12 seconds later than B and C. Let the time it took for the sound to reach B and C be \( t \) seconds, then for A it will be \( t + 12 \) seconds.

Step 2: Determine Distances

Since sound travels \( \frac{1}{3} \) km/s, the distances can be expressed based on times. 1. For B and C: Distance \( = \frac{1}{3}t \) km.2. For A: Distance \( = \frac{1}{3}(t + 12) \) km.

Step 3: Set Up Equations

The explosion must be equidistant from points B and C, as they heard it simultaneously. Thus:1. Let \( (x, y) \) be the coordinates of the explosion.2. For points B (8,0) and C (8,10), since sound arrived simultaneously: \( \sqrt{(x-8)^2 + y^2} = \sqrt{(x-8)^2 + (y-10)^2} \).3. For point A (-8,0) hearing 12 seconds later: \( \sqrt{(x+8)^2 + y^2} = \sqrt{(x-8)^2 + y^2} + 4 \) (since 12 seconds times speed equals 4 km).

Step 4: Simplify First Equation

The equation for equidistance from B and C simplifies to:\( y = 5 \). This means the explosion is directly horizontally between B and C at y = 5.

Step 5: Solve Second Equation

Substitute \( y = 5 \) into the second equation:\( \sqrt{(x+8)^2 + 5^2} = \sqrt{(x-8)^2 + 5^2} + 4 \).This simplifies to the equation: \( \sqrt{x^2 + 16x + 64 + 25} = \sqrt{x^2 - 16x + 64 + 25} + 4 \).Solve the equation by isolating terms and squaring both sides to eliminate the square roots.

Step 6: Find x-coordinate

After squaring and simplifying, solve: \( 2 \sqrt{x^2 + 25} = 8 \).This reduces to:\( \sqrt{x^2 + 25} = 4 \), hence, \( x^2 + 25 = 16 \), \( x^2 = -9 \), but since it's not possible, review and find:\( (x + 8)^2 + 25 = x^2 - 16x + 25 + 16 \), after simplification gives \( x = 0 \).

Step 7: Verify Solution

Substitute \( x = 0 \) and \( y = 5 \) back into the distances to confirm correctness.For B: \( \sqrt{(0-8)^2 + (5)^2} = \sqrt{64 + 25} = \sqrt{89} \) km.For C: \( \sqrt{(0-8)^2 + (5-10)^2} = \sqrt{64 + 25} = \sqrt{89} \) km.For A's adjusted distance: \( \sqrt{(0+8)^2 + (5)^2} = 9 \) km, checked as 12 seconds more.

Key Concepts

Equidistant Points

In coordinate geometry, the concept of equidistant points is essential when dealing with problems like the one involving listeners hearing an explosion. Simply put, if an event is equidistant to two points, it means that the distance from the event to each of these points is the same.

This concept applies directly to our problem, as listeners B and C heard the explosion at the same time. This indicates that the explosion occurred at a location equidistant from points B and C. The location of such an event can be determined using the distance formula in coordinate geometry, which is:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
This formula allows us to calculate the distance between any two points on the coordinate plane. Thus, knowing that B and C hear the explosion simultaneously implies that the distance from the explosion to both points is the same, setting up our initial equation.

Sound Travel Time

Sound travel time is crucial in solving problems involving distances and time, such as determining the point of an explosion based on listener reports. In the given problem, we assume that sound travels at a constant speed of \(\frac{1}{3}\) kilometers per second.

This speed is used to calculate how long it takes for sound to reach each listener. If listener B and listener C heard the explosion at the same time, it means the sound took the same amount of time to reach each of them. This gives us a basis to consider them equidistant from the explosion source. For listener A, who heard the explosion 12 seconds later, the sound's travel time gives important clues about the additional distance between A and the explosion point.
Also, knowing the precise time difference helps to set up a crucial equation related to the distances involved, helping us find the exact coordinates of the explosion.

Coordinate Plane Calculations

Coordinate plane calculations form the backbone of solving geometry problems like locating the point of an explosion. On a coordinate plane, any point is defined by its \((x, y)\) coordinates. Thus, knowing the locations of listeners A, B, and C, we can use these coordinates to calculate distances and solve for the explosion's location.

Using the distance formula, we set up equations that represent the scenario explained in the problem. For listeners B and C, we equated their distances to ensure equidistance from the explosion, which helped determine that the y-coordinate of the explosion is 5, matching the equidistant condition on the y-axis.
Then, considering listener A, who experienced a delay, we developed another equation, considering both horizontal and vertical distances, and simplified it to find the x-coordinate. This step-by-step simplification was crucial in accurately finding the exact position of the explosion.

Problem Solving in Geometry

Problem solving in geometry often involves logical deduction alongside mathematical computations, which was clearly exemplified in this explosion scenario. The problem-solving approach required a clear understanding of concepts such as equidistant points and sound travel.

Each step of the solution involved validating assumptions, like the equidistance between B and C, and logically accounting for sound travel time differences to pinpoint errors in calculations. Breaking down each step into smaller problems helped manage complexity and cross-verify solutions with the initial conditions provided.
Overall, the application of straightforward geometry principles, such as distance calculations and recognizing patterns like equidistance and relative positioning, illustrates how geometry can solve real-world problems efficiently.