Problem 66 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 65-66: If an earthquake has a total horizontal displacement of \(S\) meters along its fault line, then the horizontal movement \(M\) of a point on the surface of Earth \(d\) kilometers from the fault line can be estimated using the formula $$ M=\frac{S}{2}\left(1-\frac{2}{\pi} \tan ^{-1} \frac{d}{D}\right) $$ where \(D\) is the depth (in kilometers) below the surface of the focal point of the earthquake. Approximate the depth \(D\) of the focal point of an earthquake with \(S=3 \mathrm{~m}\) if a point on the surface of Earth 5 kilometers from the fault line moved \(0.6\) meter horizontally.

Step-by-Step Solution

Answer

The depth \( D \) of the focal point is approximately 2.886 km.

Step 1: Define the Known Values

We are given the total horizontal displacement \( S = 3 \) meters, the horizontal movement \( M = 0.6 \) meters, and the distance from the fault line \( d = 5 \) kilometers.

Step 2: Set Up the Equation

Use the formula to describe the relationship: \[M = \frac{S}{2}\left(1-\frac{2}{\pi} \tan ^{-1} \frac{d}{D}\right)\]Substituting the given values, we get \[0.6 = \frac{3}{2}\left(1 - \frac{2}{\pi}\tan^{-1} \frac{5}{D}\right)\]

Step 3: Simplify the Equation

First, calculate \( \frac{3}{2} = 1.5 \), then the equation becomes:\[0.6 = 1.5\left(1 - \frac{2}{\pi}\tan^{-1} \frac{5}{D}\right)\]

Step 4: Isolate the Inverse Tangent Expression

Divide both sides by 1.5:\[\frac{0.6}{1.5} = 1 - \frac{2}{\pi}\tan^{-1} \frac{5}{D}\]Simplifying the left side, we have:\[0.4 = 1 - \frac{2}{\pi}\tan^{-1} \frac{5}{D}\]

Step 5: Solve for the Inverse Tangent

Subtract 0.4 from 1:\[0.6 = \frac{2}{\pi}\tan^{-1} \frac{5}{D}\]

Step 6: Calculate Tangent Expression

Multiply both sides by \( \frac{\pi}{2} \):\[\frac{\pi}{2} \times 0.6 = \tan^{-1} \frac{5}{D}\]\[0.3 \pi = \tan^{-1} \frac{5}{D}\]

Step 7: Use the Tangent Function to Solve for D

Take the tangent of both sides:\[\tan(0.3 \pi) = \frac{5}{D}\]Calculate \( \tan(0.3 \pi) \), using a calculator:\[\tan(0.3 \pi) \approx -\sqrt{3} \approx -1.732\]

Step 8: Find D

Now solve for \( D \):\[D = \frac{5}{-1.732}\]\[D \approx -2.886 \text{ km}\] Given the context, consider absolute value as depths cannot be negative: \(D \approx 2.886 \text{ km}\).

Key Concepts

Horizontal Movement Formula

The Horizontal Movement Formula provides a way to understand the displacement caused by an earthquake. It is important because it helps us calculate how much land has moved along a fault line during an earthquake event.
The formula used is:

• \[ M = \frac{S}{2}\left(1-\frac{2}{\pi} \tan^{-1}\frac{d}{D}\right) \]

Here is what each part means:

• \(S\) is the total horizontal displacement in meters.
• \(M\) is the movement at a point on the Earth's surface.
• \(d\) is the distance from the fault line in kilometers.
• \(D\) is the depth of the earthquake's focal point in kilometers.

This formula models the mathematical relationship between these quantities, allowing us to estimate the shift of land depending on various factors. This is crucial for understanding the impact and adjusting construction practices near fault lines.

Approximation of Earthquake Depth

To find out how deep the earthquake's focal point is, we can use the information from surface movement during the quake. This process helps us determine the earthquake's origin depth, crucial for both academic research and public safety.
When given values like total displacement \(S\), horizontal movement \(M\), and distance \(d\), we solve the formula for \(D\). Understanding this depth can aid in assessing earthquake risk and designing better infrastructure.
In our specific example, substituting the known values allows us to rearrange and solve the equation for \(D\), which offers an approximation of the depth based on how a surface point moved. Approximations might be required due to assumptions within the model, but they provide a good estimate for practical considerations.

Inverse Tangent Function

The inverse tangent function, represented as \( \tan^{-1} \), is essential in solving our equation for an earthquake's depth. It allows us to reverse-calibrate the relationship between the horizontal displacement and the earthquake’s depth.
In our problem, \( \tan^{-1} \left( \frac{d}{D} \right) \) is adjusted to simplify solving for \(D\). This function helps link angles to ratios of sides in a right triangle, essential in calculating how seismic waves behave in the ground.
Understanding how to manipulate \( \tan^{-1} \) forms a core part of solving this earthquake displacement problem. It transforms the problem from a trigonometric context into a more easily managed algebraic form, bringing clarity to the depth approximation.

Trigonometric Problem Solving

Trigonometry is incredibly useful in solving problems pertaining to earthquake displacement. By involving functions like tangent and its inverse, it helps connect the angles with the actual distance and depth shifts.
We used trigonometry to understand the spatial dynamics of an earthquake. Multiplying and isolating specific terms eases the complexity of calculations, eventually allowing you to solve for unknown quantities.

• Start with known values and adjust the trigonometric formula.
• Simplify step by step, focusing on isolating the key variable \(D\).
• Use calculated trigonometric values to uncover relationships between distance, displacement, and depth.

These steps demonstrate the value of trigonometric problem solving in geoscience and highlight the importance of the mathematical link between observable ground movement and deeper seismic events.