Chapter 18, Problem 14PEB

Earthquake data from a subduction zone are shown in the following table. What is the slope of the subducting plate in m/km?

	Distance from Trench (km)	Earthquake Depth (km)
1	12.1	4.2
2	25.4	9.1
3	33.2	11.8
4	48.6	17.6
5	75.7	28.4

To determine

The slope of the subduction zone in m/km.

Answer to Problem 14PEB

Solution:

The average slope of the subduction zone is $\underset{\_}{-360\text{ m/km}}$.

Explanation of Solution

Given data:

The following data has been provided:

	Distance from Trench (km)	Earthquake Depth (km)
1	12.1	4.2
2	25.4	9.1
3	33.2	11.8
4	48.6	17.6
5	75.7	28.4

Formula used:

Write the expression for slope.

$\text{slope}=\frac{\Delta y}{\Delta x}$

Here, $\Delta y$ and $\Delta x$ are the changes in $y$ and $x$ axis respectively.

Write the expression for the average slope.

${\text{slope}}_{\text{average}}=\frac{{\text{slope}}_1+{\text{slope}}_2+{\text{slope}}_3+{\text{slope}}_4+{\text{slope}}_5}{5}$

Here, ${\text{slope}}_1$, ${\text{slope}}_2$, ${\text{slope}}_3$, ${\text{slope}}_4$ and ${\text{slope}}_5$ are the slope for the case $1$, $2$, $3$, $4$ and $5$ respectively.

Explanation:

The upper surface of the subduction plate is delineated by the earthquake foci and thus the distance of the earthquake from the trench and the depth can be used to determine the slope of the subducting plate. Consider the depth of earthquake on the $y$ axis and the distance from trench on the $x$ axis. There are five cases mentioned in the problem. Consider the case $1$.

Case 1: Convert $-4.2\text{ km}$ in meters.

$\begin{array}{ccc}\hfill \Delta y& =& -4.2\text{ km}\left(\frac{1000\text{ m}}{1\text{ km}}\right)\hfill \\ \hfill & =& -4.2\times {10}^3\text{ m}\hfill \end{array}$

Here, $\Delta y$ is negative because it represents the depth.

Calculate the slope when the earthquake depth is $4.2\text{ km}$.

${\text{slope}}_1=\frac{\Delta y}{\Delta x}$

Substitute $-4.2\times {10}^3\text{ m}$ for $\Delta y$ and $12.1\text{ km}$ for $\Delta x$.

$\begin{array}{ccc}\hfill {\text{slope}}_1& =& \frac{-4.2\times {10}^3\text{ m}}{12.1\text{ km}}\hfill \\ \hfill & =& -347.10\text{ m/km}\hfill \end{array}$

Consider case $2$.

Case 2: Convert $-9.1\text{ km}$ in meters.

$\begin{array}{ccc}\hfill \Delta y& =& -9.1\text{ km}\left(\frac{1000\text{ m}}{1\text{ km}}\right)\hfill \\ \hfill & =& -9.1\times {10}^3\text{ m}\hfill \end{array}$

Calculate the slope when the earthquake depth is $9.1\text{ km}$.

${\text{slope}}_2=\frac{\Delta y}{\Delta x}$

Substitute $-9.1\times {10}^3\text{ m}$ for $\Delta y$ and $25.4\text{ km}$ for $\Delta x$.

$\begin{array}{ccc}\hfill {\text{slope}}_2& =& \frac{-9.1\times {10}^3\text{ m}}{25.4\text{ km}}\hfill \\ \hfill & =& -358.26\text{ m/km}\hfill \end{array}$

Consider case $3$.

Case 3: Convert $-11.8\text{ km}$ in meters.

$\begin{array}{ccc}\hfill \Delta y& =& -11.8\text{ km}\left(\frac{1000\text{ m}}{1\text{ km}}\right)\hfill \\ \hfill & =& -1.18\times {10}^4\text{ m}\hfill \end{array}$

Calculate the slope when the earthquake depth is $11.8\text{ km}$.

${\text{slope}}_3=\frac{\Delta y}{\Delta x}$

Substitute $-1.18\times {10}^4\text{ m}$ for $\Delta y$ and $33.2\text{ km}$ for $\Delta x$.

$\begin{array}{ccc}\hfill {\text{slope}}_3& =& \frac{-1.18\times {10}^4\text{ m}}{33.2\text{ km}}\hfill \\ \hfill & =& -355.42\text{ m/km}\hfill \end{array}$

Consider case $4$.

Case 4: Convert $-17.6\text{ km}$ in meters.

$\begin{array}{ccc}\hfill \Delta y& =& -17.6\text{ km}\left(\frac{1000\text{ m}}{1\text{ km}}\right)\hfill \\ \hfill & =& -1.76\times {10}^4\text{ m}\hfill \end{array}$

Calculate the slope when the earthquake depth is $17.6\text{ km}$.

${\text{slope}}_4=\frac{\Delta y}{\Delta x}$

Substitute $-1.76\times {10}^4\text{ m}$ for $\Delta y$ and $48.6\text{ km}$ for $\Delta x$.

$\begin{array}{ccc}\hfill {\text{slope}}_4& =& \frac{-1.76\times {10}^4\text{ m}}{48.6\text{ km}}\hfill \\ \hfill & =& -362.14\text{ m/km}\hfill \end{array}$

Consider case $5$.

Case 5: Convert $-28.4\text{ km}$ in meters.

$\begin{array}{ccc}\hfill \Delta y& =& -28.4\text{ km}\left(\frac{1000\text{ m}}{1\text{ km}}\right)\hfill \\ \hfill & =& -2.84\times {10}^4\text{ m}\hfill \end{array}$

Calculate the slope when the earthquake depth is $28.4\text{ km}$.

${\text{slope}}_5=\frac{\Delta y}{\Delta x}$

Substitute $-2.84\times {10}^4\text{ m}$ for $\Delta y$ and $75.6\text{ km}$ for $\Delta x$.

$\begin{array}{ccc}\hfill {\text{slope}}_5& =& \frac{-2.84\times {10}^4\text{ m}}{75.6\text{ km}}\hfill \\ \hfill & =& -375.66\text{ m/km}\hfill \end{array}$

Now, calculate the average slope of the subducting plate.

${\text{slope}}_{\text{average}}=\frac{{\text{slope}}_1+{\text{slope}}_2+{\text{slope}}_3+{\text{slope}}_4+{\text{slope}}_5}{5}$

Substitute $-347.10\text{ m/km}$ for ${\text{slope}}_1$, $-358.26\text{ m/km}$ for ${\text{slope}}_2$, $-355.42\text{ m/km}$ for ${\text{slope}}_3$, $-362.14\text{ m/km}$ for ${\text{slope}}_4$, and $-375.66\text{ m/km}$ for ${\text{slope}}_5$.

$\begin{array}{ccc}\hfill {\text{slope}}_{\text{average}}& =& \frac{\left(-347.10\text{ m/km}\right)+\left(-358.26\text{ m/km}\right)+\left(-355.42\text{ m/km}\right)+\left(-362.14\text{ m/km}\right)+\left(-375.66\text{ m/km}\right)}{5}\hfill \\ \hfill & =& -359.71\text{ m/km}\hfill \\ \hfill & =& -360\text{ m/km}\hfill \end{array}$

Conclusion:

The average slope of the subduction zone is $\underset{\_}{-360\text{ m/km}}$.