Q. 61 - Calculus

Question

Emmy is charting a layer of basalt beneath a Hanford tank farm. She has determined that on the south end of the tank farm the basalt lies at 

\begin{equation}b_{1}(x,y)=-40 -\frac{1}{350}x+\frac{1}{725}y\end{equation}

while on the north the plane of the top of the basalt seems to be at 

\begin{equation}b_{2}(x,y)=-39.5-\frac{1}{150}x+\frac{1}{300}y\end{equation}

(a) If the surface of the basalt layer is continuous, what is the line of intersection of these planes on the Earth’s surface? 

(b) If the surface of the basalt layer is continuous, where should it be when x = 150 at the intersection of the planes? 

(c) Emmy drills a test hole at that point and finds that the basalt lies at −41.2 feet. What conclusions might she draw from this result? 

Step-by-Step Solution

Answer

(a) The Line of intersection of these planes on the Earth's surface is 

\begin{equation}y=\frac{232}{119}x-\frac{4350}{17}\end{equation}

(b) The height of the basalt layer is 36.55ft.

(c)  It tells that the surface of the basalt layer is not continuous.

Step 1. Given Information

The given equations are 

\begin{equation}b_{1}(x,y)=-40 -\frac{1}{350}x+\frac{1}{725}y\end{equation}

\begin{equation}b_{2}(x,y)=-39.5-\frac{1}{150}x+\frac{1}{300}y\end{equation}

Step 2. Solution

If the surface of the basalt layer is continuous. To find the line of intersection of the two planes, we can set the equations for the two planes equal to each other and solve for the variables x and y. This gives us: 

\begin{equation}b_{1}(x,y)=b_{2)(x,y)\end{equation}

\begin{equation}-40 -\frac{1}{350}x+\frac{1}{725}y =-39.5-\frac{1}{150}x+\frac{1}{300}y \end{equation}

Combining like terms and rearranging the equation gives us: 

\begin{equation}\frac{1}{150}x-\frac{1}{350}x =0.5 +\frac{1}{300}y -\frac{1}{725}y \end{equation}

\begin{equation}\frac{200}{150*350}x=1/2+\frac{425}{300*725}y\end{equation}

\begin{equation}\frac{2}{15*35}x=1/2+\frac{17}{300*29}\end{equation}

\begin{equation}\frac{1}{75}(\frac{2x}{7}-\frac{17y}{4*29})=1/2\end{equation}

\begin{equation}\frac{1}{75}(\frac{232x-119y}{14*29})=1\end{equation}

\begin{equation}119y=232x-30450\end{equation}

\begin{equation}y=\frac{232}{119}x-\frac{4350}{17}\end{equation}

Step 3. Given information

The surface of the basalt layer is continuous, we have to find the intersection of planes when x=150.

Step 4. Solution

If the surface of the basalt layer is continuous, to know where it is we can substitute x=150 in the equation of intersection of the lines found above i.e

\begin{equation}119y=232x-30450\end{equation}

\begin{equation}119y=232(150)-30450\end{equation}

\begin{equation}119y=34800-30450\end{equation}

\begin{equation}119y=4350\end{equation}

\begin{equation}y=\frac{4350}{119}\end{equation}

\begin{equation}y=36.55\;(approx)\end{equation}

The height of the basalt layer is 36.55

Step 5. Given information

A test hole is drilled by Emmy at that point and finds that the basalt lies at −41.2 feet. 

Step 6. Solution

Emmy's measurement of -41.2 at the intersection of the planes suggests that the surface of the basalt layer is not continuous at that point. This could be due to a discontinuity in the basalt layer itself, or it could be due to some other factor such as erosion or tectonic activity.