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\markright{Chapter 1: Worksheet 4}
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\Bf{\Star\ Chapter 1: Worksheet 4  \hfill Jack K. Cohen \hfill Colorado School of Mines}

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\Cb{Return of the Asteroid Problem, Lines, Lines, Lines}
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\noindent \Bf{Suggested Problems}
Section 1.2: 11, 15, 17, 21

\begin{enumerate}

\item \Star\  (Asteroid problem continued) Recall that you landed on a spherical asteroid. Your partner walked 1000 feet away along the smooth surface carrying a 10 feet rod, and thereby vanished over the horizon.  When she placed one end on the ground and held  the rod straight up and down, you (lying on your stomach to rest from the arduous journey from Earth) could just barely see the tip of the rod.    You now know that the correct equation for $r$ is $\cos (s/r) = r/(r + h)$, or in numerical form:  $\cos(1000/r) = r/(r + 20)$.  Using \Tt{Plot} and/or \Tt{Table}, find the physically sensible solution to two decimal places.

\noindent \Bf{Comment}:  How much would you contribute to the Indigent Math Professor's Support Fund for an easily derived estimate of the solution to within 1 foot?  Actually, this fee is covered in your tuition for this course---stay tuned.  Without questioning that \Tt{Plot} and \Tt{Table} are wonderful tools, your experience with the current problem may cause you to wonder if there aren't some better methods of solving equations.  If so, you will be glad to hear that you will, indeed, be learning about more advanced solution methods this semester.


\item (1.2.24) Find the perpendicular distance from the point (2, 1) to the line with equation $y = x + 1$.

\item A ``well-known'' formula from computational geometry gives the perpendicular distance $d$ from an arbitrary point $P0:(x_0, y_0)$ to an arbitrary line $L: Ax + By = C$ as \[d = \frac{|Ax_0 + By_0 - C|}{\sqrt{A^2 + B^2}}. \]  Use this formula to check your answer for the previous problem.

\item \Bf{Honor Problem}:  You know everything needed to derive the formula cited in the previous problem.  However, it's a nasty computation by hand.  Solutions accepted for the next two weeks---a prize to the one with the nicest proof!

\item (1.2.25) Find the perpendicular distance between the parallel lines $y = 5x + 1$ and $y = 5x + 9$.

\item Another ``well-known'' formula from computational geometry gives the perpendicular distance between the parallel lines $L_1: Ax + By = C_1$ and $L_2: Ax + By = C_2$ as \[d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}}. \] 
        \begin{enumerate}
        \item Use this formula to check your answer for the previous problem.
        \item Explain why the formulas given for $L_1$ and $L_2$ accurately characterize an \Em{arbitrary} pair of parallel lines (e.g., why not write them as $y = mx +b_1, y = mx +b_2$?).
        \end{enumerate}

\item In analogy with the usage in the geometry of a circle, a \Em{secant line} to an arbitrary curve  is defined as a line that intersects the curve in two (or more) points. Consider the curve $y = x^2$ and two points on it, say $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$.
        \begin{enumerate}
        \item When $x_1 = 1$ and $x_2 = 2$, find the secant line that passes through $P_1$ and $P_2$.
        
        \item Repeat for $x_1 = 1$ and $x_2 = 1.1$.
        
        \item Repeat for $x_1 = 1$ and $x_2 = 1.01$.
        
        \item Make an intuitive estimate for the slope of the \Em{tangent line} to $y = x^2$ at $x = 1$.  Use your estimate to write a formula for the tangent line at $x = 1$.
        
        \item Graph $y = x^2$ in a small region near $x = 1$ (i.e., ``zoom in'') and verify your tangent slope estimate (look out for possibly different scales on the two axes).
        
        \item Notice that the phrase \Em{tangent line} hasn't been given a definition in part (d) above.  Would you care to fill that gap?  Be sure to think about situations like that shown in Figure 1---we \Em{do} want to define the straight line as being a tangent line (at $x \approx 0.2$).  Is this line also a secant line?
        \end{enumerate} 
        
\begin{figure}[htb]
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\centerline{\epsffile{ws4p6.eps}}
\caption{Curve and a tangent line} 
\end{figure}


\end{enumerate}
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