1.4: Summary Problems
- Page ID
- 20246
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\dsum}{\displaystyle\sum\limits} \)
\( \newcommand{\dint}{\displaystyle\int\limits} \)
\( \newcommand{\dlim}{\displaystyle\lim\limits} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\(\newcommand{\longvect}{\overrightarrow}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\ket}[1]{\left| #1 \right>}\)
\(\newcommand{\bra}[1]{\left< #1 \right|}\)
\(\newcommand{\braket}[2]{\left< #1 \vphantom{#2} \right| \left. #2 \vphantom{#1} \right>}\)
\(\newcommand{\braopket}[3]{\left< #1 \vphantom{#2}\vphantom{#3} \right| #2 \vphantom{#1}\vphantom{#3} \left| #3 \vphantom{#1}\vphantom{#2} \right>}\)
\(\newcommand{\qmvec}[1]{\mathbf{\vec{#1}}}\)
\(\newcommand{\op}[1]{\hat{\mathbf{#1}}}\)
\(\newcommand{\expect}[1]{\langle #1 \rangle}\)
\(\newcommand{\dfn}[1]{\emph{\textbf{#1}}}\)
1.4 Summary Problems
1. Using the measurements provided, determine the %slope of the following slopes between Points A and B.
2. On a 60%slope, Todd wants to walk up a slope a distance equivalent to 100 feet horizontal distance. How far should he walk from Point A?
3. On the contour maps below, determine the average slopes between Points A and B. The scale is 1”=2000’. The contour interval is 80’.
Answers to Summary Questions
1.
%slope
1a.
111%
1b.
35%
2. On a 60% slope, we know that the rise is 60% of the run. Therefore, the rise here should be 60% of 100 feet or 60 feet. Using the Pythagorean Theorem, we can solve for the hypotenuse.
a2 + b2 = c2 where:
1002 + 602 = c2
13,600 = c2
ft.
3. The answers to these questions will depend upon how you measured the horizontal, or map distance – hard to do on a screen. My measurements are shown on the maps below:
At left. Point A is ≈ 3440’. Point B is ≈ 3720’. The rise is 280’. The run is ≈ 2200’. Therefore, the average slope is (280)(100)/2200 = 13%.
At right. Point A is ≈ 4040’. Point B is ≈ 3280’. The rise is 760’. The run is 1300’. Therefore, the average slope is (760)(100)/2200 = 58%


