1 // Example of a clearly worked problem and reasoning behind it.
2 // Command-line options to enable stepping and/or reverse video:
3 // asy [-u stepping=true] [-u reverse=true] slidedemo
8 settings.tex = "pdflatex";
11 // Commands to generate optional bibtex citations:
16 bibliographystyle("alpha");
17 //texpreamble("\usepackage{problempack}"); // optionclash
18 texpreamble("\usepackage{amsmath}");
19 texpreamble("\newcommand{\U}[1]{\text{ #1}} % units shortcut");
20 texpreamble("\newcommand{\E}[1]{\ensuremath{\cdot 10 ^{#1}}} % exponent shortcut");
21 texpreamble(bulletcolor("black"));
23 string eqnstring(string s, real width=100bp)
25 return minipage("\begin{align*}"+s+"\end{align*}",width);
27 // Override slide.equations to use the prettier align* env from amsmath
28 void equations(string s, pen p=itempen)
30 vbox("\begin{align*}"+s+"\end{align*}",p);
32 // Wrap textblock from the textpos package
34 // Generated needed files if they don't already exist.
35 asy(nativeformat(),"drawing");
37 // Optional background color or header:
39 // fill(background,box((-1,-1),(1,1)),Azure);
40 // label(background,"Header",(0,startposition.y));
43 titlepagepen = colorless(titlepagepen);
44 authorpen = colorless(authorpen);
45 institutionpen = colorless(authorpen);
46 pen questionPen = fontsize(18pt);
47 pen genEqnPen = 0.5*red + fontsize(18pt);
48 pen partEqnPen = 0.5*blue + fontsize(18pt);
49 pen solnEqnPen = 0.5*green + fontsize(18pt);
50 pair solutionposition = startposition-(0,0.5); // paper (-1,-1) to (1,1)
52 titlepage(title="Solving physics problems:\\ a detailed solution style",
53 author="W. Trevor King",
54 institution="Drexel University",
56 url="http://www.physics.drexel.edu/$\sim$wking");
58 title("Presenting solutions in an organized manner");
60 "Physics isn't just about getting the ``right answer,'' but also about
61 demonstrating to others why your answer is right and how you came to
62 that conclusion. This solution framework makes your argument clearer,
63 which will help you as you develop the argument, and others when you
64 try and teach/convince them.");
66 "This layout is along the lines of the homework solution format from my
67 high-school physics class, and I like it a lot. However, if you have
68 another format that works for you, go ahead and use that. Remember,
69 the purpose is to make the solution look obvious, with very clear
70 reasoning, both for you and any potential readers.");
73 item("Copy down the assigned problem");
74 item("Draw a picture of the problem, labeling useful quantities");
75 item("Make a list of relevant formulas");
76 item("Apply the formulas to the particular cases in the problem");
77 item("Solve for the unknown quantities");
78 item("Are your answers reasonable?");
80 title("Copy down the assigned problem");
81 remark("This step ensures you read the problem carefully");
82 currentposition=solutionposition;
87 A $10.0\U{m}$ length of wire consists of $5.0\U{m}$ of copper followed
88 by $5.0\U{m}$ of aluminum, both of diameter $1.0\U{mm}$. A voltage
89 difference of $85\U{mV}$ is placed across the composite wire.
90 (a) What is the total resistance (sum) of the two wires?
91 (b) What is the currnent through the wire?
92 (c) What are the voltages across the aluminum part and across the copper part?
96 title("Draw a picture of the problem");
98 "This step doublechecks your understanding of the problem, helps you
99 translate it into your own terms, and makes it clear what quantities
100 your various symbols refer to.");
102 currentposition=solutionposition;
107 A $10.0\U{m}$ length of wire consists of $5.0\U{m}$ of copper followed
108 by $5.0\U{m}$ of aluminum, both of diameter $1.0\U{mm}$. A voltage
109 difference of $85\U{mV}$ is placed across the composite wire.
110 (a) What is the total resistance (sum) of the two wires?
111 (b) What is the currnent through the wire?
112 (c) What are the voltages across the aluminum part and across the copper part?
118 title("Make a list of relevant formulas");
120 "This is the key step, digging deep into your physics knowledge to
121 dredge up any appropriate formulas. :p");
123 currentposition=solutionposition;
128 A $10.0\U{m}$ length of wire consists of $5.0\U{m}$ of copper followed
129 by $5.0\U{m}$ of aluminum, both of diameter $1.0\U{mm}$. A voltage
130 difference of $85\U{mV}$ is placed across the composite wire.
131 (a) What is the total resistance (sum) of the two wires?
132 (b) What is the currnent through the wire?
133 (c) What are the voltages across the aluminum part and across the copper part?
138 pair genEqnPos = currentposition+(-.1,.2);
140 R &= \rho \frac{L}{A} \\
142 R_{1,2} &= R_1 + R_2 \\
143 A &= \pi r^2 = \pi (d/2)^2
144 ", minipagewidth/3), genEqnPos, E, genEqnPen);
146 title("Apply formulas to patricular cases");
148 "Here we simply apply the general equations to the problem at hand, and
149 look up any constants we don't know.");
151 currentposition=solutionposition;
156 A $10.0\U{m}$ length of wire consists of $5.0\U{m}$ of copper followed
157 by $5.0\U{m}$ of aluminum, both of diameter $1.0\U{mm}$. A voltage
158 difference of $85\U{mV}$ is placed across the composite wire.
159 (a) What is the total resistance (sum) of the two wires?
160 (b) What is the currnent through the wire?
161 (c) What are the voltages across the aluminum part and across the copper part?
166 R &= \rho \frac{L}{A} \\
168 R_{1,2} &= R_1 + R_2 \\
169 A &= \pi r^2 = \pi (d/2)^2
170 ", minipagewidth/3), genEqnPos, E, genEqnPen);
172 pair partEqnPos = genEqnPos+(0,-.4);
174 R_T &= R_c + R_a & R_c &= \rho_c \frac{l}{A} & R_a &= \rho_a \frac{l}{A} \\
175 V_T &= I R_T & V_c &= I R_c & V_a &= I R_a \\
176 \rho_c &= 1.68\E{-8}\U{$\Omega$m} \\
177 \rho_a &= 2.65\E{-8}\U{$\Omega$m}
178 ", minipagewidth/1.8), partEqnPos, E, partEqnPen);
180 title("Solve for the unknown quantities");
181 remark("From this point out it's just math.");
182 currentposition=solutionposition;
187 A $10.0\U{m}$ length of wire consists of $5.0\U{m}$ of copper followed
188 by $5.0\U{m}$ of aluminum, both of diameter $1.0\U{mm}$. A voltage
189 difference of $85\U{mV}$ is placed across the composite wire.
190 (a) What is the total resistance (sum) of the two wires?
191 (b) What is the currnent through the wire?
192 (c) What are the voltages across the aluminum part and across the copper part?
197 R &= \rho \frac{L}{A} \\
199 R_{1,2} &= R_1 + R_2 \\
200 A &= \pi r^2 = \pi (d/2)^2
201 ", minipagewidth/3), genEqnPos, E, genEqnPen);
203 R_T &= R_c + R_a & R_c &= \rho_c \frac{l}{A} & R_a &= \rho_a \frac{l}{A} \\
204 V_T &= I R_T & V_c &= I R_c & V_a &= I R_a \\
205 \rho_c &= 1.68\E{-8}\U{$\Omega$m} \\
206 \rho_a &= 2.65\E{-8}\U{$\Omega$m}
207 ", minipagewidth/1.8), partEqnPos, E, partEqnPen);
209 pair solnEqnPos = genEqnPos+(0.9,-.1);
211 R_c &= \rho_c \frac{l}{A} = \rho_c \frac{4l}{\pi d^2} = 0.10654\U{$\Omega$}\\
212 R_a &= \rho_a \frac{l}{A} = \rho_a \frac{4l}{\pi d^2} = 0.16870\U{$\Omega$}\\
213 R_T &= R_c + R_a = 0.28\U{$\Omega$} \\
214 V_T &= IR_T \qquad I = V_T/R_T = 0.31\U{A} \\
215 V_c &= IR_c = 33\U{mV} \qquad V_a = IR_a = 52\U{mV}
216 ", minipagewidth/1.8), solnEqnPos, SE, solnEqnPen);
218 title("Are your answers reasonable?");
220 "Yes. The wire is made out of copper and aluminum, both good
221 conductors, and has a reasonable length and diameter, so we expect low
222 resistance. We also expect the more resistive aluminum to have a
223 higher resistance and a greater voltage drop.");
225 currentposition=solutionposition;
230 A $10.0\U{m}$ length of wire consists of $5.0\U{m}$ of copper followed
231 by $5.0\U{m}$ of aluminum, both of diameter $1.0\U{mm}$. A voltage
232 difference of $85\U{mV}$ is placed across the composite wire.
233 (a) What is the total resistance (sum) of the two wires?
234 (b) What is the currnent through the wire?
235 (c) What are the voltages across the aluminum part and across the copper part?
240 R &= \rho \frac{L}{A} \\
242 R_{1,2} &= R_1 + R_2 \\
243 A &= \pi r^2 = \pi (d/2)^2
244 ", minipagewidth/3), genEqnPos, E, genEqnPen);
246 R_T &= R_c + R_a & R_c &= \rho_c \frac{l}{A} & R_a &= \rho_a \frac{l}{A} \\
247 V_T &= I R_T & V_c &= I R_c & V_a &= I R_a \\
248 \rho_c &= 1.68\E{-8}\U{$\Omega$m} \\
249 \rho_a &= 2.65\E{-8}\U{$\Omega$m}
250 ", minipagewidth/1.8), partEqnPos, E, partEqnPen);
251 pair solnEqnPos = genEqnPos+(0.9,-.1);
253 R_c &= \rho_c \frac{l}{A} = \rho_c \frac{4l}{\pi d^2} = 0.10654\U{$\Omega$}\\
254 R_a &= \rho_a \frac{l}{A} = \rho_a \frac{4l}{\pi d^2} = 0.16870\U{$\Omega$}\\
255 R_T &= R_c + R_a = 0.28\U{$\Omega$} \\
256 V_T &= IR_T \qquad I = V_T/R_T = 0.31\U{A} \\
257 V_c &= IR_c = 33\U{mV} \qquad V_a = IR_a = 52\U{mV}
258 ", minipagewidth/1.8), solnEqnPos, SE, solnEqnPen);
260 title("Some thoughts");
262 "Obviously this formal approach is not neccessary for simple problems
263 that you can almost do in your head. However, for more complicated
264 problems, the extra work of drawing a labeled figure and explicitly
265 writing out the general equations you use will help you solve the
266 problem faster by making it very easy to remember what each symbol
267 means, see where you have information that you haven't used in your
268 solution yet, and get a good overview of your line of reasoning.");
269 label("This document was produced using Asymptote.
270 {\tt http://asymptote.sourceforge.net/}", (-0.8,-0.8), NE, fontsize(12pt));