tag:blogger.com,1999:blog-84684715993935246782014-03-18T22:39:15.703-05:00A Suitable Collection of ExamplesRoy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comBlogger28125tag:blogger.com,1999:blog-8468471599393524678.post-10217687498101173052014-01-11T08:33:00.000-06:002014-01-11T08:41:44.959-06:00Jumping Off a CircleWhere is the best point to jump off a circle? Someone posed this question a few months back — I'm not sure who — and I've been able to resist thinking about it too deeply until this morning. As so often is the case with interesting questions, there is a bit of room for interpretation. Here's how I've chosen to interpret the question:<br><br><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-5x_ZtLp6jjg/UtFRS1mP0sI/AAAAAAAAAPw/bZ9Ti12K2cg/s1600/jumpoffcircle0.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-5x_ZtLp6jjg/UtFRS1mP0sI/AAAAAAAAAPw/bZ9Ti12K2cg/s1600/jumpoffcircle0.png" height="137" width="640"></a></div><br>Let's say an object is launched from a circle with radius 1 meter, centered on the ground, and the object is launched tangentially to the circle (see the red flight path shown above, for example). What should the angle (shown in green above) be so that the object travels as far as possible before landing?<br><br>If you think about this question a little bit — no calculation required — I bet you could come up with a pretty good rough answer. If you think about it a lot more and do quite a bit of calculation and/or simulation, you could refine it some more.<br><br><a href="http://sucolex.blogspot.com/2014/01/jumping-off-circle.html#more">Think it through, then click to go on.</a>Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-87915243442534941222013-12-01T15:59:00.000-06:002013-12-01T16:20:50.284-06:00Middles and halves and alternative definitionsMy almost-7-year-old son Blaise and I recently had the following enlightening conversation.<br /><br /><b>Blaise:</b> <i>(approaching me out of the blue)</i> Did you know that every number has a middle or a half?<br /><b>Me:</b> Huh. What about 7?<br /><b>Blaise:</b> Yeah, its middle is 4.<br /><b>Me:</b> Ohh, I see what you mean. You know, there's a name for numbers with a middle. They're called "odd." And numbers with a half are called "even."<br /><b>Blaise:</b> Oh, so 1 isn't even or odd!<br /><b>Me:</b> <i>(alarmed pause)</i> Aw, crud. <i>(longer pause)</i> Wait a minute! Isn't 1 the middle of 1?<br /><b>Blaise:</b> Oh yeah! So 1 <u>is</u> odd.<i> (pause)</i> Hey, I know why they call even numbers "even." <i>(holding up two fingers on each hand)</i> Like the number 4 — you can split it into two even parts.<br /><br />Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-69253379529628348962013-11-02T06:15:00.000-05:002013-12-01T16:00:12.585-06:00Predicting the futureEvery ten years, the U.S. Census counts the approximate number of people in the United States. The table below shows the number of people in the years 1800 through 1900. Imagine you were alive just after 1900, and you wanted to predict how many people there would be in 1920. How would you do it?<br /><br />What if you wanted to figure out how many people there were back in 1883? What about this year? If you were alive just after 1900, and you saw this list of numbers, how many people would you predict for 2013?<br /><br /><center><table border="1"><tbody><tr><td>Year </td><td>People </td></tr><tr><td>1800 </td><td>5 million </td></tr><tr><td>1810 </td><td>7 million </td></tr><tr><td>1820 </td><td>10 million </td></tr><tr><td>1830 </td><td>13 million </td></tr><tr><td>1840 </td><td>17 million </td></tr><tr><td>1850 </td><td>23 million </td></tr><tr><td>1860 </td><td>31 million </td></tr><tr><td>1870 </td><td>39 million </td></tr><tr><td>1880 </td><td>49 million </td></tr><tr><td>1890 </td><td>63 million </td></tr><tr><td>1900 </td><td>76 million </td></tr></tbody></table></center>Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-22487092205056991342013-11-02T06:08:00.001-05:002013-12-01T16:00:31.516-06:00Colorful toysLet’s say I want to make seven plastic toys (an airplane, a boat, a camera, a dinosaur, an elephant, a firetruck, and a gun), and I can make the toys in three different colors: red, green, and blue. How many different ways can I pick the colors for the toys? <br /><br />Now, what if I want to use each color at least once? How many different ways can I pick the colors for the toys then?Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-73778296592842719292013-09-28T07:36:00.000-05:002013-09-28T07:56:26.275-05:00The speed of an approaching objectThe visual angle (AKA angular size) of an object depends on how far away the object is. The passenger jet shown below has a wingspan of 200 feet. Click the image for an interactive version, where you can move the jet around.<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://www.geogebratube.org/student/m50748" target="_new"><img border="0" height="223" src="http://4.bp.blogspot.com/-33O7yrllBj4/UkbMEWS5zlI/AAAAAAAAAMQ/v6lEx_Fww30/s400/plane.png" width="400" /></a></div>Suppose a jet with a wingspan of 200 feet is approaching you. Its visual angle is 1.5°, and the angle is increasing by 0.1° per second. How far away is the jet, and how fast is it moving?Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-90900441022113769602013-09-11T10:08:00.002-05:002013-09-11T10:22:22.644-05:00Squares and cubes and primesThere's something a little strange about the square numbers (e.g. 1, 4, 9, 16, 25, 36, etc.) — the only time a square number comes right after a prime number is when the square number is 4, which follows the prime number 3. If you test this out for a few other square numbers, you may begin to suspect that square numbers (other than 4) rarely, if ever, come right after a prime number.<br><br>What I'm saying, though, is that no square number aside from 4 will ever come immediately after a prime number. It can never happen. <i style="font-weight: bold; text-decoration: underline;">Ever.</i> What an all-encompassing claim! How can I be so sure of this? Well, let's say I pick some square number. Let's call the number "<i><span style="font-family: Times, Times New Roman, serif;">n</span></i>", just for the sake of having a name for it. What does it mean when I say <i><span style="font-family: Times, Times New Roman, serif;">n</span></i> is square? It means <i><span style="font-family: Times, Times New Roman, serif;">n</span></i> is the square of some other number, like 1, 2, 3, 4, etc. In other words, let's say <i><span style="font-family: Times, Times New Roman, serif;">n</span></i> is the square of some positive whole number, which we might refer to as "<i><span style="font-family: Times, Times New Roman, serif;">m</span></i>" (or whatever letter suits you). So <span style="font-family: Times, Times New Roman, serif;"><i>n</i> = <i>m</i><sup>2</sup></span>. Now, what is the number that comes right before <i><span style="font-family: Times, Times New Roman, serif;">n</span></i>? It would be the number <span style="font-family: Times, Times New Roman, serif;"><i>n</i> - 1</span>, which is really <span style="font-family: Times, Times New Roman, serif;"><i>m</i><sup>2</sup> - 1</span>.<br><br>Now, why on Earth couldn't <span style="font-family: Times, Times New Roman, serif;"><i>m</i><sup>2</sup> - 1</span> be a prime number? Well, another way to write the number <span style="font-family: Times, Times New Roman, serif;"><i>m</i><sup>2</sup> - 1</span> is this: <span style="font-family: Times, Times New Roman, serif;">(<i>m </i>- 1)(<i>m </i>+ 1)</span>. And unless I'm mistaken, <span style="font-family: Times, Times New Roman, serif;"><i>m </i>- 1</span> and <span style="font-family: Times, Times New Roman, serif;"><i>m </i>+ 1</span> are both whole numbers, so my previous sentence means is that no matter what <i><span style="font-family: Times, Times New Roman, serif;">m</span></i> is, <span style="font-family: Times, Times New Roman, serif;"><i>m</i><sup>2</sup> - 1</span> can always be "factored" into two whole numbers (specifically, whatever the numbers <span style="font-family: Times, Times New Roman, serif;"><i>m</i> - 1</span> and <span style="font-family: Times, Times New Roman, serif;"><i>m</i> + 1</span> are).<br><br>Since <span style="font-family: Times, Times New Roman, serif;"><i>m</i><sup>2</sup> - 1</span> can always be written as the product of the numbers <span style="font-family: Times, Times New Roman, serif;"><i>m</i> - 1</span> and <span style="font-family: Times, Times New Roman, serif;"><i>m</i> + 1</span>, the only way <span style="font-family: Times, Times New Roman, serif;"><i>m</i><sup>2</sup> - 1</span> might be prime is if one of those two numbers turns out the be<span style="font-family: inherit;"> 1. B</span>ut since <i><span style="font-family: Times, Times New Roman, serif;">m</span></i> is a positive whole number, <span style="font-family: Times, Times New Roman, serif;"><i>m</i> + 1</span> has to be at l<span style="font-family: inherit;">east 2. C</span>an <span style="font-family: Times, Times New Roman, serif;"><i>m</i> - 1</span> be equ<span style="font-family: inherit;">al to 1? Su</span>re it can! But only if <i><span style="font-family: Times, Times New Roman, serif;">m</span></i> i<span style="font-family: inherit;">s 2. In s</span>hort, the only way <span style="font-family: Times, Times New Roman, serif;"><i>m</i><sup>2</sup> - 1</span> might be prime is if <i><span style="font-family: Times, Times New Roman, serif;">m</span></i> i<span style="font-family: inherit;">s 2. In ot</span>her words, the only way a prime number could come right before a square is if the square number is 4.<br><br>Isn't it astounding, and beautiful, to realize that human beings can know something so infinite in scope with such absolute certainty? No matter how many square numbers I check, I know that the only one that can follow a prime number is 4.<br><br>Can a similar statement be made about the cube numbers (e.g. 1, 8, 27, 64, etc.)? If so, can you prove the statement is definitely true, no matter what cube number I might check?<br><br>What about the fourth-power integers (e.g. 1, 16, 81, 256, etc.)?<br><br><a href="http://sucolex.blogspot.com/2013/09/squares-and-cubes-and-primes.html#more">Think it through, then click to go on.</a>Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-30864130765707005312013-09-10T04:52:00.000-05:002013-09-28T08:26:43.624-05:00Working together<blockquote class="tr_bq">My garden hose can fill my kids' wading pool in 14 minutes. My neighbor's garden hose can fill it in only 10 minutes. If we use both hoses at once, how long should it take to fill the pool?</blockquote>I posed this question to some students the other day, and I was taken aback at an approach that one student came up with, seemingly spontaneously: Find the average of the filling times, and divide it by 2 (since the two hoses are working together).<br><br>In the situation at hand, the student's approach gives a nice round answer of 6 minutes. Unfortunately, this isn't quite right. But doesn't the student's approach seem reasonable? What's the easiest way to see that the approach can't be right?<br><br><a href="http://sucolex.blogspot.com/2013/09/working-together.html#more">Think it through, then click to go on.</a>Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-80037021781751442722013-02-20T06:53:00.002-06:002013-09-10T04:35:41.124-05:00Cooperation and small probabilitiesWatch this <a href="http://www.youtube.com/watch?v=BaAa9hJlA0A&t=0m23s" target="_blank">clip from the TV show Friends</a> (a popular late-90's sitcom). Joey, one of the six "friends," announces his intention to enter the <a href="http://www.ctlottery.org/Modules/Games/default.aspx?id=5" target="_blank">Connecticut Powerball</a> lottery. Most of the rest of the group decide that it would be a good idea to pool their money to buy many tickets and split the payoff if any of the tickets wins. However, Ross (a college professor in a scientific discipline, I might add) opts out, citing the incredibly small probability (or "odds") of winning. Chandler jokingly replies that with six times as many tickets, the group will have six times the probability of winning.<br><br>Is Chandler right about that?<br><br><a href="http://sucolex.blogspot.com/2013/02/cooperation-and-small-probabilities.html#more">Think it through, then click to go on.</a>Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-67370029629483103012012-06-09T11:07:00.001-05:002013-07-11T08:37:12.516-05:00Hide and PeekThere is a game for the Nintendo Wii called <a href="http://wiiparty.nintendo.com/minigames/" target="_blank">Wii Party</a> (<a href="http://en.wikipedia.org/wiki/Wii_Party" target="_blank">see also Wikipedia</a>), which largely consists of a set of four-player "minigames." In one of the minigames, shown below, three players hide on a playground and a fourth player tries to find them.<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-HzBTJwNyn8U/T9NxHjkDVVI/AAAAAAAAAKU/gnko8tz2jS0/s1600/hidepeek.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://1.bp.blogspot.com/-HzBTJwNyn8U/T9NxHjkDVVI/AAAAAAAAAKU/gnko8tz2jS0/s1600/hidepeek.jpg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Hide-and-Peek minigame (see the <a href="http://wiiparty.nintendo.com/minigames/#tab-one-vs-three" target="_blank">Wii Party website</a>, under "All Minigames: 1 vs 3," for a video)</td></tr></tbody></table>The non-hiding player has five chances to find all three of the others. There are seven hiding places, and any number of players may hide in each one. However, strangely enough, anyone who chooses the small hiding place in the center, behind the blue and pink spring horses, will actually be visible there (so unless the non-hiding player is completely inexperienced, he or she will definitely find that person).<br /><br />The non-hiding player wins if he or she finds all three of the others. If not, all of the others win (regardless of which of them were found). Should someone hide behind the spring horses?Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-49941880572209740462012-05-01T08:59:00.000-05:002012-06-16T11:34:08.519-05:00Geometry of the sextantAs I mentioned <a href="http://sucolex.blogspot.com/2012/04/backyard-trigonometry.html" target="_blank">before</a>, I've been working with a sextant lately. Not surprisingly, Wikipedia has an <a href="http://en.wikipedia.org/wiki/Sextant" target="_blank">informative article on sextants</a>, with many details about their operation. The diagram below shows the most basic aspects of the geometry by which a sextant works.<br /><div class="separator" style="clear: both; text-align: center;"></div><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://www.geogebratube.org/student/e8559?w=600&h=588" target="_blank"><img border="0" src="http://3.bp.blogspot.com/-d8UiMaOTKa4/T5_reUfvaTI/AAAAAAAAAJ8/nBujB1agvT4/s1600/sextant.png" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><i>(Click for the interactive version, in which the red dots can be dragged.)</i></td></tr></tbody></table>The fundamental idea is that you look directly at one object while adjusting the angle between the sextant's two mirrors so that your view of a second object coincides precisely with your view of the first object. That is, the path of light from object 1 to your eye overlaps exactly the path of light from object 2 to your eye after reflecting off the two mirrors, as shown above. When this is accomplished, the scale at the bottom of the sextant shows the visual angle between the two objects.<br /><br />The sextant derives its name from the fact that its scale forms a physical arc of about 60° — or one sixth of a full circle. But one of the first things that struck me about the scale is that its marks run from 0° to about 120°. So, since the sextant's mirrors are parallel when the sextant is set to 0°, it seems that the scale shows <b>two times</b> the angle between the sextant's mirrors. <br /><br />So here are my questions: Is the visual angle between object 1 and object 2 really two times the angle between the mirrors? Can we prove it?Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-8945554806033788742012-04-23T14:24:00.006-05:002012-06-16T11:37:42.007-05:00Backyard trigonometry<div class="separator" style="clear: both; text-align: center;"></div><a href="http://4.bp.blogspot.com/-Uq1UvX2Cpi0/T5a51-S2f_I/AAAAAAAAAJE/yf0s_ixw4U8/s1600/tree.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" height="237" src="http://4.bp.blogspot.com/-Uq1UvX2Cpi0/T5a51-S2f_I/AAAAAAAAAJE/yf0s_ixw4U8/s400/tree.png" width="400"></a>I recently got a hold of an old <a href="http://en.wikipedia.org/wiki/Sextant" target="_blank">sextant</a> (it looks about as old as the <a href="http://sucolex.blogspot.com/2012/03/sound-waves-part-2.html" target="_blank">tuning forks</a>, but the very same model I have is available for sale <a href="http://www.davisnet.com/marine/products/marine_product.asp?pnum=00011" target="_blank">here</a>). To see what sort of practical measurements I could make with it, I decided to try to figure out the height of a tall tree in my yard.<br><br>From some distance away, the visual angle of the tree was 40°18' (or 40.3°). From 12 feet closer, it was 46°1' (or about 46.02°). From another 12 feet closer, it was 53°33' (or 53.55°). The angles were each viewed from about 6 feet above the ground. How tall is the tree?<br><br><a href="http://sucolex.blogspot.com/2012/04/backyard-trigonometry.html#more">Think it through, then click to go on.</a>Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-52104164149879541572012-04-18T13:05:00.011-05:002012-04-19T15:59:22.526-05:00ExpofinityConsider the equation<br><div style="font-family: Times,"Times New Roman",serif; text-align: center;"><span style="font-size: x-large;"><i> x<sup> x<sup> x<sup> x<sup> x<sup> x<sup> x<sup> x<sup> x<sup> x<sup> ·<sup> ·<sup> ·</sup></sup></sup></sup></sup></sup></sup></sup></sup></sup></sup></sup></i>= 3.</span></div><br>Can you solve it? Can you check your answer?<br><br><a href="http://sucolex.blogspot.com/2012/04/expofinity.html#more">Think it through, then click to go on.</a>Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-52195190177602752182012-03-13T12:18:00.000-05:002012-06-16T11:46:29.653-05:00Sound waves (part 2)There's a room near my office full of physics demonstration equipment, including a box of rusty old tuning forks. I picked out two that looked and sounded similar.<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-zGFyeY2LLOM/T19HQsHYnhI/AAAAAAAAAIA/DlRom4psKec/s1600/forks.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="232" src="http://2.bp.blogspot.com/-zGFyeY2LLOM/T19HQsHYnhI/AAAAAAAAAIA/DlRom4psKec/s400/forks.jpg" width="400" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><i>These are probably older than me.</i></td></tr></tbody></table><br />In the following recording, you can hear me strike one of the forks, and then the other, and then both at the same time. As in my <a href="http://sucolex.blogspot.com/2012/03/sound-waves-part-1.html" target="_blank">previous post</a>, you may need to use headphones to hear the low-pitch (i.e. low-frequency) part of the sounds, which is where the interesting stuff happens.<br /><br /><object height="81" width="100%"> <param name="movie" value="https://player.soundcloud.com/player.swf?url=http%3A%2F%2Fapi.soundcloud.com%2Ftracks%2F39622218&show_comments=true&auto_play=false&color=00afff"> </param><param name="allowscriptaccess" value="always"> </param><embed allowscriptaccess="always" height="81" src="https://player.soundcloud.com/player.swf?url=http%3A%2F%2Fapi.soundcloud.com%2Ftracks%2F39622218&show_comments=true&auto_play=false&color=00afff" type="application/x-shockwave-flash" width="100%"></embed> </object> <br /><br />Here are some close-up shots of the graph produced by a sound editing program:<br /><br /><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: left; margin-right: 1em; text-align: left;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-vlp3L2zrRz8/T19N9dq4IXI/AAAAAAAAAII/ScnDUtnNeWE/s1600/forks1_.png" imageanchor="1" style="clear: left; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" height="105" src="http://4.bp.blogspot.com/-vlp3L2zrRz8/T19N9dq4IXI/AAAAAAAAAII/ScnDUtnNeWE/s320/forks1_.png" width="280" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><i>Vibrations of first tuning fork.</i></td></tr></tbody></table><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody><tr><td style="text-align: center;"><a href="http://3.bp.blogspot.com/-X1PyUs4wo1M/T19N-XoNRdI/AAAAAAAAAIQ/ecMMqVfq0js/s1600/forks2_.png" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" height="105" src="http://3.bp.blogspot.com/-X1PyUs4wo1M/T19N-XoNRdI/AAAAAAAAAIQ/ecMMqVfq0js/s320/forks2_.png" width="280" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><i>Vibrations of second tuning fork.</i></td></tr></tbody></table><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-Q65EcHwYpE0/T19N_WBD2zI/AAAAAAAAAIY/SLdr-z9E8Ag/s1600/forks3_.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="109" src="http://4.bp.blogspot.com/-Q65EcHwYpE0/T19N_WBD2zI/AAAAAAAAAIY/SLdr-z9E8Ag/s640/forks3_.png" width="600" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><i>Both tuning forks vibrating at the same time.</i><i><br /></i></td></tr></tbody></table>What is this devilry? Well, judging carefully by the first two graphs above, it appears to me that the tuning forks vibrate with frequencies of about 258.8 and 265.2 cycles per second (or <a href="http://en.wikipedia.org/wiki/Hertz" target="_blank">Hz</a>), respectively. If we graph the sum of two sine functions with those frequencies, each with an amplitude of, say, 0.2, we obtain the following graph.<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-D5aN6h4W6wc/T191GjYRGKI/AAAAAAAAAIg/en_QuKtP2S4/s1600/forkfuncs.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="166" src="http://2.bp.blogspot.com/-D5aN6h4W6wc/T191GjYRGKI/AAAAAAAAAIg/en_QuKtP2S4/s640/forkfuncs.png" width="640" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><i>Sum of two sine waves with slightly different frequencies.</i></td></tr></tbody></table>As mentioned in my <a href="http://sucolex.blogspot.com/2012/03/sound-waves-part-1.html#more" target="_blank">previous post</a>, we can use Octave/MATLAB to listen to this function. Here are the sounds of a 258.8-Hz sine wave, a 265.2-Hz sine wave, and the sum of the two (graphed above):<br /><br /><object height="81" width="100%"> <param name="movie" value="https://player.soundcloud.com/player.swf?url=http%3A%2F%2Fapi.soundcloud.com%2Ftracks%2F39622594&show_comments=true&auto_play=false&color=00afff"> </param><param name="allowscriptaccess" value="always"> </param><embed allowscriptaccess="always" height="81" src="https://player.soundcloud.com/player.swf?url=http%3A%2F%2Fapi.soundcloud.com%2Ftracks%2F39622594&show_comments=true&auto_play=false&color=00afff" type="application/x-shockwave-flash" width="100%"></embed> </object><br /><br />Now, some questions occur to me at this point:<br /><ol><li>What is the simplest, most straightforward explanation for this weird audio phenomenon?</li><li>Is there a simple relationship between the very low frequency of the pattern seen/heard here and the higher frequencies of the individual tuning forks?</li><li>The mathematical graph above matches up very well with part of the graph of the actual sound of the two tuning forks (from about 25.8 seconds onward), but the earlier part of the sound graph is a bit different. How can we explain this?</li></ol>Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-12025780756035838562012-03-13T00:23:00.004-05:002012-06-16T11:49:32.298-05:00Sound waves (part 1)I recently opened the audio from <a href="http://www.youtube.com/watch?v=iEPTlhBmwRg" target="_blank">this music video</a> in a sound editing program. Such programs (e.g. <a href="http://audacity.sourceforge.net/" target="_blank">Audacity</a>) produce a visual representation of the audio — essentially a graph of sound pressure versus time.<br><br>At certain moments during the song, the graph looks something like this: <br><br><div class="separator" style="clear: both; text-align: center;"></div><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-kd6vw4lYYXY/T14_nyY5bsI/AAAAAAAAAHo/Gf9CAFXiIrk/s1600/jagger.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://1.bp.blogspot.com/-kd6vw4lYYXY/T14_nyY5bsI/AAAAAAAAAHo/Gf9CAFXiIrk/s1600/jagger.png"></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><i>A 0.06-second interval around 70 seconds into the video.</i></td></tr></tbody></table><br>There is a clear sine wave pattern here.<br><br><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-nRQI4Va_6LA/T16-jeeCmHI/AAAAAAAAAHw/PglFQZImRyQ/s1600/jagger2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-nRQI4Va_6LA/T16-jeeCmHI/AAAAAAAAAHw/PglFQZImRyQ/s1600/jagger2.png"></a></div><br>But on top of those sine waves, there are other, less dramatic but faster sine waves. How can we combine sine functions in such a way as to produce a graph like the original one above? And where did these two sets of waves come from? <br><br><a href="http://sucolex.blogspot.com/2012/03/sound-waves-part-1.html#more">Think it through, then click to go on.</a>Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-75573063920497078312012-02-29T08:12:00.011-06:002012-02-29T13:10:33.420-06:00Leap daysOne year — that is, exactly one cycle of the Earth's seasons — is approximately 365¼ days long. So if our calendars only counted 365 days in every year, for every four years we'd fall one day behind. Our seasons would shift; after many years, North America would receive snow in June and summer heat in December. <br /><br />Unfortunately, a year is not <b>exactly</b> 365¼ days long. In reality, a better estimate for the average length of a year is <a href="http://en.wikipedia.org/wiki/Tropical_year" target="_blank">365.2425 days</a>. Because of this, the established rules for "leap days" in our calendars are a bit more complicated than simply adding one extra day (February 29) every four years. <br /><br />I wonder: Do we have the best possible system for adjusting to a 365.2425-day year? To answer that question, assuming you're not already familiar with all the details of our system, I would recommend working out a system of adjustments on your own first, and then checking to see if you come up with the same rules.Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-2760380637932365972012-02-10T23:40:00.008-06:002012-06-16T11:51:27.463-05:00Three CirclesOkay, now I am officially obsessed with spheres and circles.<br /><br />A while back, I introduced my trigonometry students to the beautiful <a href="http://mathdl.maa.org/mathDL/46/?pa=content&sa=viewDocument&nodeId=2063" target="_blank">problem of finding the area between three congruent and mutually-tangent circles</a> (though naturally I didn't pose the problem in such technical terms). Since then, I've been thinking about the more general situation in which three circles of different sizes are mutually tangent. Here's a diagram.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://www.geogebra.org/en/upload/files/rwwright/three_circles.html" target="_new"><img border="0" height="338" src="http://1.bp.blogspot.com/-J4TjzSOw9HQ/TzX7UUWNffI/AAAAAAAAAHc/t4g7jN3NawM/s400/circles.png" width="400" /></a></div><br />(Click the image for a nice interactive version.)<br /><br />Now, what questions does this diagram raise for you? In creating various incarnations of the figure, some things that I found myself asking were:<br /><ol><li>If I have two mutually-tangent circles of different sizes, how can I construct a third circle tangent to both of them? Or, to put it another way, where must the center of the third circle be? What sort of curve must it lie along?</li><li>If I choose three points that do not lie on the same line together, can they always be used as the centers of three mutually-tangent circles? If so, how do I construct those circles? In other words, how do I find their radii? </li><li>And of course, the standard question — given three mutually-tangent circles, what is the area between them (shown in blue above)?</li></ol>Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-75592335438387074412012-01-20T20:22:00.003-06:002012-06-16T12:01:40.399-05:00Spheres again<div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/--KgHvowPhpY/TxogrdpLGlI/AAAAAAAAAHM/vta9PEC6Dss/s1600/balls.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="300" src="http://1.bp.blogspot.com/--KgHvowPhpY/TxogrdpLGlI/AAAAAAAAAHM/vta9PEC6Dss/s400/balls.jpg" width="400" /></a></div>When two balls lie on a flat surface, in contact with each other, there's a pretty neat relationship between their sizes and the distance between the points where they touch the surface. There's maybe a slightly less nifty relationship between their sizes and the height of the point where they touch (the "point of tangency"). See if you can find one or both of these relationships.Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-34480566946037165652011-12-13T08:35:00.009-06:002012-06-16T12:30:33.169-05:00Focus!If you hold a lens some distance away from a bright object, and place some kind of "screen," like a piece of paper or light-colored wall, on the other side of the lens at just the right distance, you will see a clear, upside-down image of the object.<br><br><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-GAlnBQwWXvM/TxVzM93ntJI/AAAAAAAAAHA/fXA3S_ZTppA/s1600/focus.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-GAlnBQwWXvM/TxVzM93ntJI/AAAAAAAAAHA/fXA3S_ZTppA/s1600/focus.png"></a></div><br><a href="http://3.bp.blogspot.com/-CbG33vREBug/TudPyzKFdtI/AAAAAAAAAGs/tx4ZSUXBAYY/s1600/lens.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" height="131" src="http://3.bp.blogspot.com/-CbG33vREBug/TudPyzKFdtI/AAAAAAAAAGs/tx4ZSUXBAYY/s320/lens.png" width="320"></a>Some variation of this basic principle is used in many practical situations (for example, in cameras, and when projecting a movie onto a large screen). Armed with a dollar store magnifying glass for a lens, and some spare time, I tried this at various distances from a lamp in my house and, as carefully as I could, recorded the distances needed to produce a clear image. You might want to try this out yourself, but my results are given below, measured in inches.<br><br><table cellpadding="4" rules="all"><tbody><tr><td>Distance to lamp (<i>p</i>) </td><td width="30px">25 </td><td width="30px">40 </td><td width="30px">55 </td><td width="30px">70 </td><td width="30px">155 </td></tr><tr><td>Distance to image (<i>q</i>) </td><td>12 </td><td>10 </td><td>9<sup>1</sup>/<sub>4</sub></td><td>9 </td><td>8<sup>3</sup>/<sub>8</sub></td></tr></tbody></table><br>Here's a graph of the data.<br><br><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-MU3LzseqP-I/TudVRh5mHlI/AAAAAAAAAG4/oz8wphUYubY/s1600/lens_graph.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="275" src="http://3.bp.blogspot.com/-MU3LzseqP-I/TudVRh5mHlI/AAAAAAAAAG4/oz8wphUYubY/s400/lens_graph.png" width="400"></a></div><br>If you're as fascinated by this as I am, there are many questions you might have at this point. For example, what would happen if I held the lens closer to, or farther away from, the lamp? This and many other potential questions can be boiled down to the following: What kind of mathematical function / equation could describe the relationship between object distance and image distance?<br><br><a href="http://sucolex.blogspot.com/2011/12/focus.html#more">Think it through, then click to go on.</a>Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-78674895389496075172011-11-27T20:15:00.014-06:002012-06-16T12:40:28.931-05:00Human height and weightBelow is a graph showing the heights and weights of a large group of people, including infants (each dot represents a person). Based on this, what is a normal weight for someone with a given height? In other words, if <i>h</i> and <i>w</i> represent height and "average" weight, can you find a formula for <i>w</i> as a function of <i>h</i>? <br><br><a href="http://www.geogebra.org/en/upload/files/rwwright/bodies.html" imageanchor="1" style="margin-left: 1em; margin-right: 1em;" target="_new"><img border="0" height="265" src="http://3.bp.blogspot.com/-haeLCo7nt80/TtLtP_TfYpI/AAAAAAAAAGc/HuOfaO5kJT4/s400/bodies.png" width="400"></a> <br>(If you have Java installed, click the graph to load an interactive version, with which you can try plotting your function against the data. For example, one not-very-good possibility is <i>w</i>(<i>h</i>) = 3.86<i>h</i> - 110.42, which you can plot by typing it into the input box. Also, if you're a GeoGebra whiz, you might like to know that the data points are stored as <span style="font-family: "Courier New",Courier,monospace;">list1</span>.)<br><br><a href="http://sucolex.blogspot.com/2011/11/human-height-and-weight.html#more">Think it through, then click to go on.</a>Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-56850271711966350302011-10-24T15:02:00.022-05:002012-06-16T12:47:26.353-05:00Modeling with the normal distributionAccording to a 2008 study, the height and weight of American men are distributed as follows.<br /><table cellpadding="4" rules="all"><tbody><tr><td>Percentile:</td><td>5th</td><td>10th</td><td>15th</td><td>25th</td><td>50th</td><td>75th</td><td>85th</td><td>90th</td><td>95th</td></tr><tr><td>Height (inches):</td><td>64.4</td><td>65.6</td><td>66.3</td><td>67.4</td><td>69.4</td><td>71.5</td><td>72.6</td><td>73.2</td><td>74.3</td></tr><tr><td>Weight (pounds):</td><td>137.1</td><td>147.0</td><td>154.6</td><td>165.7</td><td>188.8</td><td>216.8</td><td>234.5</td><td>245.8</td><td>270.3</td></tr></tbody></table><br />Can you tell from this information whether the height and/or weight of American men are normally distributed? If either is, can you estimate the standard deviation?Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-9434353929795382382011-09-02T12:57:00.000-05:002012-06-16T13:00:12.899-05:00DerangedIn how many ways can I rearrange the letters ABCDEFGH so that no letter is in its original position?Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-15242302167843151602011-08-08T12:09:00.005-05:002012-06-16T13:15:28.315-05:00How far can you see? — revisitedHere is a photograph of the Earth:<br><br><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-fmSG5adUi3g/TkAXAind3FI/AAAAAAAAAFY/evpLnCc49AE/s1600/earth.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="http://2.bp.blogspot.com/-fmSG5adUi3g/TkAXAind3FI/AAAAAAAAAFY/evpLnCc49AE/s320/earth.jpg" width="320"></a></div><br>From how far away was this photo taken? <br><br>(Alternatively, as <a href="http://blog.mrmeyer.com/?cat=93">Dan Meyer</a> might say, "Here is a photograph of the Earth. Any questions?")<br><br><a href="http://sucolex.blogspot.com/2011/08/how-far-can-you-see-revisited.html#more">Think it through, then click to go on.</a>Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-21005596547999137522011-06-08T11:50:00.004-05:002011-06-08T15:13:01.671-05:00,000 as a unitI happened to see this the other day during an episode of <a href="http://channel.nationalgeographic.com/series/prehistoric-predators/">Prehistoric Predators</a> on the National Geographic Channel:<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-5Y_znace-u4/Te-mySQ0XKI/AAAAAAAAADw/WmsOw2blQ9w/s1600/554.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="180" src="http://3.bp.blogspot.com/-5Y_znace-u4/Te-mySQ0XKI/AAAAAAAAADw/WmsOw2blQ9w/s400/554.jpg" width="400" /></a></div>Click to enlarge. Sorry for the poor quality; I literally took a picture of the TV screen, in a state of mixed amusement and dismay.Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-41231282989917259442011-05-20T01:27:00.004-05:002011-05-20T11:05:21.306-05:00Modeling temperatureThe <a href="http://www.noaa.gov/">NOAA</a> keeps close track of various kinds of weather data at a number of locations around the U.S. For example, <a href="ftp://ftp.ncdc.noaa.gov/pub/data/uscrn/products/soilsip01/README.txt">here</a> is a description (including the web location) of some data on hourly soil temperatures for the past couple of years. The NOAA station closest to me is in Chillicothe, MO (home of sliced bread!), and a graph of the hourly near-surface soil temperature in Chillicothe, starting on July 30, 2009, is shown below.<br><br><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-X1jF1GuE__4/TdX_1cLLipI/AAAAAAAAADo/ME4a-6Ar5Tc/s1600/chillweb2.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="300" src="http://2.bp.blogspot.com/-X1jF1GuE__4/TdX_1cLLipI/AAAAAAAAADo/ME4a-6Ar5Tc/s400/chillweb2.gif" width="400"></a></div><br>A good exercise in modeling is to find a function that "fits" a set of data like this one. But for me, the graph above also inspires a number of other questions that can lead into further topics. What sort of questions does it raise for you?<br><br><a href="http://sucolex.blogspot.com/2011/05/modeling-temperature.html#more">Think it through, then click to go on.</a>Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-62647736637553230062011-04-13T10:26:00.006-05:002011-04-27T12:33:34.415-05:00The derivative as a topic in algebraIs there something wrong with giving college algebra students the following as a prompt for group discussion? <br /><blockquote>On Earth, gravity causes most objects to accelerate downward at about 32 feet per second per second. Because of this, according to physicists, if we were to throw such an object upward at a starting velocity of <eq>v</eq><sub>0</sub> feet per second from a starting height of <eq>h</eq><sub>0</sub> feet, the height <eq>h</eq> (in feet) of the object after <eq>t</eq> seconds can be modeled by <eq>h</eq>(<eq>t</eq>) = -16<eq>t</eq><sup>2</sup> + <eq>v</eq><sub>0</sub><eq>t</eq> + <eq>h</eq><sub>0</sub>.<br /><br />Suppose I throw a ball upward at 45 feet per second from 20 feet above the ground. Based on the model function, <br /><ol><li>What will the height of the ball be after 1 second? And after 3 seconds? After 5 seconds?</li><li>When will the ball hit the ground?</li><li>How high will the ball reach?</li><b></b><li><b>How fast will the ball be moving 1 second after it is thrown?</b></li><li><b>Is the ball really accelerating downward at 32 feet per second per second?</b></li></ol></blockquote>Keep in mind that the function <eq>h</eq> given above is almost always introduced in college algebra courses as a prime example of quadratics. Also note that the idea of average rates of change (and in particularly bad textbooks, difference quotients) is not uncommon in college algebra.<br /><br /><b>Update: </b>Around the same time I wrote the above, <a href="http://coxmath.blogspot.com/2011/04/teachable-moments.html">David Cox</a> reported a somewhat related (and very exciting) experience introducing the derivative to an eager group of students.Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.com