{"id":595,"date":"2014-05-05T16:06:33","date_gmt":"2014-05-05T16:06:33","guid":{"rendered":"http:\/\/onlineteachered.mit.edu\/?page_id=595"},"modified":"2024-12-06T22:57:59","modified_gmt":"2024-12-06T22:57:59","slug":"module-3","status":"publish","type":"page","link":"https:\/\/onlineteachered.mit.edu\/courses\/mathlets\/module-3\/","title":{"rendered":"Mathelets in Homework, Module 3"},"content":{"rendered":"<p><!--slice begin--><!-- For use on mathlets.org\/training --><\/p>\n<div id=\"tocs\"><\/div>\n<p>Each segment of this module consists of sample homework assignments created around one or another of the MIT Mathlets.\u00a0Each segment is followed by questions, my remarks about the features of the Mathlets, and examples that illuminate their use.\u00a0At the end there are a couple of exercises.<\/p>\n<hr \/>\n<h2>1. In this module&#8230;<\/h2>\n<p>In this module, you will watch video segments where you will be instructed to pause and engage in a variety of activities, as well as think about the questions posed.<\/p>\n<ul>\n<li>Download Mathlets in Homework Slides [ <a href=\"http:\/\/onlineteachered.mit.edu\/files\/2014\/08\/Mathlets-in-Homework-Slides.pdf\" target=\"_blank\" rel=\"noopener\">PDF<\/a>, 1MB ]<\/li>\n<li>Watch <a href=\"https:\/\/vimeo.com\/1036484435\">Mathlets in Homework, Module 3 Video<\/a><br \/>\n<!-- YouTube\nWatch <a href=\"https:\/\/youtu.be\/6Wx-Ast0p0Q\">Mathlets in Homework, Module 3 Video<\/a>\n-->\n<\/li>\n<\/ul>\n<hr \/>\n<h2>2. Learning Objectives<\/h2>\n<p>After completing this module, the participant will be able to use Mathlets to:<\/p>\n<ul class=\"mathlet-course-ul\">\n<li>Mix experiment with computation in homework.<\/li>\n<li>Improve student understanding regarding the significance computations in homework.<\/li>\n<li>Increase student enjoyment of homework exercises.<\/li>\n<\/ul>\n<hr \/>\n<h2>3. Mathlets in Homework, Segment 1<\/h2>\n<p>The final use we will look at during this short course is the use of Mathlets as part of homework. Let me introduce you to the idea and benefits of using Mathlets in homework during this next video segment.<\/p>\n<p><!-- ** Vimeo Video Source, hrm@math.mit.edu --><\/p>\n<div style=\"padding:56.25% 0 0 0;position:relative;\"><iframe src=\"https:\/\/player.vimeo.com\/video\/1036486415?title=0&amp;byline=0&amp;portrait=0&amp;badge=0&amp;autopause=0&amp;player_id=0&amp;app_id=58479\" frameborder=\"0\" allow=\"autoplay; fullscreen; picture-in-picture; clipboard-write\" style=\"position:absolute;top:0;left:0;width:100%;height:100%;\" title=\"MIT Mathlets in Homework, Module 3, Segment 1\"><\/iframe><\/div>\n<p><script src=\"https:\/\/player.vimeo.com\/api\/player.js\"><\/script><br \/>\n<!-- ** Backup YouTube Video Source\n<iframe loading=\"lazy\" width=\"560\" height=\"315\" src=\"https:\/\/www.youtube.com\/embed\/ht0ERHO_3HM?si=YTK54QdF0lWv-11C\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n--><br \/>\nTranscript [ <a href=\"https:\/\/onlineteachered.mit.edu\/files\/2014\/08\/Module-3-Mathlets-in-Homework-Segment-1-Transcript.pdf\" target=\"_blank\" rel=\"noopener\">PDF<\/a> ]<\/p>\n<hr \/>\n<h3>3.1. First Assignment: Vector Addition<\/h3>\n<p><em>This exercise would be appropriate in a class discussing parametrized curves.<\/em><\/p>\n<div class=\"shaded\">\n<h4><em class=\"fa fa-rocket\"><\/em> Launch<\/h4>\n<p>Mathlet:\u00a0<a title=\"Wheel Mathlet\" href=\"http:\/\/mathlets.org\/mathlets\/wheel\/\" target=\"_blank\" rel=\"noopener\"><em>Wheel<\/em><\/a>\n<\/div>\n<p>You see a wheel with a light at the end of a\u00a0spoke. Animate the rolling wheel using the [&gt;&gt;] key. You can return the wheel\u00a0to the start position using the [&lt;&lt;] key. You can also control <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Ctheta&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;theta\" class=\"latex\" \/>\u00a0directly using the slider.<\/p>\n<ol class=\"loweralpha mathlet-course-ol\">\n<li>Adjust the <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=a&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"a\" class=\"latex\" \/> and <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=b&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"b\" class=\"latex\" \/> sliders. What does <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=a&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"a\" class=\"latex\" \/> represent? What does\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=b&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"b\" class=\"latex\" \/> represent?<\/li>\n<li>You can cause the trajectories of the center of the wheel and the light to\u00a0be marked by selecting the [trace] key.\u00a0Give a parametric formula for the location of the center of the wheel\u00a0(in terms of these parameters) as a function of <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Ctheta&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;theta\" class=\"latex\" \/>.<\/li>\n<li>Give a parametric formula for the location of the yellow light as a function of <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Ctheta&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;theta\" class=\"latex\" \/>.<\/li>\n<li>You can cause the velocity vector to be shown by selecting the [velocity]\u00a0key.\u00a0Are there settings for <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=a&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"a\" class=\"latex\" \/>, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=b&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"b\" class=\"latex\" \/>, and <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Ctheta&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;theta\" class=\"latex\" \/> for which the velocity is zero? You may want to discover them using the Mathlet; but then verify your\u00a0observation mathematically.<\/li>\n<li>By experimenting with the Mathlet, identify situations in which the\u00a0horizontal component of the velocity vector vanishes. Then verify this observation mathematically.<\/li>\n<li>In fact, what can you say about how the sign of the horizontal\u00a0component of the velocity vector is related to the position of the light?\u00a0Again, verify this observation mathematically.<\/li>\n<\/ol>\n<h4>Comments<\/h4>\n<p>(a) Just as in a\u00a0lecture, you have to lead students through the elements of a Mathlet.\u00a0It is a good idea to encourage students\u00a0to express the meaning of elements of the Mathlet themselves:<\/p>\n<p>(b)-(c) Do you think that having the parametric expression for the light\u00a0present on the screen lessens the value of these questions?<\/p>\n<p>(d)-(f) Now we start experimenting. Homework in Mathematics classes often\u00a0involves somewhat random computations, designed to give the student practice\u00a0at carrying out some manipulation or algorithm. The Mathlet provides a graphical\u00a0<em>reason<\/em>\u00a0for wanting to do such calculations.<\/p>\n<div class=\"shaded\">\n<h4><em class=\"fa fa-lightbulb-o\"><\/em> Think About<\/h4>\n<p>What are your comments about the\u00a0<em>Wheel<\/em>\u00a0project? Explain how some part of this may be useful in one of the classes you teach.\n<\/div>\n<p><\/p>\n<hr \/>\n<h3>3.2 Second Assignment: Taylor Polynomials<\/h3>\n<div class=\"shaded\">\n<h4><em class=\"fa fa-rocket\"><\/em> Launch<\/h4>\n<p>Mathlet:\u00a0<a title=\"Taylor Polynomials Mathlet\" href=\"http:\/\/mathlets.org\/mathlets\/taylor-polynomials\/\" target=\"_blank\" rel=\"noopener\"><em>Taylor Polynomials<\/em><\/a>\n<\/div>\n<p>There is a lot to play with here!\u00a0When you have explored the Mathlet a little, settle down on the menu item\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%28x%29%3D1%2F%281%2Bx%5E2%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f(x)=1\/(1+x^2)\" class=\"latex\" \/>.<\/p>\n<ol class=\"lower-alpha mathlet-course-ol\">\n<li>Compute the full MacLaurin series (the Taylor series at <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=a%3D0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"a=0\" class=\"latex\" \/>).\u00a0Then select <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=a%3D0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"a=0\" class=\"latex\" \/> on the Mathlet and use the [Terms] key and a large value of\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=n&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"n\" class=\"latex\" \/> to check your answer.<\/li>\n<li>Set <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=n%3D2&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"n=2\" class=\"latex\" \/> and animate the second order Taylor series by setting <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=a%3D-3&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"a=-3\" class=\"latex\" \/>\u00a0and pressing the [&gt;&gt;] key. You see a family of parabolas.\u00a0Sometimes they are opening up and sometimes they are opening down.\u00a0Find (by calculus)\u00a0the values of <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=x&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"x\" class=\"latex\" \/> at which these transitions occur. Do these points have a name?<\/li>\n<li>Now set <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=n%3D3&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"n=3\" class=\"latex\" \/> and animate the third order Taylor series. From the\u00a0Mathlet, observe where the coefficient of the third order term vanishes.\u00a0What is the name of the graph\u00a0of the Taylor polynomial at those points? Then compute where those points are.<\/li>\n<li>Now, using the same function, select <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=a%3D1&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"a=1\" class=\"latex\" \/>. The [Terms] readout will\u00a0display the coefficients in the Taylor series. From these data, predict what\u00a0the entire Taylor polynomial will be. Then prove your prediction.<\/li>\n<\/ol>\n<h4>Comments<\/h4>\n<p>(a)\u00a0I thought it was important to focus on\u00a0the Taylor series at a single point, maybe beginning at the point <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=a%3D0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"a=0\" class=\"latex\" \/>, before\u00a0moving to the animation offered by this Mathlet.<\/p>\n<p>(b) Perhaps I should have interposed an animation with <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=n%3D1&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"n=1\" class=\"latex\" \/>, to\u00a0let students see the more familiar tangent line before they see the best\u00a0approximating parabola. The parabola flattens out to a straight line at\u00a0the points of inflection. This is what we envision in teaching about points\u00a0of inflection, but without a tool like this it is hard to convey to the\u00a0student, and my guess is that identifying the transition from concave to\u00a0convex with the points of inflection will come as something of a revelation.<\/p>\n<p>(c) Luckily the zeros of <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%5E%7B%283%29%7D%28x%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f^{(3)}(x)\" class=\"latex\" \/> come out nicely, but this is still\u00a0a substantial computation for beginners, one which might be resented without\u00a0the payoff of verifying an observation.<\/p>\n<p>(d) This was a surprise to me! You can substitute <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=x%3D1%2Bu&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"x=1+u\" class=\"latex\" \/> into <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%28x%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f(x)\" class=\"latex\" \/>, but\u00a0the result, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=1%2F%282%2B2u%2Bu%5E2%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"1\/(2+2u+u^2)\" class=\"latex\" \/>, does not expand in any very obvious way.\u00a0With enough insight you can realize that\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%282%2B2u%2Bu%5E2%29%282-2u%2Bu%5E2%29%3D4%2Bu%5E4&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"(2+2u+u^2)(2-2u+u^2)=4+u^4\" class=\"latex\" \/>, so <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%28x%29%3D%5Cfrac%7B2-2u%2Bu%5E2%7D%7B4%2Bu%5E4%7D%3D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D-%5Cfrac%7Bu%7D%7B2%7D%2B%5Cfrac%7Bu%5E2%7D%7B4%7D%5Cright%29%5Csum_%7Bk%3D0%7D%5E%5Cinfty%5Cleft%28-%5Cfrac%7Bu%5E4%7D%7B4%7D%5Cright%29%5Ek&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f(x)=&#92;frac{2-2u+u^2}{4+u^4}=&#92;left(&#92;frac{1}{2}-&#92;frac{u}{2}+&#92;frac{u^2}{4}&#92;right)&#92;sum_{k=0}^&#92;infty&#92;left(-&#92;frac{u^4}{4}&#92;right)^k\" class=\"latex\" \/><\/p>\n<p>agreeing with the coefficients on the Mathlet. It is actually easier to start from the Taylor expansion guessable from the displayed\u00a0coefficients and work back; so the Mathlet provides a useful hint.<\/p>\n<p>This choice of menu item was rather\u00a0random; similar exercises could be constructed around any of the others.\u00a0The last one is the standard example of a <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=C%5E%5Cinfty&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"C^&#92;infty\" class=\"latex\" \/> function which is not real-analytic. It is very interesting to see how the Taylor coefficients all\u00a0become zero as <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=x&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"x\" class=\"latex\" \/> decreases to 0. Quite often I find myself surprised and\u00a0puzzled by behaviors displayed by a Mathlet; this is a good example.<\/p>\n<div class=\"shaded\">\n<h4><em class=\"fa fa-lightbulb-o\"><\/em> Think About<\/h4>\n<p>What are your thoughts about the\u00a0<em>Taylor Polynomials<\/em>\u00a0project? What parts of this project may be useful in one of the classes you teach?\n<\/div>\n<p><\/p>\n<hr \/>\n<h2>4. Mathlets in Homework, Segment 2<\/h2>\n<p>Here are some examples of homework problems involving various Mathlets.<\/p>\n<hr \/>\n<h3>4.1. First Assignment: Secant Approximation<\/h3>\n<p><em>This assignment might be given in a calculus course.\u00a0The early sections would be appropriate when the derivative is being introduced.<\/em><\/p>\n<div class=\"shaded\">\n<h4><em class=\"fa fa-rocket\"><\/em> Launch<\/h4>\n<p>Mathlet:\u00a0<em><a title=\"Secant Approximation Mathlet\" href=\"http:\/\/mathlets.org\/mathlets\/secant-approximation\/\" target=\"_blank\" rel=\"noopener\">Secant Approximation<\/a><\/em>\n<\/div>\n<p>Accept the\u00a0default menu choice <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%28x%29%3D%5Csin%28x%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f(x)=&#92;sin(x)\" class=\"latex\" \/>. Play with the two sliders and the two\u00a0check boxes.<\/p>\n<ol class=\"mathlet-course-ol\">\n<li>When <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5CDelta+x%3D0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;Delta x=0\" class=\"latex\" \/>, the yellow secant line disappears. Please explain\u00a0why.<\/li>\n<\/ol>\n<p>Set <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=a%3D%5Cpi%2F2&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"a=&#92;pi\/2\" class=\"latex\" \/>. (Notice that you can set a slider at a value marked by\u00a0a hashmark buy clicking on the hashmark.)<\/p>\n<ol class=\"mathlet-course-ol\">\n<li>What is the slope of the tangent line at this point?<\/li>\n<li>Move the <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5CDelta+x&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;Delta x\" class=\"latex\" \/> slider left and right, and then leave it at\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5CDelta+x%3D-0.50%5Cpi&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;Delta x=-0.50&#92;pi\" class=\"latex\" \/>. What is the yellow read-out of the value of the slope\u00a0of this secant line? What is its actual value?<\/li>\n<li>The claim is that <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Clim_%7B%5CDelta+x%5Crightarrow0%7D%5Cfrac%7B%5CDelta+y%7D%7B%5CDelta+x%7D%3D0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;lim_{&#92;Delta x&#92;rightarrow0}&#92;frac{&#92;Delta y}{&#92;Delta x}=0\" class=\"latex\" \/><\/p>\n<p>Using the Mathlet to gather empirical data, make a table showing how small\u00a0you have to make <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%7C%5CDelta+x%7C&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"|&#92;Delta x|\" class=\"latex\" \/> to guarantee that <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%7C%5CDelta+y%2F%5CDelta+x%7C%3C%5Cepsilon&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"|&#92;Delta y\/&#92;Delta x|&lt;&#92;epsilon\" class=\"latex\" \/>\u00a0for the following values of <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Cepsilon&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;epsilon\" class=\"latex\" \/>: <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=0.6%2C0.4%2C0.2%2C0.1%2C0.05%2C0.02&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"0.6,0.4,0.2,0.1,0.05,0.02\" class=\"latex\" \/>.<\/li>\n<li>For this value of <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=a&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"a\" class=\"latex\" \/> and small <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%7C%5CDelta+x%7C&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"|&#92;Delta x|\" class=\"latex\" \/>, the slope of the secant\u00a0is less than the slope of the tangent for <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5CDelta+x%3E0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;Delta x&gt;0\" class=\"latex\" \/> and greater that the\u00a0slope of the tangent for <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5CDelta+x%3C0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;Delta x&lt;0\" class=\"latex\" \/>. Move the <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=a&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"a\" class=\"latex\" \/> slider to other positions\u00a0and observe the relationship between <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5CDelta+y%2F%5CDelta+x&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;Delta y\/&#92;Delta x\" class=\"latex\" \/> and <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=dy%2Fdx&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"dy\/dx\" class=\"latex\" \/> at other\u00a0points. What regularity do you observe? With what other features of the graph does this relationship seem to correlate?<\/li>\n<li>[For classes covering quadratic approximation or Taylor series]\u00a0Please explain what you observed in (5).<\/li>\n<li>Before moving onto the next project I encourage you to consider ways for incorporating the\u00a0<em>Secant Approximation\u00a0<\/em>Mathlet into your courses.<\/li>\n<\/ol>\n<div class=\"shaded\">\n<h4><em class=\"fa fa-lightbulb-o\"><\/em> Think About<\/h4>\n<p>Explain how parts of this may be useful in one of the classes you teach.\n<\/p><\/div>\n<hr \/>\n<h3>4.2 Second Assignment: Frequency Response<\/h3>\n<p><em>This assignment would be appropriate in an\u00a0engineering-oriented ordinary differential equations class.<\/em><\/p>\n<div class=\"shaded\">\n<h4><em class=\"fa fa-rocket\"><\/em> Launch<\/h4>\n<p>Mathlet:\u00a0<em><a title=\"Amplitude and Phase: Second Order IV\" href=\"http:\/\/mathlets.org\/mathlets\/amplitude-and-phase-second-order-iv\/\" target=\"_blank\" rel=\"noopener\">Amplitude and Phase: Second Order IV<\/a><\/em>\n<\/div>\n<p>This Mathlet deals with a spring-mass-dashpot system. The input signal is a\u00a0force acting directly on the mass, given by <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=F%28t%29%3D%5Ccos%28%5Comega+t%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"F(t)=&#92;cos(&#92;omega t)\" class=\"latex\" \/>;\u00a0the system response is the displacement of the mass from its neutral\u00a0position. We are interested only in the periodic system response.<\/p>\n<p>Observe the various sliders and their functions.<\/p>\n<ol class=\"mathlet-course-ol\">\n<li>Set <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=m%3D0.50&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"m=0.50\" class=\"latex\" \/>, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=b%3D0.33&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"b=0.33\" class=\"latex\" \/>, and <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=k%3D0.50&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"k=0.50\" class=\"latex\" \/>.\u00a0Sweep the angular frequency from\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega%3D0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega=0\" class=\"latex\" \/> to <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega%3D4&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega=4\" class=\"latex\" \/>. Use the Mathlet to\u00a0give a qualitative description of the system response, as it\u00a0relates to the input signal: is its amplitude greater than or less than that of\u00a0the input signal? is it in phase or does it lag behind? These observations will depend upon the value of <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega\" class=\"latex\" \/>. What happens when <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega\" class=\"latex\" \/> is small? large?<\/li>\n<li>Still with these settings, it appears that there is a particular value of\u00a0the input angular frequency <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega\" class=\"latex\" \/> for which the amplitude of the system\u00a0response is maximal. This is the <em>near-resonant frequency<\/em>, written\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega_r&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega_r\" class=\"latex\" \/>. Estimate <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega_r&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega_r\" class=\"latex\" \/> from the Mathlet, and the amplitude of the\u00a0corresponding system response. It may help to invoke\u00a0the Bode plots: the upper one shows the amplitude of the system response\u00a0as a function of <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega\" class=\"latex\" \/>. Then compute <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega_r&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega_r\" class=\"latex\" \/> and the amplitude\u00a0of the near resonant system response exactly (using <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=b%3D%5Cfrac%7B1%7D%7B3%7D&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"b=&#92;frac{1}{3}\" class=\"latex\" \/>, which\u00a0was approximated by <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=b%3D0.33&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"b=0.33\" class=\"latex\" \/> in the Mathlet experiment.) Compare the results.<\/li>\n<li>Still with these settings, it appears that for a certain angular frequency the phase lag (of the system response relative to the\u00a0input signal) is exactly <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Cpi%2F2&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;pi\/2\" class=\"latex\" \/>. Use the Mathlet to estimate this value of\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega\" class=\"latex\" \/>. It may help to invoke the Nyquist plot. This displays the <em>complex gain<\/em>\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=he%5E%7B-i%5Cphi%7D&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"he^{-i&#92;phi}\" class=\"latex\" \/>, where <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=h&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"h\" class=\"latex\" \/> is the amplitude of the system response\u00a0(which, since the amplitude of the input signal is 1, is the <em>gain<\/em>)\u00a0and <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Cphi&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;phi\" class=\"latex\" \/> is the phase lag. Then compute this angular frequency exactly.\u00a0Compare the results.<\/li>\n<li>Now find the near-resonant angular frequency <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega_r&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega_r\" class=\"latex\" \/> for general\u00a0values of <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=m&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"m\" class=\"latex\" \/>, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=b&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"b\" class=\"latex\" \/>, and <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=k&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"k\" class=\"latex\" \/>.<\/li>\n<li>Your formula for <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega_r&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega_r\" class=\"latex\" \/> may not make sense for all values of\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=m&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"m\" class=\"latex\" \/>, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=b&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"b\" class=\"latex\" \/>, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=k&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"k\" class=\"latex\" \/>. When it does not make sense, there is no near-resonant peak\u00a0(or you could say that near-resonance occurs at <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega%3D0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega=0\" class=\"latex\" \/>). Give an inequality\u00a0among the parameters <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=m&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"m\" class=\"latex\" \/>, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=b&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"b\" class=\"latex\" \/>, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=k&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"k\" class=\"latex\" \/>, guaranteeing the existence of a\u00a0near-resonant peak. Find at least one marginal case (where the inequality\u00a0you discovered is replaced by an equality) on the Mathlet.<\/li>\n<li>For general values of <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=m&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"m\" class=\"latex\" \/>, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=b&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"b\" class=\"latex\" \/>, and <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=k&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"k\" class=\"latex\" \/>, find the value of <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega\" class=\"latex\" \/>\u00a0for which the phase lag is exactly <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Cpi%2F2&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;pi\/2\" class=\"latex\" \/>.<\/li>\n<\/ol>\n<h4>Comments<\/h4>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/onlineteachered.mit.edu\/files\/2014\/05\/mathlets-in-homework-300x199.jpg\" alt=\"Photo of student working on homework with Mathlet in background\" width=\"300\" height=\"199\" class=\"alignright size-medium wp-image-1263\" style=\"margin-right: 0;\" srcset=\"https:\/\/onlineteachered.mit.edu\/files\/2014\/05\/mathlets-in-homework-300x199.jpg 300w, https:\/\/onlineteachered.mit.edu\/files\/2014\/05\/mathlets-in-homework-1024x679.jpg 1024w, https:\/\/onlineteachered.mit.edu\/files\/2014\/05\/mathlets-in-homework-321x214.jpg 321w, https:\/\/onlineteachered.mit.edu\/files\/2014\/05\/mathlets-in-homework-207x136.jpg 207w, https:\/\/onlineteachered.mit.edu\/files\/2014\/05\/mathlets-in-homework-140x94.jpg 140w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/> Mathlets in Homework(1) It is very important to be explicit about what you\u00a0want students to do: is it enough for them to make observations from the\u00a0Mathlet, or do you expect them to prove (or calculate) things?<\/p>\n<p>The rest of the parts of this problem are only reasonable as homework after\u00a0you have modeled this kind of think in lecture. There are various approaches.<\/p>\n<p>Here is how I like to teach this: Write <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=p%28s%29%3Dms%5E2%2Bbs%2Bk&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"p(s)=ms^2+bs+k\" class=\"latex\" \/> for the\u00a0characteristic polynomial, so the equation is <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=p%28D%29x%3D%5Ccos%28%5Comega+t%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"p(D)x=&#92;cos(&#92;omega t)\" class=\"latex\" \/>\u00a0where <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=D&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"D\" class=\"latex\" \/> is the differentiation operator. The input signal is\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Ccos%28%5Comega+t%29+%3D+Re+e%5E%7Bi+%5Comega+t%7D&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;cos(&#92;omega t) = Re e^{i &#92;omega t}\" class=\"latex\" \/>. The equation has real coefficients,\u00a0so if we have a solution <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=z_p&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"z_p\" class=\"latex\" \/> of the complex equation <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=p%28D%29z%3De%5E%7Bi%5Comega+t%7D&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"p(D)z=e^{i&#92;omega t}\" class=\"latex\" \/>\u00a0then <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=x_p%3DRe+z_p&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"x_p=Re z_p\" class=\"latex\" \/> is a solution of the original equation. Now,\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=De%5E%7Brt%7D%3Dre%5E%7Brt%7D&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"De^{rt}=re^{rt}\" class=\"latex\" \/>, from which it follows that <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=p%28D%29e%5E%7Brt%7D%3Dp%28r%29e%5E%7Brt%7D&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"p(D)e^{rt}=p(r)e^{rt}\" class=\"latex\" \/>,\u00a0so we can take <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=z_p%3De%5E%7Bi%5Comega+t%7D%2Fp%28i%5Comega%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"z_p=e^{i&#92;omega t}\/p(i&#92;omega)\" class=\"latex\" \/>. The amplitude of\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=x_p%3DRe+z_p&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"x_p=Re z_p\" class=\"latex\" \/> is then <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%7C1%2Fp%28i%5Comega%29%7C&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"|1\/p(i&#92;omega)|\" class=\"latex\" \/>, and this is what we are asked to\u00a0maximize. Maximizing this is minimizing the magnitude of\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=p%28i%5Comega%29%3D%28k-m%5Comega%5E2%29%2Bbi%5Comega&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"p(i&#92;omega)=(k-m&#92;omega^2)+bi&#92;omega\" class=\"latex\" \/>, or of its square\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%28k-m%5Comega%5E2%29%5E2%2Bb%5E2%5Comega%5E2&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"(k-m&#92;omega^2)^2+b^2&#92;omega^2\" class=\"latex\" \/>. You can do this by setting the derivative\u00a0with respect to <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega\" class=\"latex\" \/>, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=2%28k-m%5Comega%5E2%29%28-2m%5Comega%29%2B2b%5E2%5Comega&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"2(k-m&#92;omega^2)(-2m&#92;omega)+2b^2&#92;omega\" class=\"latex\" \/>,\u00a0equal to zero.\u00a0This gives <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega%3D0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega=0\" class=\"latex\" \/> and <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Cpm%5Comega_r&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;pm&#92;omega_r\" class=\"latex\" \/> where <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega_r%3D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D-%5Cfrac%7Bb%5E2%7D%7B2m%5E2%7D%7D&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega_r=&#92;sqrt{&#92;frac{k}{m}-&#92;frac{b^2}{2m^2}}\" class=\"latex\" \/><\/p>\n<p>This solves (4), and\u00a0with our values for <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=m&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"m\" class=\"latex\" \/>, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=b&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"b\" class=\"latex\" \/>, and <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=k&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"k\" class=\"latex\" \/>, it gives the answer to (2), <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega_r%3D%5Csqrt%7B7%7D%2F3%5Csimeq0.882&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega_r=&#92;sqrt{7}\/3&#92;simeq0.882\" class=\"latex\" \/>.<\/p>\n<p>(3) and (6)\u00a0The complex gain is <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=1%2Fp%28i%5Comega%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"1\/p(i&#92;omega)\" class=\"latex\" \/>, so we are asked to find the\u00a0angular frequency for which the real part of <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=p%28i%5Comega%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"p(i&#92;omega)\" class=\"latex\" \/> is zero and the\u00a0imaginary part is positive. This occurs when <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=k%3Dm%5Comega%5E2&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"k=m&#92;omega^2\" class=\"latex\" \/>, or\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega%3D%5Csqrt%7Bk%2Fm%7D&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega=&#92;sqrt{k\/m}\" class=\"latex\" \/> (independent of <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=b&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"b\" class=\"latex\" \/>!).\u00a0With our values of <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=k&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"k\" class=\"latex\" \/> and <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=m&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"m\" class=\"latex\" \/> this gives <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega%3D1&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega=1\" class=\"latex\" \/>.<\/p>\n<p>(5) For <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega_r&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega_r\" class=\"latex\" \/> to be real, the contents of the square root have to be\u00a0non-negative. So <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=2mk%5Cgeq+b%5E2&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"2mk&#92;geq b^2\" class=\"latex\" \/>.<\/p>\n<p>This is a lot of computation and somewhat abstract. It is very reassuring to\u00a0students to see the phenomena displayed on the Mathlet.<\/p>\n<p>The <em>complex gain<\/em>\u00a0is (in this problem) given by <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=1%2Fp%28i%5Comega%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"1\/p(i&#92;omega)\" class=\"latex\" \/>.\u00a0The trajectory of this complex-valued\u00a0function of <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega\" class=\"latex\" \/> is what is displayed in the [Nyquist plot]. It contains both the gain and the phase lag, and the Nyquist plot\u00a0shows both and the relationship between them.<\/p>\n<p>Technically, we have not drawn Bode plots: engineers would draw the\u00a0log of the frequency horizontally, and the log of the gain vertically.\u00a0My experience is that using log plots is a step too far for students at\u00a0the stage at which I see them in my classes.<\/p>\n<p>Also, a Nyquist plot, properly speaking, shows the trajectory of\u00a0the complex gain for <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Comega&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;omega\" class=\"latex\" \/> negative as well as positive.<\/p>\n<p>For more Mathlets addressing these more advanced issues,\u00a0see the Mathlets <a title=\"Bode and Nyquist Plots Mathlet\" href=\"http:\/\/mathlets.org\/mathlets\/bode-and-nyquist-plots\/\" target=\"_blank\" rel=\"noopener\"><em>Bode and Nyquist Plots<\/em><\/a>\u00a0and <a title=\"Nyquist Plot Mathlet\" href=\"http:\/\/mathlets.org\/mathlets\/nyquist-plot\/\" target=\"_blank\" rel=\"noopener\"><em>Nyquist Plot<\/em><\/a>.<\/p>\n<hr \/>\n<h2>5. Mathlets in Homework, Conclusion<\/h2>\n<p>If you decide to assign Mathlets as part of homework there are a few things to consider. Watch this video segment where I discuss some of the items you need to consider.<\/p>\n<p><!-- ** Vimeo Video Source, hrm@math.mit.edu --><\/p>\n<div style=\"padding:56.25% 0 0 0;position:relative;\"><iframe src=\"https:\/\/player.vimeo.com\/video\/1036487168?title=0&amp;byline=0&amp;portrait=0&amp;badge=0&amp;autopause=0&amp;player_id=0&amp;app_id=58479\" frameborder=\"0\" allow=\"autoplay; fullscreen; picture-in-picture; clipboard-write\" style=\"position:absolute;top:0;left:0;width:100%;height:100%;\" title=\"MIT Mathlets in Homework, Module 3, Segment 2\"><\/iframe><\/div>\n<p><script src=\"https:\/\/player.vimeo.com\/api\/player.js\"><\/script><br \/>\n<!-- ** Backup YouTube Video Source\n<iframe loading=\"lazy\" width=\"560\" height=\"315\" src=\"https:\/\/www.youtube.com\/embed\/KbDFOD8yCGI?si=9C-5xh3tdPj6P33a\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n--><br \/>\nTranscript [ <a href=\"https:\/\/onlineteachered.mit.edu\/files\/2014\/08\/Module-3-Mathlets-in-Homework-Conclusion-Transcript.pdf\" target=\"_blank\" rel=\"noopener\">PDF<\/a> ]<\/p>\n<div class=\"shaded\">\n<h4><em class=\"fa fa-lightbulb-o\"><\/em> Think About<\/h4>\n<p>How might you use the Mathlets in one of the classes you teach?<\/p>\n<\/div>\n<hr \/>\n<p>These materials are Copyright &copy; 2013, Massachusetts Institute of Technology and unless otherwise specified are licensed under a <a href=\"http:\/\/creativecommons.org\/licenses\/by-nc-sa\/3.0\/\">Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported<\/a> license.<\/p>\n<hr \/>\n<p><!-- For use on mathlets.org\/training --><!--slice end--><\/p>\n<div class=\"prev-page-container\"><a href=\"http:\/\/onlineteachered.mit.edu\/courses\/mathlets\/module-2\/\"><em class=\"fa fa-hand-o-left\"><\/em> Module 2 &#8211; Mathlets in Group Work<\/a><\/div>\n<div class=\"next-page-container\"><a href=\"http:\/\/onlineteachered.mit.edu\/courses\/mathlets\/wrap-up\/\">Wrap up <em class=\"fa fa-hand-o-right\"><\/em><\/a><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Each segment of this module consists of sample homework assignments created around one or another of the MIT Mathlets.\u00a0Each segment is followed by questions, my remarks about the features of the Mathlets, and examples that illuminate their use.\u00a0At the end there are a couple of exercises. 1. In this module&#8230; In this module, you will &hellip; <a href=\"https:\/\/onlineteachered.mit.edu\/courses\/mathlets\/module-3\/\" class=\"more-link\">Continue reading <span class=\"screen-reader-text\">Mathelets in Homework, Module 3<\/span> <span class=\"meta-nav\">&rarr;<\/span><\/a><\/p>\n","protected":false},"author":32,"featured_media":0,"parent":552,"menu_order":3,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"jetpack_post_was_ever_published":false,"footnotes":""},"jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/P4sx5t-9B","_links":{"self":[{"href":"https:\/\/onlineteachered.mit.edu\/wp-json\/wp\/v2\/pages\/595"}],"collection":[{"href":"https:\/\/onlineteachered.mit.edu\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/onlineteachered.mit.edu\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/onlineteachered.mit.edu\/wp-json\/wp\/v2\/users\/32"}],"replies":[{"embeddable":true,"href":"https:\/\/onlineteachered.mit.edu\/wp-json\/wp\/v2\/comments?post=595"}],"version-history":[{"count":87,"href":"https:\/\/onlineteachered.mit.edu\/wp-json\/wp\/v2\/pages\/595\/revisions"}],"predecessor-version":[{"id":2233,"href":"https:\/\/onlineteachered.mit.edu\/wp-json\/wp\/v2\/pages\/595\/revisions\/2233"}],"up":[{"embeddable":true,"href":"https:\/\/onlineteachered.mit.edu\/wp-json\/wp\/v2\/pages\/552"}],"wp:attachment":[{"href":"https:\/\/onlineteachered.mit.edu\/wp-json\/wp\/v2\/media?parent=595"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}