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Make It New

Last year’s posts about Arc Length On The Cardiod, Vector Products, and integral-secant-cubed are relevant again. The cardiod was a recent in-class example; the “vector” note corrects a mistake I didn’t make until yesterday; the “surprisingly useful integral” is one I had students to look up in a table on the recent exam (and work with… nobody had this one all the way right but me though in principle it’s an “easy” problem when the table is available…).

What the heck. Here’s the problem: The polar equation r = \theta, for \theta \in [0, \infty), determines an “Archimedean Spiral”, S. Determine the arc length along S for \theta between 0 and \pi\over6 (exactly; a messy “formula”… check the work numerically). And the answer: {{\pi\sqrt{36+ \pi^2}}\over{72}} + {1\over2} ln({{\pi +\sqrt{36+\pi^2}}\over6}). This evaluates to about .5466… as does \int_0^{\pi\over6} \sqrt{\theta^2 + 1}d\theta… so we can be reasonably sure this is right.

In my own work messy expressions like this are almost never right the first time… on the copies of the handwritten solution I distributed yesterday, quite a bit of erased work is clearly visible. Rooting out every last little mistake until things are just right is of course a vital part of the process for a lot of problems. One student was real close.

Foundational Adventures

Busy Writing; No Time For Blogging

\bullet Carnival of Mathematics #52.
\bulletPlanar Curves at Numericana.
\bulletSumidiot on Wolfram Alpha and calculus reform.

Another Great Weekend…

When we play dress grownup

Posted in Links. 2 Comments »

Buy Conditionally

The 151 quizzes are good. I caught almost everyone with the trusty old “differentiate a constant” wheeze… y = 11e^\pi \Rightarrow {{dy}\over{dx}}= 0 of course, but you wouldn’t've known it by these guys on the day. In fact—math teacher at play— \{11, \pi 11e^{\pi-1}, 11e^\pi, 11\root7\of{e^{22}}, e^\pi, 0\} is the “answer set” (sure as heck not the solution set!)—with 11e^\pi appearing several times (once as {{11}\over{e^{-\pi}}}) and a variant (11\pi e^{(\pi - 1)}… slightly better) on one other value.

Now. If I’m going to go in claiming that anybody “should have known” something… then how come they should’ve? Let’s see. 11e^x, 13e^\pi, cos(x), 1/x^2, polynomial, radical. How come two of ‘em are alike? What’s different about ‘em? This is pre-math stuff here: mere “test-taking skills”. Okay. What else?

Am I gonna say everybody oughta be trying to imagine a graph? Like, every single time? Well, maybe it’d be a good idea… at least right in here at first… to at least consider it. The trig function everybody darn well oughta know is some wavy thing; the polynomial’s a cubic so again one has a pretty good idea. The radical and the rational function are a little more esoteric and I wouldn’t blame a Calc I student a bit for not thinking too hard about it if nothing pops up for free in instant recall; if it comes to thinking about it, maybe it’d be a good time to break out the grapher. As for the problems with e‘s in ‘em, I do feel that everybody oughta have a mental image of “exponential growth” (y = A_0 e^{Kx}, say [with A_0 and K positive constants]): it sure is a simple doggone thing to sketch…

So then. With any luck, one then asks oneself, “what’s the graph of e-to-the-pi then, eh?” and arrives somehow at the necessary insight. In fact, if one realizes only that one does not yet know the answer, one might, again, think of using the graphing feature of the calculator (or, of course, think of estimating e^pi directly on the command line ["homescreen"... whatever]). Once you see that flat line… on screen or paper or in imagination… you’ve got it.

OK, true enough. Is that all you got? Well, I suppose a drill-instructor style teacher might favor repeating over and over until blue in the face “pi and e are constants“. Might be worth a try…

Am I supposed to be, like, hanging on every bloody word then? Well, yes, ideally, but hanging on every bloody symbol would be a darn good place to start. One is always led to ask, “what does this symbol mean in this context?”—indeed, this sounds suspiciously like a description of reading itself (and not just of reading “math”—calculations and such).

One is curious at this point as to whether they’d've done any better with y= e^3… but enough.

Because the good news is that there were five otherwise-perfect papers and several other solid high scores; in particular, the class redeemed itself very nicely with the “epsilon-delta” proof. Nonetheless, here are some trouble spots.

\bulletNo “Let \epsilon \rangle 0“. For this section, this line is mandatory. (One may of course “fix” epsilon or some other very slight variant.) I’m declaring by fiat that “all epsilon-delta proofs begin by making epsilon the name of a constant“.

\bulletEqual sign for implication.

\bulletNo mention of \delta.

\bullet\delta = {\epsilon\over{|slope|}}“—right there in the calculation as if english words were appropriate in the middle of algebraic expressions. In the actual exercise, the grader is hoping to see \epsilon\over5. But hold on. I put the “mixed media” (algebra-and-plain-english) on the very blackboard myself! Ah, but it was in a marginal note about problems of this type generically… in particular problems (of this type—limits of linear functions) one will work with honest algebraic code. Should students be expected to just recognize slangy shorthand as such in its context while we’re preaching meticulous attention to detail in some other part of the work? Maybe. We can’t be explicit about everything.

\bulletThe very inequalities I most want to see… in reverse order. Or in variant orders harder to classify… “no apparent order at all” having made at least one appearance.

\bulletInequalities replaced by equations; reversed inequalities.

The thing is… at least some of the lack of clarity here ought to be considered my fault. In particular, of the calculations after fixing epsilon and writing delta as a function of epsilon… the calculations that we should be careful to present in the reverse order to the one we actually discover it in… I now feel one should explicitly say “(Then) The Following Are Equivalent” (and I’ll introduce the abbreviation TFAE for this situation in the next class). If we’re going to obsess over this definition… and we should… then it’s the logical structure that seems to present the biggest challenge; the equivalence of a certain set of inequalities is pretty close to the heart of the matter; we should be trying to spell this out as clearly as we know how.

I’ll probably have more to say about this…

A Much-Needed Gap

The Vector Calculus Bridge project at Oregon State U; spotted at Intute’s Calculus portal.

Posted in 153, Links. 1 Comment »

Parabolic Language

So sure enough, the 153 student… “Charles”, say (since that’s nobody’s name in the class)… that stayed after class last time to talk about the parabola problem had nailed it; it’s pretty doggone sweet so here goes.

Consider a parabola in the plane: the “locus” (a term with, for me, a sort of archaic feel) of points, P, such that the distance from P to a certain fixed point (the focus, F) is equal to its distance from a certain fixed line (not passing through F; the directrix [d, I suppose... though I doubt I'll need to refer to it in code]). Dropping a co-ordinate frame (with the usual (x,y) notations) having its origin at F and its x-axis parallel to the directrix, scale so that the vertex is at (0, -1); the equation of the parabola is then 4(y+1)=x^2 (by Math 150—the “generic” 4A(y-K) =(x-H)^2 has [directed, vertical] distance A between vertex and focus, and vertex (H,K)).

Now sketch the tangent line to the parabola at P. We aim to show that the angle formed by this line and the line through points F and P is equal to the angle formed by the same tangent line and a vertical line. (When vertical lines are considered as “sunlight”, this “reflecting property of the parabola” will cause the reflected beams to concentrate at the focus [and fry the bug we've sadistically put there... hsss]).

Okay. Verticals are at angle \pi\over2 (i.e., at 90 degrees—as is usual in calculus we’ll use radian measure; also as usual angles are measured from the positive x-axis ["standard position"])… so we get that for free. Following up on an example of the text, we can “parameterize” the parabola using the slope, M; since y’ = x/2 one has x = 2M; it follows that y = M^2 – 1.

Now, I’m not even sure myself at this point if this was a particularly useful thing to do… it may turn out that it’d've been easier not to parameterize… but the point of even trying this was that one also has a concise representation of another number we’ll need for our calculations: the “standard position” angle of a line with slope M is simply \tan^{-1}(M).

The other line in question, passing through P and the focus, has slope \tan^{-1}({y\over x})—this is why we put the origin at the focus. Referring to our figure (alas left to the reader; computer graphics is hard) we see that we require
{\pi\over2} - \tan^{-1}(M) = \tan^{-1}(M) - \tan^{-1}({{M^2 - 1}\over{2M}})
{\pi\over2} - 2\tan^{-1}(M)= -\tan^{-1}({{M^2 - 1}\over{2M}}).

Finally the magic: put M = \tan(\theta)… so we’re looking at
{\pi\over2} - 2\theta = -\tan^{-1}({{\tan^2(\theta) - 1}\over {2\tan(\theta)}}).
Apply the tangent function on both sides:
\cot(2\theta) = -{{\tan^2(\theta) - 1}\over {2\tan(\theta)}}.
But this is bygod just a familiar trig identity… namely, the “tangent double-angle law”
\tan(2\theta) = {{2\tan(\theta)}\over{1-tan^2(\theta)}}. Take that.

Because, while this isn’t quite the proof we should probably seek—that proof would involve careful checking that going “backward” from the identity we’ve ended with to the equation we seek (a certain pair of angles) won’t result in any snafus—nevertheless I’m convinced and I’m quitting here.

Because the real reason for the post is for me to say that Charles’s contribution in reaching this result was substantial: there were at least a couple of points where he was the first to arrive at a necessary insight.

Now. How much of this kind of really-doing-math can I reasonably expect to get with, well, say an A or a B or a C student? Ideally, quite a bit. And somewhere along the line I can imagine figuring out how. As of now I’m openly kind of flailing about. Which has to be OK. Because it’s the only way I know how to learn anything and in particular it’s essentially what I’m trying to get students to do. There’s a lot of groping around in the dark when you’re mathing.

I’d hate to tell you how long it took me to chase down the right equation for that parabola during this writeup, for example. I will go so far as to tell you that I first noticed something wrong at the double-angle law. (Charles and I had it right the first time, you understand. But one remembers ideas not formulas mostly and so I lost… oh, half-an-hour easy… finding a simple mistake. This kind of thing happens all the time and is presumably just the kind of thing that accounts for the tremendous mass appeal of math as a subject of study.)

So I flail about figuring out how to get ‘em to willingly do math with me and their peers… and they flail about with, sometimes, stuff I persist in calling “routine exercises” or even, (but always with scarequotes) ” ‘easy’ problems”. And big-picture stuff… and everything in between.

Mathematics begins in confusion… and ends in confusion. I wish I knew who said that. Anyhow, it’s true. Once you understand something, you’ve done some math: so remember where you put it in case you need it again… and find something else to think about… something you don’t understand…

Or type it up and post it on the net.

That’s The Way You Do It

Math Teachers at Play #6, at IWtTF.

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