More Of The Same

6. Find the arc length for the spiral whose polar equation is r = e^\theta over the interval \theta \in [0, 1].

The “key example”: the easiest problem of its type. Should this be on a final exam? Heck no, it should be assigned early on and then referred to often. Here it is now. The old “unchanged by differentiation” trick…

5. Find a formula for the slope of the tangent to the “epicycloid” whose parametric equations are x = 5cos(t) - cos(5t) and y=5sin(t)-sin(5t).

Routine; I passed over it in near-silence on the review day. Also… owen by the way… played ’em ILMB and For John Henry.

4.Recall that the Taylor Series for the exponential function is e^x = \sum_{n=0}^\infty {{x^n}\over{n!}}. Use the 3^{rd} Taylor polynomial P_3 to obtain a (rational number) estimate for e^2 (write out the first four terms of the series; put x=2; simplify). Find a reasonable upper bound on the “error term” R_3 for this estimate.

The class that had this final didn’t “get” the control-the-error parts of the course; this one I didn’t even really try. You should… whoever you are… but if you’re my student, I’ll go ahead and admit there won’t be an “error term” problem on this quarter’s exam. Be able to write out a Taylor Series and use it in estimating a number; the rest, for us, is gravy.

More still to come I imagine.

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Last Year’s Final. Partly. Kind of.

I asked the 153 students to tell me a problem they’d like to see on the final exam (sort of as a challenge to other students)… the three that were named aloud today in the hearing of all present were p.806#41, p807#89, and p796#11.

I passed around last years exam and looked over quite a bit of that. To wit.

10. Find the derivative of the vector-valued function \vec{r}(t) = \sec(t)\vec{i} + \tan(t)\vec{j}.

The topic is barely introduced in this course; the whole message is “termwise differentiation works” (integration too of course… pesky constants emerge… you know the drill). That, and I get an excuse to bawl out the ones who botched “differentiate tan-x three times” in the middle of a Taylor Series calculation on the last exam.

9. Find the angle between \langle - \sqrt{3}, 1\rangle and \langle 3, \sqrt{3}\rangle.

In the actual document I had the \vec{i}, \vec{j} notation; in today’s whiteboard notes, I switched as I’ve done here. It looks cleaner and easier to understand to my eyes. Anyway, one has \vec{v}\cdot\vec{w} = |\vec{v}| |\vec{w}| \cos(\theta), where theta is of course the desired angle; the “absolute value” is (again of course) the square root of the sum of the squares of the co-ordinates of the vector in question (its “norm”, “magnitude”, “length”, or what have you… ). These last two were “gift” problems offered by way of apology for not having had time to do much more than introduce certain topics. In the case of #10; it’s built into the syllabus… in place of #9 one would prefer a problem that uses the “angle” property in service of some geometric problem.

8. Find the equation of the plane through (2,2,0), (1,0,1), and (0,1,1).

I love this problem. Three points in (x,y,z)-space not all on a line; find a linear equation. The exact natural generalization of a problem we drill diligently into hundreds of Math 102 students every quarter [two points in (x,y)-space; find the linear equation]. That’s just got to be interesting. Sure enough, the picture, together with both products (“dot” and “cross”), tell the whole story; forget the messy “formula”.
See clearly how to do it and just do it this way.

Gotta go. More later.

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Caviar For The General

Wikipedia is good (as usual) on the binomial theorem; part of today’s lecture will overlap parts of this. Ideally, one would already have the “natural number exponent” case completely under control before exhibiting Newton’s mighty generalization. And yet calculus students have typically never studied this incredibly basic topic; somehow this has been shunted off to the Statistics arm of the curriculum (where, ideally, it becomes part of Exhibit A in the—necessarily handwavy—”Central Limit Theorem” [roughly, “everybody tries to be Normal”; to be any less rough I have to tell you about Pascal’s Triangle, Binomial Coefficients, Bernoulli Trials and stuff like that—the Binomial Theorem (“easy” form)]; this work is typically omitted and its results handled by “technology” rather than student understanding).

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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.

A Much-Needed Gap

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

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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.

The Smell Of Chalkdust In The Morning

Turns out an earlier edition of this text had a whole chapter on Conic Sections. One section each for Parabolas, Ellipses, and Hyperbolas and a final section on Rotations. With the “Reflective Property of a Parabola” theorem—a “Problem Solving” problem of our text… one I’ve been looking at a little bit with the class—worked out in proof-from-the-book perfection. It even fits exactly onto a single page. Construct a quick triangle; show two sides match; conclude that the angles we want to match, do match. You can bet I’ll be zapping off copies for the class.

I’d looked at a couple of “high tech” approaches that I’m still not sure can be made to work. One student stayed a little after class last time and we worked through some calculations together; the end appeared to be in sight. I was deliberately holding back of course; I’ll probably look at this approach some more this weekend.

I’d decided to see what would happen by using an idea from an early example in the text: use slope as a parameter. Because then one of the numbers in the problem comes with a very simple representation. After fiddling, I’d decided I wanted the focus, not the vertex, at the origin (the idea here is again to make a certain number appear with an “easy” algebraic representation). The student in question took it from there (I’d showed my ideas to the whole class) and had worked out a few appropriate formulas; these matched mine from the night before… and so we were “on the same page” as we worked out the next couple of moves. Just how it oughta go; I’ll bet he’s got it finished up one way or another by Monday.

Anyhow, next I guessed: oh, hey. I’ll bet this Polar Co-ordinates stuff that we’re handling so badly in this version of the course could be the best way to go. And it very well may be one very good way to go. I think I know the next “formula” I’d need to work out and have spotted an exercise earlier on that might be a really useful hint. Then… let no one else’s work evade your eyes… I looked to the books on the shelf.

And right there, first one I looked at (Edwards & Penny… another two-editions-ago freebie of course) had a sort of a neat one (with an unusual twist in the logic that I thought I’d rather avoid). And I found the proof-from-the-book ideal next in the older Larson. And now I’m about to look on the net.

Because I sure as heck put in the instructions that looking things up counts as work in this context and should be presented proudly. Then, like a fool, when I was trying to get everybody to volunteer in good order to work up front I somehow more-or-less forgot that we’ve got a computer hook-up where the whole class can see what the speaker does with the computer up front and I’ve used it—and on that day—to get up on the net for illustrations and whatnot and had meant to ask a student to look up some leads online as part of the “quiz” which consists of “everybody shows everybody what they can do”.

So Monday maybe somebody’ll show us some browsing skills. Here’s some of mine while we’re waiting. No, wait. They’ll’ve appeared as links (not yet found as I write these words; nor as I first post ’em momentarily). God bless this World Wide Web.

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