## 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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## Down At The Bottom Of The Garden

What with the outstanding early returns in Calc I, I’m even more convinced that “students working up front in the classroom” is something I should be doing in every class (at least once per quarter; I’m far from saying “every class meeting” [unless in a class of about three]). So the plan is to make part of the “quiz” grade depend on doing a presentation.

Today I drew a circle centered on a co-ordinate frame and filled it in. “Use calculus to find the area bounded by the unit circle”.

That is, though I didn’t say so, find the area of the unit disk. Since I’m feeling expansive and don’t always dare to introduce such sidetrackers in lecture—or even in “problem solving” sessions—the unit circle is known in the pros as $S^1$ (the “One-Sphere”… the n-sphere generally is the set of $\langle x_1, x_2, ... , x_n\rangle$ such that $x_1^2 + x_2^2 + ... + x_n^2 = 1$ [i.e., points at distance 1 from 0 (the zero vector]). If the unit ball has a cool symbol of its own I’ve evidently forgotten it but of course it’s the set of points at distance 1or less from zero. Back to today’s class.

Our first volunteer then wrote out $y = \sqrt{1-x^2}$ and observed that since the bit in the first quadrant was obviously a quarter of the whole, we were looking at $4\int_0^1 \sqrt{1-x^2} dx$. Okay, good work… how do you do the integral? A brave stab at a “u-substitution”; I jump in (much too soon I imagine): “We want a trig substitution here…” and, when he doesn’t know right away how next to act, I allow as how it’d be okay to pass the chalk (really whiteboard marker; ugh) to somebody else. Who works through it pretty well, though with a few gentle reminders from me (again probably premature… of course I want corrections to come from the other students… and there were some of these) about, in particular, changing the limits on the integral when integrating with respect to the new variable. So things are going swimmingly.

“Now somebody do it in polar co-ordinates!” (Along with a little mini-rant by me to the effect that I should’ve already—in the first lecture about polar co-ordinates—have mentioned that [of course] “radius equals constant” gives the equation of a circle [centered at the origin] and “theta equals constant” [theta—again of course—here denotes the angle in a point given by polar co-ordinates $(r, \theta)$] is the equation of a line through the origin [one is supposed to say “pole” here and probably soon I’ll have that habit]. I’d like to be able to take it for granted that these are glaringly obvious [when pointed out] but actually they’d make darn good quiz questions in pre-calculus and might slip few minds even around here. To continue.)

And somebody does. For better or worse, it’s the student doing much the most participation during the others’ presentations (so this is probably way too easy for him); I’d hinted that I intended to make the problems get harder to encourage early volunteers (this appears to’ve been a tactical error). Anyhow, he knocks it off flawlessly… one could wish for a teacher-like running commentary I suppose but anyhow the written work required no corrections (and he wasn’t entirely silent as students often will be with boardwork). Part of the point here is how much easier the calculation is— $\int d\theta$ could make a pretty good claim to be the world’s simplest integral ( $\int 0 d\theta$ is of course its only rival).

The work has essentially been done by choosing the right notations (and having already developed the “formula” $A = {1\over 2}\int_{\theta_1}^{\theta_2} r^2 d\theta$; I took the opportunity to remind everyone—anything you only say once in a lecture series, you might as well never have said at all [or so it sometimes seems]—that the “one-half” comes about because of triangles in a certain drawing… and that remembering the reason it’s there is a good way to remember that it is there).

Then a textbook exercise: slope of a parametric curve at a point. More good student work; some more hints from me about connections with stuff we’ve looked at before. Next exercise, the area bounded by the same curve gets no takers and I’m content to look the others in the eye and say they all owe me boardwork.

There’s plenty of other stuff I want to do in our two-hours-and-change today; more than I’ll ever be able to do… and this is progress. Still, it’s beginner’s progress indeed for me. I’m so doggone uncomfortable asking anybody to do anything they’re uncomfortable with… and it’s so easy to just wimp out and just lecture…I’d be deeply ashamed of this if so many of my colleagues weren’t even worse (at drawing students out by voice… at insisting on homework and times and dates and whatnot I’m probably one of the biggest wimps there is).

So the rest is mostly me talking about the other mid-course correction I’ve just instituted: “Project” style homeworks. I’d mentioned a couple problems as potential “project” problems; today I opened it up and said “pick any two from the ‘P.S.’ sections of chapters 9 and 10″… and went on to talk about what I‘d done with the two I’d singled out on Monday (without giving too much away; these can still be finished for credit).

Then went on and (quickly) worked on still another topic (Taylor Series… the centerpiece for half the course). But the point is that with the student boardwork and the substantial problems, I feel I’m one step closer to the course as it ought to be.

At which point there ought to be a rant about the textbook. I’ve got quite a bit of material prepared. But that’s it for tonight. Thanks for your kind attention.

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## A Drop In The Ocean

Free online Calc III notes by Paul Dawkins (spotted at Lane Vosbury’s Calc III Exam notes). This stuff is mostly Calc IV at Midstate Community College but looks like a pretty good guide for me: how to do it right.

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## Aspects Of A Topic

Our topic is “arc lengths for curves given parametrically”. Good news right away: the first display is (or would be were I not graphically challenged unto uselessness… probably oughta get some pencilpad gizmo or something) the drawing—a curve in the plane with line segments connecting a few points—that accompanies the derivation of the main “formula” (in the sense of “memorize this first of course”) for the section at hand (1): $\int_a^b \sqrt{x_t^2 + y_t^2} dt$.

The first thing I’d like to’ve said about that is that it’s easier to make sense of this formula than of something like $\int_\alpha^\beta \sqrt{1 + ({{dy}\over{dx}})^2} dx$. This may be a matter of taste but I doubt it. It appears to me that by thinking of the x and y co-ordinates of a point of the curve as functions of a parameter, we are removing an artificial asymmetry that exists in the “y is a function of x” point of view. By not even having to ask which co-ordinate is to be considered “independent” and which “dependent”, we get a simpler proof and a more symmetric, principle-revealing, easily generalized thing-of-beauty and joy-forever, formula (1).

The next thing is probably along the lines of “this derivation would be well worth looking at even if it were entirely review”. Touching various parts of the drawing with chalk or pencil (or in our case, alas, whiteboard marker) while talking about the symbols leads in the best-case scenario to thinking like a physicist… one becomes able to perform certain calculations by manipulating some visual-seeming “object” in the time-and-space of one’s imagination. This is something I’m not very good at; quite often like the Algebra major I am I’m reduced to pure symbolism and have to think of certain combinations of letters (and suchlike symbols) undergoing certain transformations according to rules.

What’s always been clear to me—to my considerable frustration—is that, once one “sees” the meaning of, say $ds = \sqrt{x_t^2 + y_t^2} dt$ (rather than thinking of it as a piece of “code” to be “plugged in” as part of some calculation), a proof like that for the Area of a Surface of Revolution becomes simply an exercise in writing down the obvious.

The “geometric” material in this course was mostly invented for the needs of physical science (the study of electromagnetic forces in particular). If you get good at it now, probably you could go on to do well in ODE and grab hold of the half of the Program that almost completely eluded me.

(Instant footnote: “ODE” means “Ordinary Differential Equations” in mathmajorese [everybody else says “diff, E, Q”; ugh]—the subject that follows our course in a typical STEM major (Science, Tech, Engineering, Math) and which is typically [but not for me] followed by PDE—Partial Differential Equations—also known, in math departments, as “Applied Mathematics”. Math for physics, in a word…)

Any new “formula” should be taken for a test drive. In the last lecture, I looked at two problems that somehow eluded me last time… the line segment and the arc of a circle. These are the simplest interesting examples; one habit of mind I’ve found very useful is to look for these. I blogged about this idea last quarter in VME. Specifically, I did x(t) = t/3, y(t) = t/4, for $t \in [0, 12]$ (the segment connecting the origin to (4, 3)… like a beginner I pulled this parametrization out of thin air in my eagerness to apply formula (1)… today I’ll probably say some more about how to find it…); then I said “here’s a problem from today’s quiz; I’m afraid the class as a whole isn’t ready; somebody do it at the board and I’ll give the whole class full credit”… and put up x =cos(t), y=sin(t), $t \in [0, \pi/2]$ and somebody stepped up and did it. Yay, this class.

As it turns out, there are only a few “types” of arc length problem where it’s practical to work out exact answers with paper-and-pencil methods so suchlike basic-example motivation might be even more crucial than necessary. As simple a curve as the parabola is already brutally tough for Calc III according to my admittedly very limited understanding: the curve $x = \sqrt{t}, y = 3t - 1, t\in [0,1]$—a textbook problem—requires a tricky substitution in the integral (supplied by a very capable student last week) even in order to transform it into another tricky integral, one I blogged about last year.

Then there’s the cycloid (and related curves). And some carefully constructed quadratics, designed so that “middle terms” will “cancel” in a convenient way. And that’s all I know. Probably we’ll do one from this last category today. The first exam is Monday…

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