56 Replies to “WitCH 63: A Separate Reality”

  1. Maybe I’m missing something here: it’s not exactly wrong as such, but a bit sloppy. One error is writing “x = ” in the second line, when of course it should be “dx = “. And this might be a purely personal beef, but I think the notation \sin^{-1}(x) for inverse sine should be abandoned; it’s too confusing with the comparable notation \sin^n(x) for powers of \sin(x)\arcsin(x) should be used instead. (I don’t so much mind the notation f^{-1} for the inverse of a generic function, but I don’t like the use of two separate conventions for trig functions.) My other beef is all those little fiddly “therefore” signs which I’ve never seen outside of school. Except once there’s an implication sign used instead. As an example of how to construct and present a mathematical argument it’s pretty poor.

      1. Wouldn’t be the first time… I’ll just wait for my betters to inform me of what I’ve missed.

        1. Don’t be so humble, Alasdair. All of us have missed things at one time or another. Personally, I’ve missed so many things it’s water off a duck’s back to me. Having said this, I’ll make a small correction to your comment:

          “One error is writing “x = ” in the second line, when of course it should be “dx = “. ”

          There is an error in the second line, but it’s not the x. They’ve divided by the sqrt and then integrated both sides wrt x, so the x is OK. However …

          btw I agree with you about the \therefore‘s, particularly in the last line. Diabolical stuff.

          1. Geez I seem to be digging myself into mathematical holes. That’s what you get for writing without thinking, and indeed I was wrong about the dx, as they had already integrated… but the more I look and learn, and read the answers, the clearer it becomes.

            Reminds me of somebody I knew a long time ago who was doing a masters degree by research, in group theory. After about a year of intense work, he proved that if a group satisfied certain conditions, it would also satisfy some others. Then somebody pointed out that the initial set of conditions were contradictory.

          2. I just noticed, should the second line also have a +c since, the author of the solution has seemingly taken the indefinite integral of the left hand side…?

            Or am I just nit-picking now?

            1. Hi RF. I’m fine with no ‘+ k’ in line two … I’m happy to assume it’s subsumed ex post facto into the ‘+c’ of line three.

  2. I think they should have chosen a different x to give the value of y at (I mean this one is contradictory). 🙂 And maybe just give the domain of the solution in the question, rather than pull it out of a hat to make sure the derivative stays positive.

    1. RF, you’ve missed the main point (as has everyone), but the solution not being defined at the initial condition is a pretty damn big point.

      1. Something to do with zero denominators when y=3 perhaps?

        I know this sounds like guessing (it partly is) – but your title about “real” got me wondering…

        1. It is always worth pondering titles for clues, but in this case there’s no such clue. Certainly the issue is with y = 3.

          1. When y=3, dx/dy = 0 which I find problematic as it (to my version of logic) means dy/dx is undefined.

            …or my version of what “undefined” means (which I will be the first to admit, I do not fully understand thanks to comments from others on a much earlier post).

  3. Thanks to everyone who has commented so far. Just a couple quick points.

    1) The text’s answer to the exercise is clearly screwed.

    2) It is not so obvious how to unscrew the answer.

    3) There are much deeper ideas at play.

    In ways, this Witch can be compared to the asymptotes WitCH. The exercise is *much* more complicated than intended, and properly sorting it out requires going well beyond school mathematics. And, as in the asymptotes case, the underlying moral will become clear: A man’s gotta know his limitations.

  4. Given dy/dx = 0 initially, y will stay at 3. So the solution is y=3. And yes this is related to the fact that the domain of the solution as given does not include the initial condition.

    PS Alasdair
    I would never claim to be your better.

  5. Hmmm… my first reaction is that y=3 is a solution. So my hunch is that there’s no unique solution to this IVP.

    Edit: I must have been reading another tab for longer than I thought, Tom made my point above.

      1. OK
        Is y=3\cos(x) also a solution? For this y' = -3\sin(x) whereas \sqrt{9-x^2} = 3|\sin(x)| which does agree on the interval [-\pi,0]. I consider this to be a “fuzzy” solution as it is unobtainable in the kinematics sense for a particle starting at t=0, even if we let time run backwards? Difference between a solution and an initial value problem, I guess.

        I was impressed at Latrobe as they taught the students to look for the envelope solution.

        Agreed, far beyond the students and even some teachers.

        1. tom, it would be far beyond pretty much all teachers. It’ll take me some serious time to sort it out, and I know this stuff. As for your dismissal of the “fuzzy” solutions, I think clutching at the kinematics straw is pretty unconvincing.

      2. I think there are multiple (once differentiable) solutions because because there is the solution that is piecewise

        y = -3 for x< - \pi and then y = 3  cos(x) for -\pi \le x < 0 and then y=3 for x \ge 0.

        So, you get a family of solutions that are constant at y=-3 and then move up using the
        y = 3 cos(x) at some point, and then become constant again? (Is that what it means to have an envelope solution – as Tom says?)

        I don't know how to show there are not even more solutions apart from these though and the constant ones. (Admittedly, I'm not very good at differential equations.)

        1. Thanks, wst. That’s one nice way to get a solution that includes an open interval around the trouble point. (I’m not quite sure what tom meant by “envelope”.)

              1. Aha ha. That’s funny, wst. The Stupid Design has lots of Bullschitz but no Lipschitz.

                1. wst, save yourself the confusion. There’s no point trying to get your head around this – students aren’t meant to see anything because there was nothing meant to be seen! Only the INcorrect obvious. The question was written in blissful ignorance and was intended to be completely benign.

                  The brighter students will realise that the given solution is flawed because of “the solution not being defined at the initial condition”. Those students are smarter than the writers of both the question and its so-called solution.

                  There will be a new edition of the textbook at some stage (probably no earlier than 2023, coinciding with the new Stupid Design) which might include an appropriate correction to this question.

                2. wst, as John said, the students aren’t meant to see anything here. The textbook writers simply screwed up (at least five years ago …)

          1. > I’m not quite sure what tom meant by “envelope”.

            When I first mentioned envelope I was saying that the solution is not unique. Looks like this is no longer standard 2nd year uni. work. Must learn to be more expansive. So here goes.

            Suppose we have a family of solutions to a DE, say y=3\sin(x-c). Graph them all together on x, y axes. The graphs occupy the entire solid strip between the envelopes y=3 and y=-3. Each envelope is also a solution to the DE. Why? Well each point on, say y=3, is tangent to one member of the family of solutions and hence has the same gradient and hence satisfies the DE. In general, if the family is F(x,y,c)=0, then an envelope is specified by this equation combined with the equation \partial F/\partial c = 0. I guess in the days of differentials, the \partial c gave an infinitesimal move to the ‘next’ member of the family. To get the envelope as a solution you also need an appropriate initial condition, in this case y(0)=3. If the question had said y(0)=2 the special solution would almost disappear.

            I am indebted to wst to learn that you can make a solution swap from the envelope to one of the original family and back again as much as you like. So there is not just a c infinity of different solutions, but even more! Maybe.

            I am a little put out Marty by your dismissal of my kinematics leaning 🙂 I sometimes find that something that appears abstruse in mathematics becomes dog’s balls when rephrased in physics.

            1. tom, I wasn’t dismissing kinematics: go talk to VCAA if you want people who really hate kinematics. I was dismissing the idea that if an IVP doesn’t make much sense in a kinematics manner therefore the IVP has no good meaning or purpose.

              1. OK But by “fuzzy” solution all I meant was it is unobtainable in the kinematics sense – not that it has no good meaning. Again my lazy writing. I do think the distinction is worth making.

            2. Thank you for your explanation of envelopes. I did differential equations a long time ago and not very well, so I wouldn’t say it’s not standard (necessarily). I probably just didn’t learn it very well.

              Actually, you can only make each solution swap once, because the derivative \frac{dy}{dx} must be non-negative, so once it goes up through the cosine function, it can’t go back.

              Also, I thought maybe a kinematic solution could mean twice differentiable, which is why the ones that aren’t constant aren’t ‘kinematic’. I think that is fair. They didn’t say which class of functions we are looking at in the question, so I guess we can make it whatever you like?

              1. The class of functions is not so much the issue here as the class of writers. They had no idea what they were asking, so of course you’re now free to interpret the question in any reasonable manner (or to ignore it entirely).

    1. I didn’t have time to answer when I saw this, and I’m glad I didnt! I think I will wait some more :).

  6. When y=3, dx/dy = 0 which I find problematic as it (to my version of logic) means dy/dx is undefined.

    …or my version of what “undefined” means (which I will be the first to admit, I do not fully understand thanks to comments from others on a much earlier post).

    1. RF, I wouldn’t criticise any teacher for not being able to untangle this mess. It is not easy, and not remotely within VCE.

      1. Yes, this would be a “star” question in our second semester first year DE course (the end of it), or, a normal question in the beginning of our second year DE subject. Not high school.

      2. Oh, far from being critical, I’m being thankful for the education I got from other commenters about the prickly nature of “undefined” on one of Woo’s posts.

        Made me realize just how many pre-conceptions I still hold incorrectly (or at least need to be a bit more careful with)

        1. RF, it’s the nature of university education. Maths departments simply don’t concern themselves with meaning in the bread and butter subjects. And, the advanced subjects are too difficult for most Pre-teachers, even on the rare occasions that the subjects exist and don’t suck. Of course the education faculties aren’t capable of teaching mathematical meaning, even if they had a mind to. The consequence is that that pretty much no one teaches pre-teachers the nature of mathematics in an accessible and helpful manner.

          Simon the Likeable came up with a very good idea to try to fix this, at least for the analysis/calculus aspects. (The number system aspects are in as much need of attention, particularly for primary teachers.) Whether Simon and I will get to do it, and whether anyone with money and clout will care to listen, God knows.

          1. Without wanting to go off on too much of a tangent (pun not intended; happy accident) – can you elaborate on what you mean by number system aspects here – especially as it relates to Primary/Early Secondary?

            Thanks.

            1. Thanks, RF. I probably don’t mean anything you don’t already think about. What I mean is the foundations of number systems: what is a natural number? What is a fraction? How do they work, and why?

              To give the starkest example, defining a fraction is really an advanced 1st year or 2nd year task, involving equivalence classes and the like. Of course few pre-teachers get to see that stuff, and fewer understand it. The material is presented in a legalistic, pure manner, not in the needed “what the Hell does this really mean?” manner. I’m convinced that this lack of meaning, the lack of gut instinct understanding of “fraction” and so on is one of the reasons it is taught so poorly.

              So, the point is to present the basics of numbers in a useful and accessible manner.

  7. I was looking in Lucas and James (1976 Edition). One of their exercises contains a technical infelicity at the Specialist Maths level that is related to this discussion:

    Ex 10B Q5 (e) offers \displaystyle \frac{d^2 x}{dt^2} = -x with initial conditions x = 5 and \frac{dx}{dt} = 0 when t = 0 leads to \displaystyle \frac{dx}{dt} = \sqrt{25 - x^2}.
    The given answer is \displaystyle x = 5 \cos{t}.
    The snag is that you can’t use x = 5 when t = 0 because \displaystyle x \neq \pm 5 is required in order to get \displaystyle \frac{dt}{dx} = \frac{1}{\sqrt{25 - x^2}}.
    Nevertheless the answer is correct and readily follows without trouble if you solve the 2nd order DE with constant coefficients \displaystyle \frac{d^2 x}{dt^2} + x = 0 (which requires expertise beyond Specialist Maths).
    The issue is using the \displaystyle \frac{d^2 x}{dt^2} = v \frac{dv}{dx} technique where \displaystyle v = \frac{dx}{dt}.

    The previous question (Ex 10B Q5 (d)) has no snags:

    \displaystyle \frac{d^2 x}{dt^2} = -4x with initial conditions x = 0 and \frac{dx}{dt} = 2 when t = 0 leads to \displaystyle \frac{dx}{dt} = 2\sqrt{1 - x^2} which in turn leads without incident to \displaystyle x = \sin{2t}.

    I also had a look in Fitzpatrick and Galbraith Applied Mathematics. They have an example in Chapter 4 (Example 26, pp 108 – 109 in the 3rd Edition) with the same technical infelicity at the Specialist Maths level:

    Solve the differential equation \displaystyle \frac{d^2 y}{dx^2} = -y, given that y = 4 and \displaystyle \frac{dy}{dx} = 0 when x = 0.

    Their solution gets to the following:

    \displaystyle \frac{dy}{dx} = \sqrt{16 - y^2}, -4 \leq y \leq 4

    from which they obtain

    \displaystyle \frac{dx}{dy} = \frac{1}{\sqrt{16 - y^2}}, -4 < y < 4.

    Recall the initial condition stated at the start of the question, namely, y = 4 when x = 0 …! They end up with the solution \displaystyle y = 4 \cos{x}.
    The snag here is that you can't use y = 4 when x = 0 because \displaystyle y \neq \pm 4.
    Nevertheless the answer is correct and readily follows without trouble when solving the 2nd order DE with constant coefficients \displaystyle \frac{d^2 y}{dx^2} + y = 0 (which requires expertise beyond Specialist Maths).
    Again, the issue is using the \displaystyle \frac{d^2 y}{dx^2} = v \frac{dv}{dy} technique where \displaystyle v = \frac{dy}{dx}.

  8. This is a try at sending the message again, with LaTeX errors fixed!

    I think you’d get round it by solving the characteristic equation, given that the DE is linear, and which is \lambda^2+1=0. This leads to \lambda=\pm i and the solution being y=A\cos x+B\sin x , for which the coefficients can be found from the initial conditions. This completely side-steps potential problems with the function or its derivatives being undefined. In fact y and dy/dx are defined for all x, but the relation

        \[ dx = \frac{dy}{\sqrt{16-y^2}} \]

    requires that -4<y<4. This introduces an unnecessary restriction, since y is indeed defined at both \pm 4. My take is that the solution given by Fitzpatrick and Galbraith is incorrect.

    If the LaTeX is still stuffed, please remove both replies until I can work out how to use LaTeX properly here.

    1. Hi, Alasdair. Sorry about the “God knows what works” LaTeX thing. I’ve removed your garbled comment. In future, remember you have a half hour after posting to edit your comment. And, after that window is closed, you can always request I fix a TeX error, or whatever.

      1. Thanks, Marty – no ’twas my mistake entirely; leaving out a closing brace. As my kids used to say, it’s now “all fixled”.

      2. Hi Marty. When I posted my ‘latex-intense’ comment last night the 30 min edit window did not appear. Luckily my latex was OK! (but there were some some small typesetting changes I might have made). The same thing might have happened to Alasdair, and he was less lucky than I.

    2. The only snag is that this technique for solving a 2nd order DE with constant coefficients is not within the scope of the Specialist Mathematics course.
      Within Specialist Maths, it has to be done in the two stages I outlined and this necessarily introduces the tighter restriction - 4 < y < 4. The way to work around this restriction is to use limits, but Fitzpatrick and Galbraith did not do this. They simply substituted y = 4 despite the restriction y /neq 4 being only 1 cm away.

      1. An even better “solution” would be to use only problems that can be correctly solved by the techniques and methods available to the students without introducing errors and inconsistencies.

        1. They probably thought they had! But as Marty said: “They had no idea what they were asking …”

          And Cambridge’s Example 9 c ignores the existence of the equilibrium solutions y = \pm 1. It gives the solution to \displaystyle \frac{dy}{dx} = \sqrt{1 - y^2} as simply \displaystyle y = \sin{(x-c)}, which does not contain the solutions y = \pm 1. Furthermore, it ignores y = \pm 1 altogether when solving the DE.

          The best solution, one which VCAA should also adopt (but won’t because of nepotism), is to use vettors that are competent.

          We can whack Cambridge (and, to a lesser extent, excellent textbooks from yesteryear) for these errors, but I’ll wager every single textbook currently on the market contains the same errors.

          1. Most of the exercises in that section included the equilibrium solution if there was one. They’re basically fine on that. It’s the non uniqueness that tripped them up here.

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