WitCH 138: Second Rate

Not counting the Witchfests on Nelson‘s Proof and Complex chapters, I don’t think we’ve had a textbook WitCH for quite a while. This one is hardly in the same league but it’s been on my to do list for a long time and it’s definitely a source of irritation. The topic came up in discussions with a colleague today, and I’m too busy and too tired to do the posts I should be doing, so this’ll do for now. It is an example from the differential equations chapter of Cambridge’s Specialist Mathematics 3 & 4, and it is related to this previous WitCH from the same text and chapter. As was the case with the previous WitCH, it is a representative, one of a type.

10 Replies to “WitCH 138: Second Rate”

  1. x = 0 when t = 0 is used as the boundary condition but x = 0 is excluded from the domain. Perhaps a limit should be used.

    1. Why is x = 0 excluded? In any case the endpoint thing is a fussy aspect that VCE never does well. Not one of my major concerns here.

  2. I guess they really want a DE for x(t) not dx/dt

    Of course, they make an overly big show of deriving and solving it too…
    Much easier to use similarity on the volume instead of the side:

        \[V( x) =\left(\frac{x}{h}\right)^{3} V( h) =\frac{\pi r^{2}}{3h^{2}} x^{3}\]

    Then get the DE from the chain rule

        \[\frac{dV}{dt} =\frac{dx}{dt}\frac{dV}{dx} =\frac{dx}{dt}\frac{\pi r^{2}}{h^{2}} x^{2} =1000k\]

        \[\Longrightarrow \frac{dx}{dt} x^2=\frac{1000kh^{2}}{\pi r^{2}}\]

    Integrate up as usual to get their answer.

    1. Thanks, Simon. Your similarity thing is nice. But even going directly, I think the text’s derivation is a mess. And good point about “differential equation for dx/dt”. I hadn’t noticed, but that is sloppy.

      1. Funny! I thought your second rate title referred to the “DE for dx/dt” statement…

        And yeah, the rest of their derivation is “Flippin’ Ridiculous” – they should just treat the DE as a separable one / use a substitution to integrate from 0 to t in one clean move.

            \[\int\frac{dx}{dt} x^2dt=\int x^2dx=\tfrac13 x^3=\frac{1000kh^{2}}{\pi r^{2}}t+c\]

        use the initial condition to get c=0 and then solve for x.

        1. Yes, this section comes after Separation. So even on the text’s own silly terms, the flipping here is ridiculous.

  3. Perhaps the solution should come with the disclaimer that it’s only valid for

    \displaystyle 0 < t \leq \frac{\pi r^2 h}{3000 k}.

    Otherwise it leads us to believe that x increases without bound. As Dirty Harry said when he saw this question: "A model's got to know its limitations."

Leave a Reply

Your email address will not be published. Required fields are marked *

The maximum upload file size: 128 MB. You can upload: image, audio, video, document, spreadsheet, interactive, text, archive, code, other. Links to YouTube, Facebook, Twitter and other services inserted in the comment text will be automatically embedded. Drop file here