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Mathematical modelling question

  1. Mar 23, 2006 #1
    Hey,

    So, In a question about finding volume in a dome we were

    Given that V = pi.k(k^2/r - k^2/3r^3 - r/8)
    And K = m/2pi.p.t

    Firstly we are asked to obtain an expression for the value of r that maximises the volume of air and rearrange to obtain an equation of the form ar^4 + br^2 + c = 0

    V' => r^4/8 + kr^2 - k^2 = 0

    So far im confident that is correct, but next it asks you to solve the equation for r^2 and hence find r, approximate squareroot of 96 to be 10.

    Therefore pi^2 = 10

    and substituting the value of k into this equation i obtained,

    r^4/8 + mr^2/2pi.p.t - m^2/4pi^2.p^2.t^2 = 0

    That is where i get stuck and i hope someone can help me out, thnx :\
     
  2. jcsd
  3. Mar 23, 2006 #2

    HallsofIvy

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    Is that [itex]V= \pi k(\frac{k^2}{r}-\frac{k^2}{3r^3}- \frac{r}{8}[/itex] and [itex]k= \frac{m}{2\pi pt}[/itex]? If you don't use LaTex, use lots of parentheses. Also do not use "k" and "K" to mean the same thing.

    What?? Not the [itex]\pi[/itex] I know! "Approximate square root of 96 to be 10"? Why???

    You were told to solve for r2 first- use the quadratic formula. That's fairly straight forward and gives a relatively simple formula for r2. Since you still have unknowns m, p, t in the formula, I see no reason to "approximate" [itex]\pi[/itex] by 10.
     
  4. Mar 23, 2006 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Is that [itex]V= \pi k(\frac{k^2}{r}-\frac{k^2}{3r^3}- \frac{r}{8}[/itex] and [itex]k= \frac{m}{2\pi pt}[/itex]? If you don't use LaTex, use lots of parentheses. Also do not use "k" and "K" to mean the same thing.

    What?? Not the [itex]\pi[/itex] I know! "Approximate square root of 96 to be 10"? Why???

    You were told to solve for r2 first- use the quadratic formula that gives a fairly straight forward expression for r. Since you still have unknowns m, p, t, I see no reason for "approximating" [itex]\pi[/itex] by 10. You aren't going to get a numerical answer anyway.
     
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