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anonymous

  • one year ago

university 1st year math question

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  1. anonymous
    • one year ago
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    @IrishBoy123 can u plz help me and also how to use owl bucks ?

  2. IrishBoy123
    • one year ago
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    start by finding a way to describe the volume of the trough with h as the variable. where h is the height above the base of the trough. width of the trough changes as h increases, the length is constant.

  3. IrishBoy123
    • one year ago
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    |dw:1443176643865:dw|

  4. anonymous
    • one year ago
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    can u plz delete the question of the photo legally i am not allowed to post it anywhere its from our university and after this i was going to delete my attachment hope u dont mind

  5. anonymous
    • one year ago
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    thank you

  6. IrishBoy123
    • one year ago
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    |dw:1443176712781:dw|

  7. IrishBoy123
    • one year ago
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    w = width does that w = 4 + h/4 make sense to you?

  8. IrishBoy123
    • one year ago
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    if so, what is the shaded area in terms of h? |dw:1443177106816:dw|

  9. anonymous
    • one year ago
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    Yea

  10. baru
    • one year ago
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    the shape of the drain isn't mentioned...?

  11. IrishBoy123
    • one year ago
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    soz that drawing is the trough, a trapezoidal prism

  12. anonymous
    • one year ago
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    its a tricky question

  13. baru
    • one year ago
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    soo... assuming the entire bottom surface is the drain?

  14. IrishBoy123
    • one year ago
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    no leave the drain alone for now!

  15. IrishBoy123
    • one year ago
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    the fudge factor k seems to deal with that

  16. IrishBoy123
    • one year ago
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    we just need a formula for the shaded area in terms of h. it is a trapezoidal prism so basic geometry

  17. anonymous
    • one year ago
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    i am going to post this question again with ask qualified help thing hope u dont mind

  18. IrishBoy123
    • one year ago
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    go for it, that's completely not a problem! and good luck to you!

  19. IrishBoy123
    • one year ago
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    you should probs close this one though....as there will be confusion:p

  20. anonymous
    • one year ago
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    i did

  21. baru
    • one year ago
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    anyway...for arguments sake... you would write volume(V) in terms of 'h' and solve the diff equation by of variables dV/dt= -k*h^(1/2) right?

  22. baru
    • one year ago
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    *by separation of variables

  23. IrishBoy123
    • one year ago
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    no separation had we proceeded we would have got for the trough something like \(V = 10(4h + \frac{h^2}{8})\) [i think the length was 10, the pic's gone now] so \(\dot V = 10(4 + \frac{h}{4})\dot h\) and for the drain from \(v = \sqrt{2gh}\) we would have got \(\dot V = k \sqrt{2gh}\) [never got down to nailing the k but i think it would have been a proxy for the area of the drain] so \(10(4 + \frac{h}{4})\dot h = k \sqrt{2gh}\) that's why it is a plot,

  24. baru
    • one year ago
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    ok...got it :)

  25. IrishBoy123
    • one year ago
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    @baru oh my bad, yes you can do \(\frac{10(4 + \frac{h}{4})}{\sqrt{2gh} } \, dh = k \, dt\) etc if you wanted an analytic solution but i recall they wanted a plot.

  26. baru
    • one year ago
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    yep... i don't remember the question specifics either. i was just look looking for the underlying idea :)

  27. IrishBoy123
    • one year ago
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    fluid flow and continuity. i'm sure you know that anyways:p

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