anonymous
  • anonymous
university 1st year math question
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@IrishBoy123 can u plz help me and also how to use owl bucks ?
IrishBoy123
  • IrishBoy123
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.
IrishBoy123
  • IrishBoy123
|dw:1443176643865:dw|

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anonymous
  • anonymous
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
anonymous
  • anonymous
thank you
IrishBoy123
  • IrishBoy123
|dw:1443176712781:dw|
IrishBoy123
  • IrishBoy123
w = width does that w = 4 + h/4 make sense to you?
IrishBoy123
  • IrishBoy123
if so, what is the shaded area in terms of h? |dw:1443177106816:dw|
anonymous
  • anonymous
Yea
baru
  • baru
the shape of the drain isn't mentioned...?
IrishBoy123
  • IrishBoy123
soz that drawing is the trough, a trapezoidal prism
anonymous
  • anonymous
its a tricky question
baru
  • baru
soo... assuming the entire bottom surface is the drain?
IrishBoy123
  • IrishBoy123
no leave the drain alone for now!
IrishBoy123
  • IrishBoy123
the fudge factor k seems to deal with that
IrishBoy123
  • IrishBoy123
we just need a formula for the shaded area in terms of h. it is a trapezoidal prism so basic geometry
anonymous
  • anonymous
i am going to post this question again with ask qualified help thing hope u dont mind
IrishBoy123
  • IrishBoy123
go for it, that's completely not a problem! and good luck to you!
IrishBoy123
  • IrishBoy123
you should probs close this one though....as there will be confusion:p
anonymous
  • anonymous
i did
baru
  • baru
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?
baru
  • baru
*by separation of variables
IrishBoy123
  • IrishBoy123
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,
baru
  • baru
ok...got it :)
IrishBoy123
  • IrishBoy123
@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.
baru
  • baru
yep... i don't remember the question specifics either. i was just look looking for the underlying idea :)
IrishBoy123
  • IrishBoy123
fluid flow and continuity. i'm sure you know that anyways:p

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