university 1st year math question

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- anonymous

university 1st year math question

- chestercat

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- anonymous

@IrishBoy123 can u plz help me and also how to use owl bucks ?

- 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

|dw:1443176643865:dw|

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- 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

thank you

- IrishBoy123

|dw:1443176712781:dw|

- IrishBoy123

w = width
does that w = 4 + h/4 make sense to you?

- IrishBoy123

if so, what is the shaded area in terms of h?
|dw:1443177106816:dw|

- anonymous

Yea

- baru

the shape of the drain isn't mentioned...?

- IrishBoy123

soz
that drawing is the trough, a trapezoidal prism

- anonymous

its a tricky question

- baru

soo... assuming the entire bottom surface is the drain?

- IrishBoy123

no
leave the drain alone for now!

- IrishBoy123

the fudge factor k seems to deal with that

- IrishBoy123

we just need a formula for the shaded area in terms of h. it is a trapezoidal prism so basic geometry

- anonymous

i am going to post this question again with ask qualified help thing hope u dont mind

- IrishBoy123

go for it, that's completely not a problem!
and good luck to you!

- IrishBoy123

you should probs close this one though....as there will be confusion:p

- anonymous

i did

- 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

*by separation of variables

- 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

ok...got it :)

- 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

yep... i don't remember the question specifics either. i was just look looking for the underlying idea :)

- IrishBoy123

fluid flow and continuity. i'm sure you know that anyways:p

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