## anonymous one year ago university 1st year math question

1. anonymous

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

2. 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.

3. IrishBoy123

|dw:1443176643865:dw|

4. 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

5. anonymous

thank you

6. IrishBoy123

|dw:1443176712781:dw|

7. IrishBoy123

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

8. IrishBoy123

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

9. anonymous

Yea

10. baru

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

11. IrishBoy123

soz that drawing is the trough, a trapezoidal prism

12. anonymous

its a tricky question

13. baru

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

14. IrishBoy123

no leave the drain alone for now!

15. IrishBoy123

the fudge factor k seems to deal with that

16. IrishBoy123

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

17. anonymous

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

18. IrishBoy123

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

19. IrishBoy123

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

20. anonymous

i did

21. 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?

22. baru

*by separation of variables

23. 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,

24. baru

ok...got it :)

25. 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.

26. baru

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

27. IrishBoy123

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