Question

university 1st year math question

Answer 1

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

Answer 2

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.

Answer 3

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

Answer 4

thank you

Answer 5

w = width

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

Answer 6

if so, what is the shaded area in terms of h?

Answer 7

Yea

Answer 8

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

Answer 9

soz

that drawing is the trough, a trapezoidal prism

Answer 10

its a tricky question

Answer 11

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

Answer 12

no

leave the drain alone for now!

Answer 13

the fudge factor k seems to deal with that

Answer 14

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

Answer 15

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

Answer 16

go for it, that's completely not a problem!

and good luck to you!

Answer 17

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

Answer 18

i did

Answer 19

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?

Answer 20

*by separation of variables

Answer 21

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,

Answer 22

ok...got it :)

Answer 23

@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.

Answer 24

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

Answer 25

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