university 1st year math question.

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- anonymous

university 1st year math question.

- jamiebookeater

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- welshfella

i looked up torricelli's law and only found a formula for a cylindrical tank..

- anonymous

- anonymous

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## More answers

- anonymous

- anonymous

i am finding it very difficult

- ganeshie8

@ParthKohli try this

- ParthKohli

are you mathmath333?

- anonymous

what ?

- ganeshie8

No, he is not

- ParthKohli

I haven't studied fluids yet, but I looked at the Wikipedia page and it sorta makes sense.

- ParthKohli

This is what we're really looking at though. https://en.wikipedia.org/wiki/Torricelli%27s_law#Application_for_time_to_empty_the_container

- ganeshie8

same here, it seems torcelli's law gives the speed of fluid of height \(h\), leaving tank from bottom :
\[v = \sqrt{2gh}\]

- anonymous

noo \[v=\sqrt{2gh}\]

- ytrewqmiswi

What does that k represents?

- anonymous

i was thinking of an approach where i find A(h) and use integration

- ParthKohli

Is that the area of cross-section as a function of height? And are you sure this is the equation?\[A(h) \frac{dh}{dt} = -k \sqrt{h}\]

- anonymous

yes 100% because we havent used other v=sqrt2gh yet and my teacher wouldnt expect us to use that

- anonymous

the only problem i am having is how to find A(h)

- ytrewqmiswi

That wuld mean that the area of cross section varies with height will that make any sense?

- ParthKohli

But in your equation, \(\sqrt{2g}\) or \(g\) won't appear.
We can find the area of the cross-section as a function of height.

- ParthKohli

If my eyes work, we're looking at a rectangular cross-section, right?

- anonymous

yes

- anonymous

and yes i wan to find the area as a function of height plz

- ParthKohli

Width increases in a linear manner from 4 to 5. At h = 0, it is 4 and at h = 4, it is 5.
Thus in general it should be \(w(h) = h/4 + 4\).
Meanwhile the length remains 10 so \(l(h) = 10\).
Multiply...

- anonymous

\[\frac{ 10h }{ 8 }\]

- ParthKohli

noooo I didn't write h/8
h/4+4 means (1/4)*h + 4

- anonymous

\[\frac{ 10(h+16) }{ 4 }\]

- ganeshie8

40----> 50
A(y) = 40 + (y/4)*10
does that work ?

- ParthKohli

looks good

- ParthKohli

it only works because one of them remains constant
can't do the same thing when both width and length increase
which is why it's a good practice to do them separately

- ganeshie8

yeah 10cm is fixed, only the other dimension is changing, so it is linear

- anonymous

just a question srry interpreting ur thought process how did u get width as (1/4) times h +4

- ganeshie8

\[A(h)\frac{ dh }{ dt }=-k \sqrt{h}\]
plugin \(A(h)\), the de becomes
\[(40+(\frac{h}{4})*10)\frac{ dh }{ dt }=-k \sqrt{h}\]

- ytrewqmiswi

the liquid escapes with a velocity =sqrt(2gh) which varies with the level of liquid. Hence, we have to use integral. Let y be the height of the liquid at an instant.
This height changes by dy in time dt.
Volume of water leaving out per secound=-Ady.dt
at the hole volume escaping per second is av=a[sqrt(2gy)]
a[sqrt(2gy)] =-Ady/dt ---->\[\int\limits_{H}^{0}\frac{ -dy }{ \sqrt{y} } =\frac{ a \sqrt{2g} }{ A }\int\limits_{t}^{0}dt\]
\[t=\frac{ A }{ a \sqrt{2g}}2\sqrt{H}\] a<

- ganeshie8

maybe think of it in reverse :
A(h) = 40 + (h/4)*10
plugin h=0, what do you get ?
plugin h=4, what do you get ?

- ParthKohli

|dw:1443179501695:dw|

- ytrewqmiswi

well...that thing can be used to find time tho..?

- anonymous

@ytrewqmiswi we cant use v=sqrt2gh because well i want the area as a function of h

- ytrewqmiswi

A = cross sectional area of tank
H=height till water is filled
a=cross sectional area of hole

- ParthKohli

but didn't you assume A to be constant?

- ytrewqmiswi

yes i took it constant

- ParthKohli

yes that solution matches ours so far

- anonymous

good because i am not sure about my final answer it doesnt make sense so i need to be assured

- ParthKohli

you shouldn't have any problem with integration

- anonymous

noo

- anonymous

but would love it if you solve it step by step plzzz

- anonymous

it will make my life so easy

- ganeshie8

you don't have teamviwer ?

- anonymous

again what is teamviewer?

- anonymous

do i have to buy it ?

- ganeshie8

no, it is a free screen sharing app

- anonymous

for the solution u an see i am stuck in the end

- anonymous

ahm do i have to get it for this question

- ParthKohli

\[A(h) \frac{dh}{dt} = -k\sqrt{h}\]We've got all messy differentials here. What we do is we take all the \(h\)-terms to to one side and \(t\)-terms to the other. Then integrate both sides with proper limits.

- anonymous

ok

- ParthKohli

\[\frac{A(h)}{\sqrt h} dh=-kdt\]

- anonymous

\[\frac{ A(h) }{ \sqrt{h} } dh=-k dt\]

- ParthKohli

Great. Now when \(t= 0\), \(h=4\). When \(t = t_0\), \(h = 0\). So you integrate both sides with those limits corresponding to each other.

- anonymous

how?

- ParthKohli

OK, first put the value of \(A(h) \) in the equation and simplify. You don't have to type it out here. Keep doing that on paper. I'll guide you through each step.

- anonymous

which A(h) value must i put?

- ParthKohli

The one we found out earlier?

- ParthKohli

We found the value of area of the cross-section in terms of height earlier, yes?

- anonymous

\[\frac{ 10(h+16) }{ 4 }\ /\sqrt{h} =-kdt\]

- ParthKohli

Good. I know you're having trouble with typing everything out. So don't. Just keep doing it on paper alright?

- anonymous

no i need to show u , i can be writting jubrish

- ganeshie8

//ahm do i have to get it for this question
Absolutely not, but considering the fact that you already have a detailed solution which you don't understand it, the jibber jabber that I do here only confuses you more. If you have teamviewer, we could use voice also and engage more efficiently.

- anonymous

\[\frac{ h+16 }{ \sqrt{h}} dh = - \frac{ 2 }{ 10 } \times 0.03 dt\]

- anonymous

@ganeshie8 i am on computer , will that work on this and how do i add uz

- ganeshie8

Hey it is really not required, let try and finish it off here

- anonymous

ok

- ganeshie8

have you deleted the attachment ?

- anonymous

noo

- ganeshie8

i couldn't find the solution above

- ganeshie8

your solution..

- anonymous

ok u see i did the A(h) wrong then towards the end i thought when t=0 , h=3

- anonymous

my solution is wrong

- anonymous

i did A(h) wrong

- ganeshie8

your A(h) looks fine to me

- anonymous

and i dont seem to get what that refers to as 3 quarter filled

- anonymous

so plz ignore the solution

- ganeshie8

dude you're good,
don't you see what you have is same as `A(h) = 40 + (h/4)*10`

- anonymous

ok so my solution makes sense all of it ok what about part c

- anonymous

i just gave up on that

- anonymous

i mean iii

- anonymous

and also i dunno how to go furthur from here

- ganeshie8

so whats you final answer for part i

- anonymous

i dunno what is h and t if i dunno t how am i suppose to find h

- anonymous

what about h

- ganeshie8

Notice that the depth of water, h, is 0 when the tank is empty.
so, for part iii, i think you simply plugin h=0 and solve \(t\)

- anonymous

is this what u saying?

- ganeshie8

I am asking you to simply replace \(h\) by \(0\)
because, empty tank means that the depth of water is 0

- anonymous

ohhh

- ganeshie8

\[\frac{ 2 }{ 3 }\sqrt{h} (h+48)=-0.012t+58.89\]
\[0=-0.012t+58.89\]
solve \(t\) and you're done!

- anonymous

t=4907.5

- anonymous

just one question did uz consider the three quarter filled thing

- ganeshie8

Oops! we missed that, my mistake, sorry, lets start over..

- anonymous

ok so part i is finding an expression we have done it

- anonymous

ur being sarcastic right

- ganeshie8

I'm never sarcastic

- ganeshie8

we need to start over, our work for all parts is wrong because we assumed that the tank was full

- anonymous

ok

- ganeshie8

gimme some time, il type in the complete solution

- anonymous

ok

- Jhannybean

nice comforter I mean. Lol.

- ganeshie8

Hey, it turns out our previous solution works out perfectly fine because we have worked it from bottom.

- ganeshie8

No need to redo anything. we are good.

- ganeshie8

@ParthKohli please double check if psble

- anonymous

srry last thing what about part b

- anonymous

as u said make h=0 and solve for T

- anonymous

thnx guys cheers i got it

- ganeshie8

try
http://www.wolframalpha.com/input/?i=plot+%5Cfrac%7B+2+%7D%7B+3+%7D%5Csqrt%7Bh%7D+%28h%2B48%29%3D-0.012t%2B58.89%2C+0%3Ct%3C5000

- ganeshie8

deleted, let me know if you want me delete any other stuff

- anonymous

no its ok thak uso much

- ganeshie8

np

- anonymous

how can i manupilate above equation so the subject is h because thats what it is ultimately asking

- anonymous

- anonymous

- anonymous

but the question is asking me to make h the subject

- dan815

ok

- dan815

it says you can use wolfrad alpha so use that

- anonymous

noo i cant because its for graph only h vs t

- dan815

yeah?

- anonymous

when u read the question b (i) last line determine the depth of the milk at any instant after the drain has opened

- anonymous

so we need to find expression where h is the subject and i did use wolfram alpha it didnt work there were always two h

- anonymous

did u understand i am after i want an expression where h = .....

- dan815

\[\frac{ 2 }{ 3 }\sqrt{h} (h+48)=-0.012t+58.89\]

- dan815

you just specify in wolfram plot h vs t

- anonymous

look i am not talking about ii use wolfram alpha to obtain plot h verus t

- dan815

what do u want then

- anonymous

i am talking about question above (i)

- dan815

oh i thought u are done that

- anonymous

nooo

- dan815

do u want me to solve b(i)

- anonymous

how do i make h the subject

- anonymous

in this expression

- anonymous

how do i get rid of sqrt h

- anonymous

so i have an expression like h =..... when i manipulate above eq

- dan815

it's not going to be pretty

- dan815

yes.. this 0.03 for k means nothing to me

- anonymous

k is just viscosity its a constant

- dan815

doesnt help

- anonymous

see the solution u will realise

- dan815

k im assuming is a function of the radius of the opening

- dan815

yep see that equation u are using there is an application of the bernoullis principle, where that k in there would be some factor with the radius of the opening at the bottom accounted in for it

- anonymous

yea can u see any way where h is the subject

- dan815

what do u mean h is the subject

- anonymous

ok h is the depth

- anonymous

at 4 sec

- anonymous

h= depth in equation and t = time

- dan815

what you have is good enough

- dan815

i dont see where it says to make h the subject in your question

- anonymous

ok at 10 sec what would be the depth of tank can u give me answer for that by substituting 4 into the final eq

- anonymous

substituing 10*

- dan815

you gotta use nonlinear methods lol

- anonymous

how would i use non linear method

- dan815

just use matlab or something for now

- anonymous

ok thnx for ur help btw before u go mind deleting this whole question plz

- dan815

i have no powers

- dan815

you know as far as applications and solutions go solving for t=f(h) is good enough

- anonymous

yea i hope so

- dan815

as you can graph that, and rotating it returns h=f(t)

- anonymous

yea true

- dan815

if you really must solve for h i can show u a way

- anonymous

can u plz delete the expression above

- dan815

it's not pretty yo uwill need to use complex numbers

- dan815

and it might be too early for you right now in first year

- anonymous

no dont i have solved its not pretttyyy

- dan815

okay

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