# Torrcellis law differential equations

1. Feb 24, 2014

### mikky05v

1. The problem statement, all variables and given/known
This is a group project for differential equations. I ended up without a group, lucky me. Ive been trying to work through this on my own but im stuck. Sorry about the pictures, typing it all out would of taken ages.

http://imgur.com/qeOkl3n
http://m.imgur.com/uLZWRCX
http://imgur.com/z3SC4st

2. Relevant equations
He sent us the email
Project C:  you can get an approximate answer for part (d) by using the formula r = 3h/5 in part (c)  [so A(h) = Pi*(3h/5)^2]  you really should be able to solve that diff eq in (c) with that info.  (note that the "actual" time is a bit longer if you solve it the "correct" way using r = .59h + .5).  For part (e)--you first need the formula r = -3/5 h +30 which of course gives you the radius at any given height.  This makes your A(h) = Pi* (-3/5 h + 30)^2.  When you put this into the differential equation and divide by the square root of h and then integrate you should get an equation with h^5/2 and h^3/2 and h^1/2 and t (and of course coefficients on all of those!).  Once you find your constant by inputting your initial condition [h(0)=50] you can solve the resulting implicit equation BUT you can't do it by hand, you have to use technology!  (I used desmos by inputting "x" for my "t" and "y" for my "h")  the graph then showed me the time to drain the other tank is around 10 minutes (600 seconds).  I'll leave it to you to find the exact value.

3. The attempt at a solution

l did part a by integrating twice.

b. Im not entirely sure what this section wanted. I put A (h) dh/dt= -a(sqrt(2gh))

C. r=3h/5 according to email
A (h) = pi (3h/5)^2
a= pi (1/2) ^2 = pi/4
g= 98.1 cm/s^2
Giving the seperable differential equation
Pi (3h/5)^2 dh/dt = -pi/4 sqrt (2×98.1h)
That simplifies to .0257012h^(3/2) dh=dt

D. Integrating both sides I got t= .0102805h^(5/2) + c
I thought to solve for c by taking t=0 and h=50 but I get c=-181.7352816 so I know I did something wrong.
Thats as far as ive gotten.

2. Feb 24, 2014

### SteamKing

Staff Emeritus
The top of the second page got cut off. Could you make a different copy of this page and post it?

http://m.imgur.com/uLZWRCX

Try turning off the toolbar at the top of the page before making the image.

3. Feb 24, 2014

### mikky05v

The 2nd page is just the bottom of the first page, to see the stuff above it look at the first page. :)

4. Feb 24, 2014

### mikky05v

has anyone had a chance to look at this yet, i keep doing circles around the same thing. I know my equation is wrong but I can't figure out what to do differently