# Torrcelli's law of fluids

#### mikky05v

##### New member
I am working on this project for Diff eq's and I seem to be stuck. i think I am messing something up but I need a fresh pair of eyes to show me where it is.

imgur: the simple image sharer
imgur: the simple image sharer (this is the same page as ^ just scrolled down)
imgur: the simple image sharer

Extra information: He sent us this email which really lost me
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.

my work so far:

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.
That's as far as i've gotten.

#### MarkFL

Staff member
For part b), we are being told to consider that:

$$\displaystyle \frac{dV}{dt}=A(h)\frac{dh}{dt}=a\cdot v(t)$$

Now, from part a), we found:

$$\displaystyle v(t)=-\sqrt{2gh}$$

And so the differential equation given follows.

For part c) if $r$ is the radius of the circular surface $A(h)$ and $h$ is the depth of the water, both of these at time $t$, we know by similarity, we must have:

$$\displaystyle \frac{r}{h}=\frac{3}{5}\implies r=\frac{3}{5}h$$

And so we have:

$$\displaystyle A(h)=\pi\left(\frac{3}{5}h \right)^2=\frac{9\pi}{25}h^2$$

$$\displaystyle a=\pi\left(\frac{1}{2} \right)^2=\frac{\pi}{4}$$

Hence, the ODE becomes:

$$\displaystyle \frac{9\pi}{25}h^2\,\frac{dh}{dt}=-\frac{\pi}{4}\sqrt{2gh}$$

Multiplying through by $$\displaystyle \frac{100}{\pi\sqrt{h}}$$ we may now write the IVP:

$$\displaystyle 36h^{\frac{3}{2}}\,\frac{dh}{dt}=-25\sqrt{2g}$$ where $$\displaystyle h(0)=50$$

I would suggest solving this without using decimal approximations, or substituting for $g$.