You are not logged in.
Pages: 1
dV/dt = -a*sqrt(2gh)
Suppose a tank is cylindrical with height 6 ft and radius 2 ft and a hole at the bottom of the tank is circular with radius 1 inch. If we take g = 32 ft/s^2, show that h satisfies the differential equation dh/dt = -sqrt(h)/72.
This first part is easy so I don't need help on that.
The next question is: solve this equation to find the height of water at time t.
Integrating we get: 2sqrt(h) = -t/72 + C
and eventually: h(t) = t²/144² - Ct/72 + C²
C² must equal 6 since the height is 6 at t = 0 so we get:
h(t) = T²/20736 - sqrt(6)t/72 + 6
Is this correct? Thx.
If this is correct, could we also go like this when integrating?
∫h-½ dh = -1/72 ∫dt
(from 6 to h) (0 to t)
I end up getting the exact same thing.
Last edited by fusilli_jerry89 (2008-04-08 13:40:26)
Offline
Pages: 1