Toricellis proof??wierd math question?

Question

Let y(t) and V(t) be the height (in m) and the volume of the water (m^3), respectively, in a resevoir at time t (in s). If the water leaks out through a hole of area a ( in m^2) at the bottom of the resevoir, Toricelli's Law states that dV/dt= -a(squareroot(2gy)) where g is the acceleration due to gravity.

Suppose that the resevoir is a cylinder of height 5m and radius 50cm and that the hole in the bottom is circular with radius 2.5cm. If we take g=10m/s^2, show that y satisfies dy/dt= -1/200(squareroot(5y))

Well i figured the volume of the cylinder is (5/4)pi and area of small circle is 0.00196m^2 i dunno what to do next ?? please help .but im stuck... plz plz help me with this !! ive asked everyone and no1 can help :(

Answer 1

the base of the reservoir is a constant value so dy/dt is easy to calculate.
read ;'s as comments in the following.

call the base of the reservoir b , then since V = b * y,

dy/dt = dV/dt * 1/b
'
= -a/b(sqrt(20y) ; (substituted the g~10m/s^2)
= pi(.025m)^2/pi(.5m^2)*2sqrt(5y) ; (2 = sqrt(4))
=.000625m^2/.25m^2 * 2sqrt(5y) ;
=.0025 * 2sqrt(5y)
=.005sqrt(5y)
QED