How to find the rate of decrease of water height in a tank

[psymple]

Homework Statement

Water flows from a tank of constant cross-sectional area 54 ft2 through an orifice of constant cross-sectional area 1.7 ft2 located at the bottom of the tank.
Initially the height of the water in the tank was 20 and its height t sec later is given by the following equation.

How fast was the height of the water decreasing when its height was 9 ft? (Round your answer to two decimal places.)____________1 ft/sec

Homework Equations

2(sqrt H) +1/24t-2(sqrt 20)=0 (0<=t<=50(sqrt 20))

The Attempt at a Solution

First let me say that this is my very first post-so forgive if i did not do everything up to par. TIA for all the help this site will bring.

Im almost clueless. I think I have to write it out;
dH/dt 2H^2 + (1/24t)^2 -2(20)^2=0
then we have to minus the dH/dt chain rule out;
2(1/24t)(1/24) -2(20)^2=(1) dH/dt (4H) = 1/288t=4H dH/dt
then divide out
dH/dt = (1/288)/4h

Am I even close??
thanks again and please advise if i am posting equations wrote!

 

[gabbagabbahey]

Im almost clueless. I think I have to write it out;
dH/dt 2H^2 + (1/24t)^2 -2(20)^2=0
then we have to minus the dH/dt chain rule out;
2(1/24t)(1/24) -2(20)^2=(1) dH/dt (4H) = 1/288t=4H dH/dt
then divide out
dH/dt = (1/288)/4h

Am I even close??
thanks again and please advise if i am posting equations wrote!

I'm not sure what you are doing here;

\frac{d}{dt}\left(2\sqrt{H(t)}+\frac{1}{24}t-2\sqrt{20}\right)=\frac{d}{dt}\left(2\sqrt{H(t)}\right)+\frac{d}{dt}\left(\frac{1}{24}t\right)-\frac{d}{dt}\left(2\sqrt{20}\right)\neq 2H^2\frac{dH}{dt}+\left(\frac{1}{24}t\right)^2-2(20)^2