# Implicit differentiation

• psymple
In summary, we are given a problem where water is flowing from a tank of constant cross-sectional area through an orifice of constant cross-sectional area located at the bottom of the tank. We are asked to find the rate at which the height of the water is decreasing when its height is 9 ft. Using the given equation, we can use the chain rule to find the rate of change of the height with respect to time. After solving for the derivative, we can plug in the given values and round the answer to two decimal places to find the rate at which the height is decreasing.

## 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??

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