Pre Calculus and Rate of Change Sacha drains the water from a hot tub. The hot tub holds $1600$ L of water. It takes $2$ h for the water to drain completely. The volume of water in the hot tub is modelled by $V(t) = 1600 - \dfrac{t^2}{9}$, where $V$ is the volume in litres at $t$ mins and $0 < t < 120$.
a) Determine the avg rate of change in volume during the second hour.
I know the answer is $20$ L/min after looking at it from the answers section of my text book, but I dont know how they got that answer.
 A: The volume at the beginning of the second hour is $1600-\frac{60^2}{9}$.  This is $1200$.
The volume at the end of the second hour is $1600-\frac{120^2}{9}$. This is $0$.
So the change in the second hour is $1200$. This took $60$ minutes. So the average rate of change in the second hour is $\frac{1200}{60}$ litres per minute.   
Remarks: $1$. Later, in calculus, one takes a somewhat different point of view. At the beginning of the second hour we have $1200$ litres, and at the end we have $0$. So the change in the second hour is $0-1200$, that is, $-1200$. The change is negative because the amount of water has decreased. We conclude that the average rate of change in the second hour is $\frac{-1200}{60}$ litres per minute, that is, $-20$ litres per minute.  That, in my opinion, is the correct answer. But since the amount is clearly decreasing, the somewhat sloppy "$20$" is sort of acceptable.  
$2$. The given formula is physically implausible. For note that the change in the first hour is $1600-1200$, so the average rate of change in the first hour is $\frac{400}{60}$ litres per minute, far less than the average rate of change in the second hour. However, one would expect that because of the higher pressure at the beginning, the average rate of change in the first hour would be greater than the average rate of change in the second hour. 
A: The average rate of change in $V$ between times $t_1$ and $t_2$ is just the slope of the line connecting $V(t_1)$ and $V(t_2)$. So here $t_1=60, t_2=120$, $V(60)=1200, V(120)=0$, and so the slope is $$\frac{1200-0}{120-60}=20$$
A: The volume of water at the beginning of the second hour is
$$
V(60) = 1200 \text{ L}.
$$
The volume of water at the end of the second hour is
$$
V(120) = 0 \text{ L}.
$$
The net change in the volume is thus
$$
0 \text{ L} - 1200 \text{ L} = -1200 \text{ L}.
$$
Since this change took 60 minutes total, the average rate of change is
$$
\frac{-1200 \text{ L}}{60 \text{ min.}} = -20 \text{ L/min}.
$$
The negative sign accounts for the fact that we are losing water over time rather than gaining it.
