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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.

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One way I guide myself through these kind of problems is to observe the units. If you know the rate is measured litres/min, how would compute L litres and divide it out by 60 minutes (= second hour)? What should be those L litres? See answers below. –  user2468 Aug 4 '12 at 2:07

3 Answers 3

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.

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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 $$\frac{1200-0}{120-60}=20$$

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Involving slopes and lines seems to be overcomplicating the problem when we can have an explicit formula to plug values into and directly compute the amount of change and the amount of time that change happens over. –  Henning Makholm Aug 4 '12 at 2:27
    
OK, I agree as far as involving the words "slope" and "line," but your explicit formula is just going to be computing a slope. –  Kevin Carlson Aug 4 '12 at 2:52
    
Only to the extent that you consider every division ever to be "computing a slope". There are no slopes, no geometry at all involved in the original problem. –  Henning Makholm Aug 4 '12 at 13:50
    
This may be an artifact of my education, but for me "average rate of change" is conceptually very closely tied to "draw a secant line." I'm sure I'm committing the fallacy of identifying a function with its graph, but I don't think this does subsume every division possible-not division by unitless quantities, for instance. –  Kevin Carlson Aug 4 '12 at 13:54

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.

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