algebra-precalculus
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.
This is a minimization problem, and you have to minimize
$$E(x) = 2\cdot |AC| + |CD|$$
where $x = |BC|$, with $C \in[B,D]$, $A\hat{B}D=90°$, $|AB|=5$ and $|BD|=13$.
That is,
$$E(x) = 2 \cdot \sqrt{5^2+x^2}+13-x$$
Take the derivative of this and find the zeros in $[0, 13]$, etc., giving the multiple choice answer (b).
I think there's a mistake in your formulation of the problem. My guess is that the correct formulation is the following:
A rocket is launched and it is observed that the rocket's altitude, as a function of time, follows the equation $$ y(t) = -4.9 t^2 + 49t + 155, \qquad t>0 $$ When will the rocket hit the ground?
Solution: When the rocket hits the ground, the altitude is zero. So we simply need to solve the equation $$ y(t) = -4.9 t^2 + 49t + 155 = 0 $$ This is easy, as we can use the quadratic formula: $$ t = \frac{-49 \pm \sqrt{49^2 -4\times (-4.9) \times 155}}{2 \times (-4.9)} $$ since time is only positive, we can take the negative-signed solution, giving $t\approx 12.525$.
Best Answer
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.