Question

a cylindrical tank is full at time t = 0 when a valve in the bottom of the tank is open. by torricellis law, the volume of water in the tank after t hours is v = 600(200 - t)^2, measured in cubic meters. a. graph the volume function. choose the correct graph. b. how long does it take for the tank to empty? c. find the rate at which water flows from the tank and plot the flow - rate function. d. at what time is the magnitude of the flow rate a minimum? a maximum? the volume of water in the tank before the valve is opened is 6000000 cubic meters. b. if it takes (square) hr for the tank to empty

Explanation:

Step1: Identify the volume - time function

The volume function is given as $V(t)=600000 - 3000t^{2}$, where $V$ is the volume of water in the tank in cubic meters and $t$ is the time in hours.

Step2: Find when the tank is empty

Set $V(t)=0$. So, $600000 - 3000t^{2}=0$. Rearrange the equation: $3000t^{2}=600000$, then $t^{2}=\frac{600000}{3000}=200$. Solving for $t$, we get $t = \sqrt{200}=10\sqrt{2}\approx14.14$ hours.

Step3: Analyze the volume - time graph

The volume of water in the tank is decreasing over time. The initial volume is $V(0) = 600000$ cubic meters. As time $t$ increases, the volume $V(t)$ decreases. The function $V(t)=600000 - 3000t^{2}$ is a parabola opening downwards.

Step4: Find the flow - rate function

The flow - rate $r(t)$ is the derivative of the volume function with respect to time. Using the power rule $\frac{d}{dt}(ax^{n})=nax^{n - 1}$, if $V(t)=600000 - 3000t^{2}$, then $r(t)=V^\prime(t)=- 6000t$. The negative sign indicates that the volume is decreasing.

Step5: Find the time when the flow - rate is maximum or minimum

The flow - rate function $r(t)=-6000t$ is a linear function. The magnitude of the flow - rate $|r(t)| = 6000|t|$. The magnitude of the flow - rate is minimum when $t = 0$ (initially, the rate starts to increase in magnitude as water starts flowing out), and it has no maximum value in the physical context of the tank emptying (as time progresses, the magnitude of the flow - rate gets larger until the tank is empty).

a. The correct graph of the volume function $V(t)=600000 - 3000t^{2}$ is a parabola opening downwards with an initial value of $V(0)=600000$ and it intersects the $t$ - axis at $t = 10\sqrt{2}$. Of the given graphs, the one that starts at $V = 600000$ and decreases down to $V = 0$ over time is the correct one.
b. To find when the tank is empty, we set $V(t)=0$.

• $600000 - 3000t^{2}=0$
• $3000t^{2}=600000$
• $t^{2}=200$
• $t = 10\sqrt{2}\approx14.14$ hours

c. The flow - rate function $r(t)=V^\prime(t)=-6000t$. The magnitude of the flow - rate is $|r(t)| = 6000|t|$.
d. The magnitude of the flow - rate is minimum at $t = 0$ (initially when the valve is just opened) and has no maximum in the context of the tank emptying as $|r(t)|$ increases with time until the tank is empty.

Answer:

a. The graph that starts at $V = 600000$ on the $V$ - axis and decreases to $V = 0$ over time (a downward - opening parabola).
b. $t = 10\sqrt{2}\approx14.14$ hours
c. $r(t)=-6000t$
d. Minimum at $t = 0$, no maximum in the context of tank emptying.