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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 = 100(200 - t)^2, measured in cubic - meters. a. graph the volume function. what is the volume of water in the tank before the valve is opened? 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 4000000 cubic meters. b. it takes 200 hr for the tank to empty. c. find the rate at which water flows from the tank. v(t)=□
Step1: Analyze the volume - function
The volume function is given as $V(t)=100(200 - t)^2$, where $V$ is in cubic - meters and $t$ is in hours.
Step2: Find the derivative of the volume function
Using the chain - rule, if $y = u^2$ and $u = 200 - t$, then $\frac{dy}{du}=2u$ and $\frac{du}{dt}=-1$. So, $V^\prime(t)=\frac{dV}{dt}=100\times2(200 - t)\times(-1)=-200(200 - t)$.
Step3: Answer part (a)
The volume function $V(t)=100(200 - t)^2$ is a parabola opening upwards. When $t = 0$, $V(0)=100\times200^2 = 4000000$ cubic meters. As $t$ increases from $0$ to $200$, $V(t)$ decreases to $0$. The correct graph is a parabola opening downwards (since the volume is decreasing over time) with $V(0) = 4000000$ and $V(200)=0$.
Step4: Answer part (b)
Set $V(t)=0$. Then $100(200 - t)^2 = 0$. Solving for $t$, we get $(200 - t)^2 = 0$, so $t = 200$ hours.
Step5: Answer part (c)
The rate at which water flows from the tank is given by $V^\prime(t)=-200(200 - t)$.
Step6: Answer part (d)
To find when the magnitude of the flow rate is a maximum or minimum:
The magnitude of the flow rate is $|V^\prime(t)|=200|200 - t|$.
The function $y = |V^\prime(t)|$ is a linear function. It is minimized when $t = 200$ (since $|V^\prime(200)| = 0$) and maximized when $t = 0$ (since $|V^\prime(0)|=200\times200 = 40000$ cubic meters per hour).
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a. The graph of $V(t)$ is a parabola opening downwards with $V(0)=4000000$ and $V(200)=0$.
b. It takes $200$ hours for the tank to empty.
c. $V^\prime(t)=-200(200 - t)$
d. The magnitude of the flow rate is maximum at $t = 0$ and minimum at $t = 200$.