QUESTION IMAGE
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(100 - t)^2, measured in cubic - meters.
a. graph the volume function. choose the correct graph.
a. graph with range 0,100 by 0,1000000
b. graph with range 0,100 by 0,1000000
c. graph with range 0,100 by 0,1000000
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?
Step1: Analyze the volume - function behavior
The volume function is \(V(t)=100(100 - t)^2\), \(V(0)=100\times100^2 = 1000000\) and as \(t\) increases towards \(100\), \(V(t)\) decreases to \(0\). The volume - function is a quadratic function opening upwards in terms of \((100 - t)\). When \(t = 0\), \(V=1000000\) and when \(t = 100\), \(V = 0\).
Step2: Determine the correct graph
The volume starts at a maximum value of \(1000000\) at \(t = 0\) and decreases to \(0\) at \(t = 100\). Graph C shows a function that starts at a high - value and decreases to \(0\) over the interval \([0,100]\).
Step3: Find the time to empty the tank
Set \(V(t)=0\). We have \(100(100 - t)^2=0\). Divide both sides by \(100\) to get \((100 - t)^2=0\). Then \(100 - t=0\), so \(t = 100\) hours.
Step4: Find the rate of change of volume
Differentiate \(V(t)=100(100 - t)^2=100(10000-200t + t^2)=1000000-20000t + 100t^2\) with respect to \(t\). Using the power rule \(\frac{d}{dt}(x^n)=nx^{n - 1}\), we get \(V^\prime(t)=-20000 + 200t\).
Step5: Find the minimum and maximum of the flow - rate function
The flow - rate function is \(r(t)=V^\prime(t)=-20000 + 200t\). This is a linear function. The derivative of \(r(t)\) with respect to \(t\) is \(r^\prime(t)=200>0\), so \(r(t)\) is an increasing function. The minimum value of \(r(t)\) occurs at \(t = 0\), \(r(0)=-20000\) (the negative sign indicates the volume is decreasing) and the maximum value occurs at \(t = 100\), \(r(100)=-20000+200\times100 = 0\).
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a. C
b. 100 hours
c. The flow - rate function is \(r(t)=-20000 + 200t\). The minimum value of the flow rate is \(- 20000\) cubic meters per hour at \(t = 0\) and the maximum value is \(0\) cubic meters per hour at \(t = 100\) hours.