Question

section 4.1 related rates
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water is leaking out of an inverted conical tank at a rate of 7500 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. the tank has height 9 meters and the diameter at the top is 6.5 meters. if the water level is rising at a rate of 18 centimeters per minute when the height of the water is 2.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
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Explanation:

Step1: Find the relationship between radius and height of water in the cone

The tank has height \(H = 900\) cm and diameter \(d=650\) cm (so radius \(R = 325\) cm). For similar - triangles in the cone, \(\frac{r}{h}=\frac{R}{H}\). Substituting \(R = 325\) and \(H = 900\), we get \(r=\frac{325}{900}h=\frac{13}{36}h\).

Step2: Write the volume formula for the cone of water

The volume of a cone is \(V=\frac{1}{3}\pi r^{2}h\). Substitute \(r = \frac{13}{36}h\) into the volume formula: \(V=\frac{1}{3}\pi(\frac{13}{36}h)^{2}h=\frac{169\pi}{3888}h^{3}\).

Step3: Differentiate the volume formula with respect to time \(t\)

Using the chain - rule \(\frac{dV}{dt}=\frac{169\pi}{3888}\times3h^{2}\frac{dh}{dt}=\frac{169\pi}{1296}h^{2}\frac{dh}{dt}\).

Step4: Substitute the given values

We are given \(h = 250\) cm and \(\frac{dh}{dt}=18\) cm/min. Then \(\frac{dV}{dt}=\frac{169\pi}{1296}\times(250)^{2}\times18\).

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Let the rate of water being pumped in be \(x\) cm³/min. We know that \(\frac{dV}{dt}=x - 7500\).

Step5: Solve for \(x\)

\(x=\frac{dV}{dt}+7500\). Substituting \(\frac{dV}{dt}\approx576060.42\), we get \(x\approx576060.42 + 7500=583560.42\)

Answer:

\(583560\) (rounded to the nearest whole number)