Question

homework6: problem 3
(3 points)
water is leaking out of an inverted conical tank at a rate of 0.0088 m³/min. at the same time water is being pumped into the tank at a constant rate. the tank has height 12 meters and the diameter at the top is 7 meters. if the water level is rising at a rate of 0.28 m/min when the height of the water is 1.5 meters, find the rate at which water is being pumped into the tank.
water is being pumped in at □ m³/min.
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Explanation:

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

For a cone, by similar - triangles, if the height of the cone is \(H = 12\) meters and the radius of the cone is \(R=\frac{7}{2}\) meters, and for the water in the cone with height \(h\) and radius \(r\), we have \(\frac{r}{h}=\frac{R}{H}\). Substituting \(R = \frac{7}{2}\) and \(H = 12\), we get \(r=\frac{7}{24}h\).

Step2: Write the volume formula of the water in the cone

The volume of a cone is \(V=\frac{1}{3}\pi r^{2}h\). Substitute \(r = \frac{7}{24}h\) into the volume formula, then \(V=\frac{1}{3}\pi(\frac{7}{24}h)^{2}h=\frac{49\pi}{1728}h^{3}\).

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

Using the chain rule \(\frac{dV}{dt}=\frac{49\pi}{1728}\times3h^{2}\frac{dh}{dt}=\frac{49\pi}{576}h^{2}\frac{dh}{dt}\).

Step4: Substitute the given values

We are given that \(h = 1.5\) meters and \(\frac{dh}{dt}=0.28\) m/min.
First, calculate \(h^{2}=(1.5)^{2}=2.25\).
Then \(\frac{dV}{dt}=\frac{49\pi}{576}\times2.25\times0.28\).
\(\frac{dV}{dt}=\frac{49\pi\times2.25\times0.28}{576}\).
\(\frac{dV}{dt}=\frac{49\pi\times0.63}{576}\approx\frac{49\times3.14\times0.63}{576}\).
\(\frac{dV}{dt}\approx\frac{97.118}{576}\approx0.169\).

Let the rate at which water is pumped in be \(x\) m³/min and the rate at which water is leaking out is \(0.0088\) m³/min.
We know that \(\frac{dV}{dt}=x - 0.0088\).

Step5: Solve for \(x\)

Since \(\frac{dV}{dt}\approx0.169\), then \(x=\frac{dV}{dt}+ 0.0088\).
\(x\approx0.169 + 0.0088=0.1778\).

Answer:

\(0.178\)