homework6: problem 3\n(3 points)\nwater is leaking out of an inverted conical tank at a rate of 0.0088…

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)