water is leaking out of an inverted conical tank at a rate of 10000.0 cubic centimeters per min at the same…

water is leaking out of an inverted conical tank at a rate of 10000.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. the tank has height 11.0 meters and the diameter at the top is 4.5 meters. if the water level is rising at a rate of 20.0 centimeters per minute when the height of the water is 5.0 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute. note: let \r\ be the unknown rate at which water is being pumped in. then you know that if v is volume of water, dv/dt = r - 10000.0. use geometry (similar triangles?) to find the relationship between the height of the water and the volume of the water at any given time. recall that the volume of a cone with base radius r and height h is given by 1/3 πr²h.

water is leaking out of an inverted conical tank at a rate of 10000.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. the tank has height 11.0 meters and the diameter at the top is 4.5 meters. if the water level is rising at a rate of 20.0 centimeters per minute when the height of the water is 5.0 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute. note: let \r\ be the unknown rate at which water is being pumped in. then you know that if v is volume of water, dv/dt = r - 10000.0. use geometry (similar triangles?) to find the relationship between the height of the water and the volume of the water at any given time. recall that the volume of a cone with base radius r and height h is given by 1/3 πr²h.

Answer

Explanation:

Step1: Find the ratio of radius to height for the cone

The cone has height $H = 11.0$ m $= 1100$ cm and diameter at the top $d=4.5$ m, so radius $R_{0}=\frac{4.5}{2}\times100 = 225$ cm. By similar - triangles, for the water - cone at any time, $\frac{r}{h}=\frac{R_{0}}{H}$, where $r$ is the radius of the water - cone and $h$ is the height of the water - cone. So $\frac{r}{h}=\frac{225}{1100}=\frac{9}{44}$, then $r = \frac{9}{44}h$.

Step2: Express the volume of water in terms of $h$

The volume of a cone is $V=\frac{1}{3}\pi r^{2}h$. Substitute $r=\frac{9}{44}h$ into the volume formula: $V=\frac{1}{3}\pi(\frac{9}{44}h)^{2}h=\frac{1}{3}\pi\frac{81}{1936}h^{3}=\frac{27\pi}{1936}h^{3}$.

Step3: Differentiate the volume with respect to time $t$

Using the chain - rule, $\frac{dV}{dt}=\frac{27\pi}{1936}\times3h^{2}\frac{dh}{dt}=\frac{81\pi}{1936}h^{2}\frac{dh}{dt}$.

Step4: Substitute the given values

We are given that $h = 5.0$ m $= 500$ cm and $\frac{dh}{dt}=20.0$ cm/min. $\frac{dV}{dt}=\frac{81\pi}{1936}\times(500)^{2}\times20$. $\frac{dV}{dt}=\frac{81\pi\times250000\times20}{1936}=\frac{40500000\pi}{1936}$.

Step5: Solve for $R$

We know that $\frac{dV}{dt}=R - 10000$. $R=\frac{dV}{dt}+10000=\frac{40500000\pi}{1936}+10000$. $R=\frac{40500000\pi + 19360000}{1936}\approx\frac{40500000\times3.14+19360000}{1936}=\frac{127170000 + 19360000}{1936}=\frac{146530000}{1936}\approx75790$ cm³/min.

Answer:

$\frac{40500000\pi + 19360000}{1936}\approx75790$