water is leaking out of an inverted conical tank at a rate of 11,500 cm³/min at the same time that water is…

water is leaking out of an inverted conical tank at a rate of 11,500 cm³/min at the same time that water is being pumped into the tank at a constant rate. the tank has height 6 m and the diameter at the top is 4 m. if the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate (in cm³/min) at which water is being pumped into the tank. (round your answer to the nearest integer.)

water is leaking out of an inverted conical tank at a rate of 11,500 cm³/min at the same time that water is being pumped into the tank at a constant rate. the tank has height 6 m and the diameter at the top is 4 m. if the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate (in cm³/min) at which water is being pumped into the tank. (round your answer to the nearest integer.)

Answer

Explanation:

Step1: Establish the ratio of radius to height

The tank has height $h = 6m=600cm$ and diameter at the top $d = 4m = 400cm$, so radius $r = 200cm$. For a cone, the ratio of radius to height is constant. So $\frac{r}{h}=\frac{200}{600}=\frac{1}{3}$, then $r=\frac{1}{3}h$.

Step2: Find the volume formula of water in the cone

The volume of a cone is $V=\frac{1}{3}\pi r^{2}h$. Substitute $r = \frac{1}{3}h$ into it, we get $V=\frac{1}{3}\pi(\frac{1}{3}h)^{2}h=\frac{1}{27}\pi h^{3}$.

Step3: Differentiate the volume with respect to time

Differentiate $V=\frac{1}{27}\pi h^{3}$ with respect to time $t$ using the chain - rule. $\frac{dV}{dt}=\frac{1}{9}\pi h^{2}\frac{dh}{dt}$.

Step4: Substitute the given values

We know that $h = 2m=200cm$ and $\frac{dh}{dt}=20cm/min$. Then $\frac{dV}{dt}=\frac{1}{9}\pi(200)^{2}\times20=\frac{800000\pi}{9}cm^{3}/min$.

Step5: Calculate the rate of water being pumped in

Let the rate of water being pumped in be $P$ and the rate of water leaking out be $L = 11500cm^{3}/min$. We know that $\frac{dV}{dt}=P - L$. So $P=\frac{dV}{dt}+L$. Substitute $\frac{dV}{dt}=\frac{800000\pi}{9}$ and $L = 11500$ into the equation: $P=\frac{800000\pi}{9}+11500\approx\frac{800000\times3.14}{9}+11500=\frac{2512000}{9}+11500\approx279111.11 + 11500=290611.11\approx290611cm^{3}/min$.

Answer:

$290611$