Question #189414

Water is being poured at the rate of 2\pi m^(3) min into an inverted conical tank that is 12-meter deep with a radius of 6 meters at the top. if the water level is rising at the rate of 1/6 m/min and there is a leak at the bottom of the tank, how fast is the water leaking the water is 6-meter deep?


1
Expert's answer
2021-05-07T13:01:53-0400

For the conical tank with radius R=6 and height h=12

Rh=12R=12h\dfrac{R}{h}=\dfrac{1}{2}\Rightarrow R= \dfrac{1}{2}\cdot h


Whatever the water level, the water in the tank is a cone

similar in shape to the tank itself, with  R=12hR= \dfrac{1}{2}h


The volume of water in the tank when the water is h meter deep is

V=13πR2h=13π(12h)2h=π12h3  m3V=\dfrac{1}{3}\pi R^2h=\dfrac{1}{3}\pi (\frac{1}{2}h)^2h=\dfrac{\pi}{12}h^3 \ \ m^3


The volume in the tank increases at a rate

dVdt=(π12)3h2(dhdt)=(π4)h2(dhdt)\dfrac{dV}{dt}=(\dfrac{\pi}{12})\cdot 3h^2(\frac{dh}{dt})=(\dfrac{\pi}{4})\cdot h^2(\frac{dh}{dt})


When the water depth is h=6 m and the rate of water volume increases is

dVdt=(π4)62(16)=3π12  m3/min\dfrac{dV}{dt}=(\dfrac{\pi}{4})\cdot 6^2\cdot (\dfrac{1}{6})=\dfrac{3\pi}{12} \ \ m^3/min


With water entering the tank at 2π\pi   m3/min\ \ m^3/min

and water leaking out at unknown rate of x  m3/min\ \ m^3/min


The water is increasing at a rate of (3π12)  m3/min(\dfrac{3\pi}{12})\ \ m^3/min


That means that

3π12=2πxx=2π3π12x=π2\dfrac{3\pi}{12}=2\pi-x\\\Rightarrow x=2\pi-\dfrac{3\pi}{12}\\\Rightarrow x=\dfrac{\pi}{2}


Answer: When the water depth is 6 meter, water is leaking at π12\boxed{\dfrac{\pi}{12}} cubic meters per minutes.

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