Water is flowing out of a cone shaped water tank with a speed = 10 cm^3. The cone is vertical with the tip down. The radius of the tank is 2 m and the height is 6 m.
i) How fast does the radius on the water surface change when the water is 4,5 m deep?
ii) How fast in cm/min does the water sink?

Question

Water is flowing out of a cone shaped water tank with a speed = 10 cm^3. The cone is vertical with the tip down. The radius of the tank is 2 m and the height is 6 m.
i) How fast does the radius on the water surface change  when the water is 4,5 m deep?
ii) How fast in cm/min does the water sink?

anonymous · Answer

letx=2,y=6m, dV/dt=10cm^3
x/y = 2/6=1/3  => x=y/3
volume V=(pix^2 y)/3
             V=pi(y/3)^2 y/3
             V=pi y^3 /27 now take dV/dt=(3pi y^2 /27)dy/dt=pi y^2/9 dy/dt
dy/dt=(9/pi y^2)dV/dt, now take y=4m=400 cm,dVdt=10 cm^3 /min
dy/dt=1.79 x10^-4 cm/min
for y=5m=500cm
dy/dt=1.146x10^-4 cm/min

anonymous · Answer

This is really complicated for me! Is there an alternative method?