A small funnel in the shape of a cone is being emptied of fluid at the rate of 12 cm cube per second. The height of the funnel is 20 cm and the radius of the top is 4 cm. How fast is the fluid level dropping when the level stands 5 cm above the vertex of the cone?

Question

anonymous · Answer

let r   be the radius and h be the height at any instant 
then 
r/h = 4/20 or r/h =1/5    or r=(1/5) h...

anonymous · Answer

now volume at any instant is given by
V=(pi)*(r^2)*h 
or 
V=(pi)*((h/5)^2)*h
V=((pi)* h^3)/25
V=((pi)/25)*h^3
now differentiating wrt t  both sides we have
dV/dt   =3*(pi/25)*h^2 *(dh/dt)
now given dV/dt =-12 and  h=5 ...we need (dh/dt)  
just solve it  to get (dh/dt)

minatsukisaya95 · Answer

thank you
i got it

anonymous · Answer

yw

goformit100 · Answer

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