How fast will the level of coffee be rising when the coffee is halfway up the cup?Given that, coffee is poured at a uniform rate of 3 cm3/s into a cup whose inside is shaped like a truncated cone....
Let the volume of the truncated cone = V
Then V = (1/3)*pi*(r1^2+r1*r2+r2^2)h
Where r1=lower radius ,r2=upper radius and h=height
The angle of the cone surface = tan^(-1)(4-2)/(6) = 18.43 deg.
Lower radius is constant always when filling the cup and only upper radius changing with height.
Then we can say r1=2 and r2=r
So at any point of the cone if the upper radius is r then;
h= (r-2)/tan18.43 = 3(r-2)
When cup is half filled h= h/2 = 3
Then 3 = 3(r-2) and we get r=2
V = (1/3)*pi*(r1^2+r1*r2+r2^2)h
V = (1/3)*pi*(2^2+2*r+r^2)* 3(r-2)
V = pi*(r^2+2*r+4)(r-2)
The rate of pouring of coffee = dV/dr
The rate at h=h/2 is (dV/dr) at r=2
dV/dr = pi*[*(2^2+2*2+4)+(2-2)(2*2+2)]
= 37.68 cm^3/s
The rate of coffee pour at half way fill is 37.68 cm^3/s