Rate measure
posted by Abhi on .
A balloon in the form of a right cone surmiunted by a hemispere,having a diameter equal to height of the cone,is being inflated.how fast is its volume changing w.r.t its total height 'h' when h

I assume that h is the height of the cone from the floor to the base of the hemisphere. Thus the diameter of the cone at the floor is 2 h and the height of the cone if it went to the tip would be 2 h
volume of cone of base diameter 2 h and height 2 h:
(1/3)(2h)(pi/4)(4 h^2)= (2/3)pi h^3
volume of cut off tip of cone:
(1/3)(h)(pi/4)h^2 = (1/12) pi h^3
so
volume of cone base = (7/12)pi h^3
now
volume of hemisphere = (1/2)(4/3)pi(h/2)^3 = pi h^3/12
so
total balloon volume = (2/3) pi h^3
dV/dh = 2 pi h^2
dV/dt = dV/dh * dh/dt
so
dV/dt = 2 pi h^2 dh/dt