SOLUTION: Fig. 4 shows a cone. The angle between the axis and the slant edge is 30°. Water is poured into the cone at a constant rate of 2 cm3 per second. At time t seconds, the radius of t

Algebra ->  Volume -> SOLUTION: Fig. 4 shows a cone. The angle between the axis and the slant edge is 30°. Water is poured into the cone at a constant rate of 2 cm3 per second. At time t seconds, the radius of t      Log On


   



Question 1090347: Fig. 4 shows a cone. The angle between the axis and the slant edge is 30°. Water is poured into the
cone at a constant rate of 2 cm3 per second. At time t seconds, the radius of the water surface is r cm and the volume of water in the cone is V cm3.
Show that V= (sqrt3/3*pi*r^3) and find dv/dr
[You may assume that the volume of a cone of height h and radius r is (1/3*pi*r^2*h)]
I worked it out but could only prove that V= sqrt3/3*pi*r^2... why would h equal (sqrt 3 r) and not just sqrt 3?
Thank you ever so much!!!

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
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h=r%2A%28sqrt%283%29%2F3%29
So then,
V=%28pi%2F3%29r%5E2%2Ah
V=%28pi%2F3%29r%5E2%2Ar%28sqrt%283%29%2F3%29
V=%28%28sqrt%283%29pi%29%2F9%29r%5E3
So then differentiating,
dV%2Fdr=%28%28sqrt%283%29pi%29%2F9%293r%5E2
dV%2Fdr=%28%28sqrt%283%29pi%29%2F3%29r%5E2