The ratio of the height to the top radius if the water is constant. You can use this to your advantage. Think "Similar Triangles".
A inverted conical vessel contains water to a height of 12cm initially. A hole is opened at the vertex and water leaks away in such a manner that after t minutes, the rate of decrease of the height, h cm, of the water in the vessel is given by dh/dt = -2t/3 cm/min
a) calculate the time taken for the vessel to empty
b)Given the volume of water in the vessel after t minutes is ( h^3) cm^3 , calculate the rate of change of the volume of water in the vessel when t =3
I am already stumped by the first part. As dh/dt is constantly changing due to the vessel not being uniform in base area as the water level decreases i can't just divide the figures. Can anyone point me in the right direction for this question?
a) Integrate -2t/3
you get : h = -t^2/3 + c. Now, since t=0, h=12, Therefore, h = -t^2 / 3 + 12
Therefore time taken to empty is to put h = 0, t = 6
b) V= 1/3 {\pi} r^2 h
put V = ( h^3) = 1/3 {\pi} r^2 h
Solve and you get, 1/2 h^2 = r^2
Therefore, V = 1/6 {\pi} h^3
From (a), t = 3, h = 9
dV/dh = 1/2 {\pi} h^2
Use the rate of change: dV/dt = dV/dh x dh/dt
= 1/2 {\pi| (9)^2 x (-2)
=81pi
I am a teacher and joined this forum recently. btw, can someone help me with how to type the codes for symbols?