Question

Consider that there is a container having the shape of an inverted cone. It has a base- diameter of 24cm and the height is twice as its diameter. The inverted cone contains the liquid and due to a small leakage at the bottom end, liquid flows out at the rate of 8ml of every second. Find the rate of change of the height of the liquid when the height of liquid is 19cm above its bottom end.

Answer 1

using similar triangles, when the depth of the liquid is h, then the radius of the surface is h/4
Thus, the volume of the liquid is
v = 1/3 V(h/4)^2 h = π/48 h^3
dv/dt = π/16 h^2 dh/dt
when h=19 cm,
-8 = π/16 * 19^2 dh/dt
dh/dt = -128/(361π) cm/s

Answer 2

can you elaborate please? i dont get how you can find the final answer

Answer 3

really? I showed every step.
Read up on implicit differentiation, and look in your text, or online, for "related rates" examples.

sorry about the typo. That V should have been π (did shift-v instead of ctrl-v)