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Question

1. A right conical vessel with base radius 2m and height 4m is being filled with water at a constant rate of 3m^3/min. At what rate is the water rising when the depth is 3m?

Please show full working as I have an exam in 2 days and need to get rid of all my mistakes. Thanks

garypopki, yes.

Answer 1

Is the vessel standing on its pointy end?

Assuming it is standing on its pointy end, the volume from the pointy end up to any height is

V = 1/3 * (radius at given height)^2 * π * given height

The radius at any height is half the height, and at the full 4 m height the radius is 2 m. So

V = 1/3 * (h/2)^2 * π * h
V = 1/3 * (h^3/4) * π

dV/dH = π * h^2/4

At a height of 3 m,

dV/dH = 9π/4

so

dV = (9π/4) * dH

You know dV per unit time, so you can find dH per unit time.

Answer 2

NEVER MIND, I assumed the cone was on its base.
If it's upside down: v=Pi*x^3/12, where x is height of water (yeah I skipped some steps, but I think previous response will help you get this far).
dv/dt = 1/4 * Pi * x^2 * dx/dt
dv/dt = 3, find dx/dt when x=3
3 = 1/4 * Pi (3^2) * dx/dt
dx/dt = 4/3/Pi = .424 m/min
-----------------------
if the cone is on its base:
the volume of water in the cone is the volume of the frustum of height x (x is also the height of water).

the volume of a cone is Vcone = Pi/3 * r^2 * h

volume of frustum = Vcone - Vtop
where Vtop is the cone of height h-x
with a base radius = (h-x)*r/h

so volume of water (V) as a function of height x:
V = Pi/3 * [(x-16)^3 + 16]

differentiate both sides with respect to time:
dV/dt = Pi * (x-16)^2 * dx/dt

we know dV/dt is 3 m^3/min and we want to find
dx/dt when x=3
3 = Pi (3-16)^2 * dx/dt
dx/dt = 3/(169Pi) m/min = 5.65 mm/min