finding rate of depth decrease of an inverted coneA funnel has a circular top of diameter 20cm and a height of 30cm. When the depth of the liquid...

We'll start by recalling the formula that gives the volume
of the cone:

V = `pi` `r^(3)`
h/3

In this case, w'ell consider V as being the volume of
the liquid remained in the funnel, h is the depth of remaining liquid and r is the
radius of the circle which represents the base of the new cone formed by the remaining
liquid.

Therefore, if h = 12, dV/dt =
-0.2

Since in the formula that gives the volume of the
cone, we have two variables, we'll express the radius in terms of the height. For this
reason, we'll use similar triangles:

r/h = (20/2)/30 =
1/3

r = h/3

We'll re-write the
formula that gives the volume of cone, in terms of h.

V =
`pi` `h^(3)` /27

We'll differentiate with respect to
t:

dV/dt = (3`pi``h^(2)`
/27)(dh/dt)

dV/dt = (`pi` `h^(2)`
/9)(dh/dt)

But dV/dt =
-0.2

-0.2 = (` `144`pi`
/9)(dh/dt)

-0.2 = (16`pi`)
(dh/dt)

(dh/dt) =
-0.2/16`pi`

(dh/dt) `~~` -0.003
cm/s

Therefore, the depth of the liquid in
the funnel is decreasing at a rate of 0.003 cm/s.