How fast is the water level rising when the water is 3 cm deep (at its deepest point) if water is poured into a conical container at the rate of 10 cm3/sec. the cone points directly down, and it has a height of 25 cm and a base radius of 15 cm?

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Answer 1 · 2015-02-25T23:06:20

Given the rate of change in volume with respect to time is
$\frac{d V}{d t} = 10$ $(d V) / dt = 10$ $\frac{c u b . c m}{sec}$ $(cub. cm)/(sec)$

We are looking for the rate of change in height with respect to time:
$\frac{d h}{d t}$ $(d h) / dt$

$\frac{d V}{d t} \cdot (?) = \frac{d h}{d t}$ $(d V)/(dt) * ( ? ) = (d h)/(dt)$

The obvious replacement for $(?)$ $( ? )$ would seem to be
$\frac{d h}{d V}$ $(d h)/(dV)$ , the rate of change in height with respect to Volume.

The volume of a cone is given by the formula:
$V = (Pi r^2 h)/3$

The ratio of the radius to the height for the given cone is
$r/h = 15/25 = 3/5$ (see diagram)

or
$r = 3/5 h$
enter image source here

So, for the given cone:
$V = (Pi * (3/5 h)^2* h)/3 = (3 Pi h^3)/25 (cub. cm)$

Therefore
$(d V)/(dh) = (9 Pi)/(25) h^2 (sq. cm)$

$rarr (d h)/(dV) = (25)/( 9 Pi h^2 (sq. cm))$

$(d h)/(dt)$
$=(d V)/(dt) * (d h)/(dV) = (10 cub. cm)/(sec) * (25)/(9 Pi h^2 (sq. cm))$
$= (250 cm)/(9 Pi h^2 sec)$

at h=3
$(d h)/(dt)$ becomes
$(250)/(9 Pi xx (9)) ((cm)/(sec))$

$= (250/(81 * Pi)) ((cm)/(sec))$