A can of soda measures 2.5 inches in diameter and 5 inches in height. If a full-can of soda gets spilled at a rate of 4 in^3/sec, how is the level of soda changing at the moment when the can of soda is half-full?

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Answer 1 · 2015-03-31T01:08:18

The level is decreasing at a rate of $(4)/(1.56pi)"in/sec"$

The volume of liquid in the full can is given by:

$V=pir^(2)h=pi(d/2)^(2)h=(pid^(2))/(4).h$ $color(red)((1))$

The rate of loss of liquid is given by:

$(dV)/(dt)=-4"in"^(3)"/s"$

We want to know how the level $h$ is changing with time when $h=5$

From $color(red)((1))$ :

$d=2.5$

So

$(dV)/(dh)=(6.25)/(4)pi=1.562pi$ $color(red)((2))$

This is constant.

$(dV)/(dt)=(dV)/(dh)xx(dh)/(dt)$ .

So from $color(red)((2))$ :

$-4=1.562pixx(dh)/(dt)$

$(dh)/(dt)=-(4)/(1.562pi)$