Question

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?

Answer 1

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)`