During the spring, water levels in rivers rise and flooding often occurs. To reduce the severity of flooding, sometimes a flood diversion (or ditch) is built to carry some of the water during this season.

Consider a diversion which is built as in the diagram below.

An engineering study has determined that the relationship between the average speed and depth of water in the diversion in metres/sec is

This relationship has the following graph.

Let us compute the maximum volume of water the diversion can
handle. Suppose excess spring runoff lasts for a month. Water
levels increase for 12 days and then decrease back to normal over
the next 18 days. Suppose at the start *(day 0)* the water
is just starting to enter the diversion (** y =
0**). From that time until

What was the total volume of water that passed through the diversion during the 30 day period?

Using calculus, the instantaneous volume passing through a point in the diversion is

as indicated in the following diagram.

The total volume passing through the diversion during the 30 day period is

Steps 2 to 6: Compute *A* and
*v* as functions of
*t*.

When the depth of water is *y*, the
cross-sectional area of water is (*recall the formula for the
area of a trapezoid, and the geometry of a triangle*)

*y* increases linearly for
**12** days starting at **0m** and
ending at **2m**. With the formula for a straight
line,

we can use the two points (*t* = 0, *y*
= 0) and (*t* = 12, *y* =
2) to determine the relationship,

where *t* is in days and
*y* is in metres. Starting at day
**12**, the depth decreases for the next
**18** days to **zero**. As above, we
obtain the formula for a straight line between the points
(*t* = 12, *y* = 2) and
(*t* = 30, *y* = 0). This line
is

Hence we have

*A* is related to *y*
by equation (3), and *y* is related to
*t* by equation (4).

Substituting equation (4) into (3), we obtain

Expanding the brackets:

*v* is related to *y*
by equation (1), and *y* is related to
*t*by equation (4). If we substitute
equation (4) into (1), we obtain

Expanding the brackets:

Notice that the unit of time in the speed,
*v*, is expressed in
**seconds** but time everywhere else is expressed in
**days**. To make the units coincide, let us express
the speed in **metres/day**.

So, the speed expressed in **metres/day** is

From equation (2), break the integral into two time periods
since there is a different relationship between
*A*, *v* and
*t* in each of these periods.

Expanding the brackets and bringing constant factors outside the integral,

Recall that for any *n* not equal to
-1,

Also, we can integrate an expression one term at a time.
Integrating the expression for *V*:

We can now substitute the limits for the unknown,
*t*, to obtain the answer.

Evaluating the above expression we get ** V =
1,368,576 m3** or about

This is equivalent to an area **1.17km** by
**1.17km** covered to a depth of
**1m**.