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Flooding in the Grand Canyon

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Tahj Burnett

Calculus can be a helpful tool for scientists when it comes to solving real-world environmental problems that could potentially have major consequences for the environment if left unresolved. One such environmental problem exists when thinking about one of the 7 Natural Wonders of the World, the Grand Canyon. Formed by the Colorado River in Arizona, the Grand Canyon is approximately 277 miles long and about a mile deep. As you might expect, throughout history the Grand Canyon has at times had to deal with flooding given how deep it is.

Thankfully, the Glen Canyon Dam at the top of the Grand Canyon helps prevent natural flooding. In 1996, scientists decided an artificial flood was necessary to restore the environmental balance. Water was released through the dam at a controlled rate shown in the figure below. The figure also shows the rate of flow of the last natural flood in 1957.

In hindsight, this graph can help us gain some valuable information into whether or not this artificial flooding was actually helpful in reducing the flooding as well as to what extent. 

Information from the graphs: comparing the 1957 natural flood to the 1996 artificial flood

Normal rate of discharge

By looking at this graph, we see that the rate of discharge of water (in m3/s) passing through the dam is displayed on the y-axis and time (in months) is on the x-axis. To approximate the water passing through the dam before both the artificial (shown as the black curve) and natural floods (shown as the blue curve) we evaluate the y-value of the curves before the curves start to sharply increase (because of flooding).

  • The y-value of the curve before the sharp increase in the 1996 artificial flood is ~400 m³/s
  • The y-value of the curve before the sharp increase in the 1957 natural flood is ~250 m³/s

These two values signify what the “normal” rate of discharge was prior to flooding.

Maximum rate of discharge

We can also determine the maximum rates of water discharge for each flood by looking at the highest y-values on each curve. The maximum rates of discharge based on the graph are:

  • ~1250 m³/s for the 1996 artificial flood
  • ~3500 m³/s for the 1957 natural flood
Duration

To find the duration of the floods, we can this time take a look at the x-axis on the graph and estimate how the total time it took for the rate of discharge to increase to the maximum and then decrease (roughly) back down to the “normal rate” of water discharge prior to the flooding. The duration of each curve based on the graphs are:

  • ~half a month = ~15 days for the 1996 artificial flood
  • ~four months = ~122 days for the 1957 natural flood

How much additional water passed down the river in 1996 as a result of the artificial flood?

Here is where the real calculus comes in! Since our graph models two different curves that represent the rates at which water is flowing into the canyon over time we can determine the amount of additional water that passed down the river (aka the total accumulated change) by calculating the area under the curve but above that “normal rate.” This area under the curve could be calculated using the definite integral, but since we only have a graph of the functions representing the two curves, we can estimate this accumulated change by looking at the graph itself.

The area under the 1996 curve can be estimated using the formula for the area of a rectangle:

\text{Area} = \text{Length} \times \text{Width}

Per the answers we found in the previous questions, we know:

\text{Length (the maximum rate - the ``normal'' rate)} \approx 1250-400 = 850~\text{m}^3/\text{s}

\text{Width (the duration of the flooding)} \approx 15~\text{days}

\text{Area} \approx ( 850~\text{m}^3/\text{s}) \times (15~\text{days})

To simplify things we can also convert the 15 days → 1,296,000 seconds so that they match the units shown on the graph. This leaves us with a final answer of:

\text{Area} \approx ( 850~\text{m}^3/\text{s}) \times (1,296,000~\text{s}) = 1.1016 \times 10^9~\text{m}^3

of additional water.

How much additional water passed down the river in 1957 as a result of the flood?

Looking at the natural flooding curve, it’s clear that the shape of the curve isn’t quite the same as the artificial flood’s rectangle. Instead we’ll do the same process except we’ll use the formula for determining the area of a triangle:

\text{Area} = \frac12 \times \text{Length} \times \text{Width}

Per the answers we found in the previous questions, we know:

\text{Length (the maximum rate - the ``normal'' rate)} \approx 3500-250 = 3250~\text{m}^3/\text{s}

\text{Width (the duration of the flooding)} \approx 4~\text{months} = 122~\text{days}

\text{Area} \approx \frac12 \times (3250~\text{m}^3/\text{s}) \times (122~\text{days})

Again we can convert the days to seconds (1.054 \times 10^7~\text{s}) which leaves us with a final answer of:

\text{Area} \approx \frac12 \times (3250~\text{m}^3/\text{s}) \times (1.054 \times 10^7~\text{s}) = 1.7128 \times 10^{10}~\text{m}^3

of additional water.

It’s clear (from the graph and numbers), that the use of artificial flooding, as opposed to just allowing natural flooding, substantially reduces the amount of water discharged into the grand canyon by a significant amount.