Problem 107 - Algebra for College Students

Question

Solve each problem. See Example 9. Depth and flow. When the depth of the water in the Tangipahoa River at Robert, Louisiana, is 9.14 feet, the flow is 1230 cubic feet per second ( \(\mathrm{ft}^{3} / \mathrm{sec}\) ). When the depth is 7.84 feet, the flow is \(826 \mathrm{ft}^{3} / \mathrm{sec} .\) (U.S. Geological Survey, www.usgs.gov). Let \(w\) represent the flow in cubic feet per second and \(d\) represent the depth in feet. a) Write the equation of the line through \((9.14,1230)\) and \((7.84,826)\) and express \(w\) in terms of \(d .\) Round to two decimal places. b) What is the flow when the depth is \(8.25 \mathrm{ft} ?\) c) Is the flow increasing or decreasing as the depth increases?

Step-by-Step Solution

Answer

a) \(w = 310.77d - 1610.83\), b) 951.02 \( \text{ft}^3/\text{sec} \), c) Increasing.

Step 1 - Finding the slope

Use the formula for the slope of a line, \(\text{slope} = \frac{w_2 - w_1}{d_2 - d_1}\), where\((d_1, w_1) = (9.14,1230)\) and \((d_2, w_2) = (7.84,826)\). Substitute the given values:\[\text{slope} = \frac{826-1230}{7.84-9.14}=\frac{-404}{-1.3}\approx 310.77\]

Step 2 - Writing the equation in point-slope form

Use the point-slope form of the equation of a line: \(w - w_1 = m(d - d_1)\), replacing \(m\) with the slope found in Step 1 and using the point \((9.14, 1230)\):\[w - 1230 = 310.77(d - 9.14)\]

Step 3 - Expressing w in terms of d

Rearrange the equation from Step 2 to solve for \(w\):\[w = 310.77(d - 9.14) + 1230\]Simplify it:\[w = 310.77d - 310.77 \times 9.14 + 1230\approx 310.77d - 2840.83 + 1230\approx 310.77d - 1610.83\]Thus, the equation is:\[w = 310.77d - 1610.83\]

Step 4 - Finding the flow at 8.25 ft depth

Substitute \(d = 8.25\) into the equation from Step 3:\[w = 310.77 \times 8.25 - 1610.83\approx 310.77 \times 8.25 - 1610.83\approx 2561.85 - 1610.83\approx 951.02\]Therefore, the flow when the depth is 8.25 ft is approximately 951.02 \( \text{ft}^3/\text{sec} \).

Step 5 - Determining the relationship between depth and flow

Since the slope is positive (310.77), it indicates that the flow increases as the depth increases.

Key Concepts

slope-intercept form

The slope-intercept form of a linear equation is one of the most useful formats for understanding linear relationships. This form is written as:
\(y = mx + b\), where:

• \(y\) represents the dependent variable
• \(m\) is the slope of the line
• \(x\) represents the independent variable
• \(b\) is the y-intercept, or the point where the line crosses the y-axis

To put an equation in slope-intercept form, you need two pieces of information: the slope and the y-intercept of the line. In our exercise, we start with two points: \((9.14, 1230)\) and \((7.84, 826)\). We use these to calculate the slope, then use one of the points to solve for the y-intercept.

From our solution, we found the slope to be approximately 310.77. After calculation, we simplified our equation to: \(w = 310.77d - 1610.83\). This equation represents the relationship between the depth \(d\) and the flow \(w\) in the Tangipahoa River, using the slope-intercept form.

point-slope form

The point-slope form of a line is particularly handy when you know one point on the line and the slope. The point-slope form is written as:
\(y - y_1 = m(x - x_1)\), where:

• \((x_1, y_1)\) is a point on the line
• \(m\) is the slope of the line

In our exercise, we use one of our given points \((9.14, 1230)\) and our calculated slope 310.77. Plugging these into the point-slope formula, we get: \(w - 1230 = 310.77(d - 9.14)\). This step is crucial as it helps transition the equation to a more familiar form.

Rewriting the equation in slope-intercept form allows us to express \(w\) directly in terms of \(d\), making it easier to calculate the flow for any given depth. This rearrangement is achieved by isolating \(w\) on one side of the equation, resulting in \(w = 310.77d - 1610.83\).

hydrological data analysis

Hydrological data analysis involves understanding and interpreting data related to water bodies, such as rivers, lakes, and reservoirs. This type of analysis is essential for various purposes including flood forecasting, water resource management, and environmental protection.

In our exercise, we're looking at the relationship between the water depth and flow in the Tangipahoa River. We used linear equations to model this relationship, which simplifies the analysis. By examining two sets of data points for depth and flow, we established a link between them using a linear model.

This method allows hydrologists to predict the flow of the river at different depths, which is valuable for understanding river behavior under different conditions. The positive slope we discovered indicates that flow increases with depth, a common trend in open channel flows. This kind of linear relationship helps in creating predictive models, crucial for making informed decisions in water management and safety planning.