Flow rates in the Spokane River The daily discharge of the Spokane River as it flows through Spokane, Washington, in April and June is modeled by the functions
r1(t) = 0.25t²+37.46t+722.47 (April) and
r2(t) = 0.90t²−69.06t+2053.12 (June), where the discharge is measured in millions of cubic feet per day, and t=0 corresponds to the beginning of the first day of the month (see figure).

c. The Spokane River flows out of Lake Coeur d’Alene, which contains approximately 0.67mi³ of water. Determine the percentage of Lake Coeur d’Alene’s volume that flows through Spokane in April and June.

Verified step by step guidance

1. Identify the discharge functions for April and June: April is given by \(r_1(t) = 0.25t^{2} + 37.46t + 722.47\) and June by \(r_2(t) = 0.90t^{2} - 69.06t + 2053.12\), where \(t\) is in days and \(r(t)\) is in millions of cubic feet per day.
2. Calculate the total volume of water discharged in each month by integrating the discharge functions over the time interval from \(t=0\) to \(t=30\) (assuming 30 days in April and June). This means computing the definite integrals: \(V_1 = \int_0^{30} r_1(t) \, dt\) for April and \(V_2 = \int_0^{30} r_2(t) \, dt\) for June.
3. Convert the volume of water discharged from millions of cubic feet to cubic miles to be consistent with the volume of Lake Coeur d’Alene. Use the conversion factor: 1 cubic mile = 147,197,952,000 cubic feet. Since the discharge is in millions of cubic feet, multiply by \$10^6$ before converting.
4. Calculate the percentage of the lake's volume that flows through Spokane by dividing the volume discharged in each month by the total volume of the lake (0.67 cubic miles) and then multiplying by 100. That is, for April: \(\text{Percentage}_1 = \frac{V_1}{0.67} \times 100\) and for June: \(\text{Percentage}_2 = \frac{V_2}{0.67} \times 100\).
5. Interpret the results to understand how much of the lake's water volume passes through Spokane in each month, which provides insight into the river's flow relative to the lake's capacity.

Key Concepts

Definite Integral as Accumulated Quantity

The definite integral of a rate function over a time interval represents the total accumulated quantity during that period. In this problem, integrating the discharge rate functions r1(t) and r2(t) over the days in April and June gives the total volume of water flowing through the river in those months.

Unit Conversion and Volume Comparison

To compare the total volume of water discharged with the volume of Lake Coeur d’Alene, it is essential to convert units consistently. The river discharge is given in millions of cubic feet per day, while the lake volume is in cubic miles. Converting cubic miles to cubic feet allows for a meaningful percentage comparison.

Quadratic Functions and Their Graphs

The discharge rates are modeled by quadratic functions, which describe how the flow changes over time. Understanding the shape and behavior of these parabolas helps interpret the flow trends and ensures correct integration limits and calculations for total discharge.