Problem 76 - Algebra and Trigonometry with Analytic Geometry

Question

The flow rate (or water discharge rate) at the mouth of the Orinoco River in South America may be approximated by $$ F(t)=26,000 \sin \left[\frac{\pi}{6}(t-5.5)\right]+34,000, $$ where \(t\) is the time in months and \(F(t)\) is the flow rate in \(\mathrm{m}^{3} / \mathrm{sec}\). For approximately how many months each year does the flow rate exceed \(55,000 \mathrm{~m}^{3} / \mathrm{sec}\) ?

Step-by-Step Solution

Answer

The flow rate exceeds 55,000 m³/sec for about 2.43 months each year.

Step 1: Set up the Inequality

To find for how many months the flow rate exceeds 55,000 m³/sec, set up the inequality: \[ F(t) = 26,000 \sin \left[ \frac{\pi}{6}(t-5.5) \right] + 34,000 > 55,000 \] Subtract 34,000 from both sides to simplify the inequality:\[ 26,000 \sin \left[ \frac{\pi}{6}(t-5.5) \right] > 21,000 \] Divide both sides by 26,000:\[ \sin \left[ \frac{\pi}{6}(t-5.5) \right] > \frac{21,000}{26,000} \] This simplifies to:\[ \sin \left[ \frac{\pi}{6}(t-5.5) \right] > 0.8077 \]

Step 2: Find the Angle Range

The sine function exceeds 0.8077 when the angle is within the range: \[ \theta_1 = \arcsin(0.8077) \quad \text{and} \quad \theta_2 = \pi - \arcsin(0.8077) \] Calculating these gives approximate values:\[ \theta_1 \approx 0.9352 \quad \text{and} \quad \theta_2 \approx 2.206 \] Thus, the angles are in the range \[ 0.9352 < \frac{\pi}{6}(t-5.5) < 2.206 \]

Step 3: Solve for t

To solve for t, first set up the inequality for t:\[ 0.9352 < \frac{\pi}{6}(t-5.5) < 2.206 \] Multiply through by 6/π to solve for (t-5.5):\[ \frac{6}{\pi} \cdot 0.9352 < t-5.5 < \frac{6}{\pi} \cdot 2.206 \] Calculate these values:\[ 1.785 < t-5.5 < 4.215 \] Add 5.5 to each part of the inequality:\[ 7.285 < t < 9.715 \]

Step 4: Determine the Duration

The solution interval for t is approximately between 7.285 and 9.715 months. Therefore, the flow rate exceeds 55,000 m³/sec for about (9.715 - 7.285) months. Calculate the duration:\[ 9.715 - 7.285 = 2.43 \] Thus, the flow rate exceeds 55,000 m³/sec for about 2.43 months each year.

Key Concepts

Inequalities

Inequalities are mathematical expressions that show the relationship between two values, where one is not equal but greater or less than the other. In this exercise, inequalities are used to determine when the flow rate from the Orinoco River exceeds 55,000 cubic meters per second.

To solve the problem, we set up an inequality based on the function given:

• Considering the flow rate function: \[ F(t) = 26,000 \sin \left[ \frac{\pi}{6}(t-5.5) \right] + 34,000 \]
• We need the value of \( F(t) \) to be greater than 55,000 m³/sec.
  So the inequality becomes: \[ 26,000 \sin \left[ \frac{\pi}{6}(t-5.5) \right] + 34,000 > 55,000 \]
• By simplifying the equation, first subtract 34,000 from both sides, leading to: \[ 26,000 \sin \left[ \frac{\pi}{6}(t-5.5) \right] > 21,000 \]
• Then, divide both sides by 26,000 to isolate the sine function: \[ \sin \left[ \frac{\pi}{6}(t-5.5) \right] > 0.8077 \]

This inequality indicates when the adjusted angle of the sine function yields values greater than 0.8077. Solving these helps us know the months when the condition is true.

Sine Function

The sine function is a periodic function that describes a smooth, repetitive oscillation. It is part of trigonometry and plays a crucial role in this exercise where it models how flow rate varies over time.

Key features of the sine function include:

• The range is \([-1, 1]\).
• Repeats every \(2\pi\) radians, making it periodic.
• Total amplitude in this context is driven by the coefficient 26,000.
• It shifts depending on horizontal phase shifts and vertical translations.

In our equation, \( \sin \left[ \frac{\pi}{6}(t-5.5) \right] \) models the seasonality of the river flow rate:

• The term \( \frac{\pi}{6} (t-5.5) \) affects the periodicity and phase shift. The phase shift centers the wave around the 5.5-month mark.
• The sine component captures the variation due to seasonal changes, showing when more or less water flows.

By evaluating \( \sin \left[ \frac{\pi}{6}(t-5.5) \right] > 0.8077 \), we find the range of months where flow rates peak beyond a high set threshold of 55,0000 m³/sec.

Flow Rate

Flow rate refers to the volume of fluid that passes through a point per unit of time; here, it's measured in \( \mathrm{m}^3/\mathrm{sec} \). This concept is key in understanding rivers and how their discharge varies through the year.

In the given exercise, the function:\[ F(t) = 26,000 \sin \left[ \frac{\pi}{6}(t-5.5) \right] + 34,000 \]
Models the flow rate of the Orinoco River over time. The function uses:

• A baseline flow rate of 34,000 m³/sec. This is the minimum or static level around which the sine variations occur.
• An oscillating component due to the sine function, which adds up to 26,000 m³/sec depending on the month to showcase seasonal variability.

By setting a threshold (in this case, 55,000 \( \mathrm{m}^3/\mathrm{sec} \)), inequalities help predict when the river flows strongly due to additional factors like rainfall.This allows calculations to ensure understanding of the time span during when these conditions persist, reflecting crucial insights for water resource management.