Problem 259 - Calculus Volume 1

Question

The depth (in feet) of water at a dock changes with the rise and fall of tides. The depth is modeled by the function \(D(t)=5 \sin \left(\frac{\pi}{6} t-\frac{7 \pi}{6}\right)+8,\) where \(t\) is the number of hours after midnight. Find the rate at which the depth is changing at 6 a.m.

Step-by-Step Solution

Answer

The depth is changing at a rate of \(\frac{5\pi\sqrt{3}}{12}\) feet per hour at 6 a.m.

Step 1: Understand the Problem

We need to find the rate of change of the water depth at 6 a.m. using the function \(D(t) = 5 \sin \left(\frac{\pi}{6} t - \frac{7\pi}{6}\right) + 8\). This involves calculating the derivative \(D'(t)\) and evaluating it at \(t = 6\).

Step 2: Differentiate the Function

Differentiate \(D(t)\) with respect to \(t\). Using the chain rule, we differentiate the sine function. The derivative of \(D(t) = 5 \sin \left(\frac{\pi}{6} t - \frac{7\pi}{6}\right) + 8\) is \(D'(t) = 5 \cdot \cos \left(\frac{\pi}{6} t - \frac{7\pi}{6}\right) \cdot \frac{d}{dt} \left(\frac{\pi}{6} t - \frac{7\pi}{6}\right)\).

Step 3: Simplify the Derivative

The inner derivative \(\frac{d}{dt} \left(\frac{\pi}{6} t - \frac{7\pi}{6}\right)\) is \(\frac{\pi}{6}\). Therefore, substituting back, \(D'(t) = 5 \cdot \frac{\pi}{6} \cdot \cos \left(\frac{\pi}{6} t - \frac{7\pi}{6}\right)\). This simplifies to \(\frac{5\pi}{6} \cos \left(\frac{\pi}{6} t - \frac{7\pi}{6}\right)\).

Step 4: Evaluate the Derivative at \(t = 6\)

Substitute \(t = 6\) into the derivative \(D'(t)\). We have \(D'(6) = \frac{5\pi}{6} \cos \left(\frac{\pi}{6} \cdot 6 - \frac{7\pi}{6}\right)\). Simplify inside the cosine: \(\frac{\pi}{6} \cdot 6 = \pi\), so the expression becomes \(\cos(\pi - \frac{7\pi}{6}) = \cos(-\frac{\pi}{6})\).

Step 5: Calculate \(\cos(-\frac{\pi}{6})\)

\(\cos(-\frac{\pi}{6})\) is equal to \(\cos(\frac{\pi}{6})\) due to the even property of cosine. \(\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\). So, \(D'(6) = \frac{5\pi}{6} \cdot \frac{\sqrt{3}}{2}\).

Step 6: Final Calculation

Multiply the values to find the rate of change: \(D'(6) = \frac{5\pi\sqrt{3}}{12}\). This represents the rate of change of the water depth at 6 a.m.

Key Concepts

Understanding the Chain Rule

The chain rule is a powerful tool in calculus for differentiating composite functions. When you have a function inside another function, like \(D(t) = 5 \sin \left(\frac{\pi}{6} t - \frac{7\pi}{6}\right) + 8\)in our example, you need to use the chain rule to find the derivative. The idea is to differentiate the outer function while multiplying by the derivative of the inner function.

In the function \(D(t)\), the outer function is the sine, \(\sin(x)\), and the inner function is the expression inside, \(\frac{\pi}{6} t - \frac{7\pi}{6}\). When differentiating, consider:

• Differentiate the outer function: The derivative of \(\sin(x)\) is \(\cos(x)\).
• Multiply by the derivative of the inner function: For \(\frac{\pi}{6} t - \frac{7\pi}{6}\), the derivative is simply \(\frac{\pi}{6}\).

Putting these together, you get: \[D'(t) = 5 \cdot \cos \left(\frac{\pi}{6} t - \frac{7\pi}{6}\right) \cdot \frac{\pi}{6}\]This is how the chain rule helps you handle such differentiated tasks simply and effectively.

Exploring Rate of Change

The rate of change is a concept that tells us how fast a quantity changes over time. In our dock example, it indicates how quickly the water depth changes as time passes. We calculate it using the derivative of the depth function, \(D(t)\).

At a specific time, the rate of change is found by evaluating the derivative at that time. In the problem, we want to find the rate of change at 6 a.m., so we substitute \(t = 6\) into the derivative function. This gives us:

• \(D'(6) = \frac{5\pi}{6} \cdot \cos \left(\pi - \frac{7\pi}{6}\right)\)
• Simplify: \(\cos(-\frac{\pi}{6}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\)
• Final result: \(D'(6) = \frac{5\pi\sqrt{3}}{12}\)

This calculated value shows us how fast the water level rises or falls at the exact time of 6 a.m., giving valuable information about tidal behaviors.

Trigonometric Functions in Action

Trigonometric functions, like sine and cosine, are essential in modeling periodic phenomena such as tides. In our task, the function \(\sin\left(\frac{\pi}{6}t - \frac{7\pi}{6}\right)\) represents the periodic nature of tides changing with time.

Sine functions are particularly useful:

• The sine function smoothly oscillates between -1 and 1, capturing natural cycles.
• In \(D(t)\), coefficients before the sine function (5 in this case) scale the amplitude, affecting the water depth range.
• The phase shift \(\frac{7\pi}{6}\) controls the starting point of the cycle, aligning the model to actual data.

These functions are integral in handling real-world tasks where understanding patterns, such as water levels or other oscillations, is crucial. Mastering how to apply and differentiate them allows deeper insights into the underlying systems.