Problem 259 - Calculus Volume 1

Question

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

Step-by-Step Solution

Answer

At 6 a.m., the rate of change of depth is \(\frac{5\pi\sqrt{3}}{12}\) feet per hour.

Step 1: Differentiate the Function

To find the rate at which the depth is changing, we need to find the derivative of the function \(D(t)\) with respect to \(t\). The given function is \(D(t) = 5 \sin \left(\frac{\pi}{6} t - \frac{7\pi}{6}\right) + 8\). Apply the chain rule to differentiate: \[ D'(t) = 5 \cos \left(\frac{\pi}{6} t - \frac{7\pi}{6}\right) \cdot \frac{d}{dt}\left(\frac{\pi}{6} t - \frac{7\pi}{6}\right) \]The derivative of \(\frac{\pi}{6} t - \frac{7\pi}{6}\) with respect to \(t\) is \(\frac{\pi}{6}\), so:\[ D'(t) = 5 \cos \left(\frac{\pi}{6} t - \frac{7\pi}{6}\right) \cdot \frac{\pi}{6} \]Simplifying gives:\[ D'(t) = \frac{5\pi}{6} \cos \left(\frac{\pi}{6} t - \frac{7\pi}{6}\right) \]

Step 2: Evaluate the Derivative at 6 a.m.

The problem asks for the rate of change at 6 a.m., which corresponds to \(t = 6\) hours after midnight. Substitute \(t = 6\) into the derivative:\[ D'(6) = \frac{5\pi}{6} \cos \left(\frac{\pi}{6} \times 6 - \frac{7\pi}{6}\right) \]Calculate the angle:\[ \frac{\pi}{6} \times 6 = \pi \]So:\[ D'(6) = \frac{5\pi}{6} \cos (\pi - \frac{7\pi}{6}) = \frac{5\pi}{6} \cos (-\frac{\pi}{6}) \]

Step 3: Calculate the Cosine Value

Now, calculate \(\cos(-\frac{\pi}{6})\). Recall that \(\cos(-\theta) = \cos(\theta)\), so:\[ \cos(-\frac{\pi}{6}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \]

Step 4: Compute the Final Rate of Change

Use the value from Step 3 to compute the rate of change:\[ D'(6) = \frac{5\pi}{6} \cdot \frac{\sqrt{3}}{2} = \frac{5\pi\sqrt{3}}{12} \]Thus, the rate at which the depth is changing at 6 a.m. is \(\frac{5\pi\sqrt{3}}{12} \text{ feet per hour}\).

Key Concepts

Chain Rule

When differentiating composite functions, the chain rule is your best friend. It enables us to differentiate a function by breaking it down into simpler parts. In practical terms, if you have a function within another function—like a trigonometric function of a linear expression—the chain rule shows us how to differentiate it correctly.

For example, in the problem above, we see the function inside another, specifically, a sine function with the argument \(\frac{\pi}{6} t - \frac{7\pi}{6}\). To differentiate this composite function, follow these steps:

• Differentiate the outer function, here sine, changing it to cosine.
• Multiply this by the derivative of the inner function, \(\frac{\pi}{6} t - \frac{7\pi}{6}\), which is simply \(\frac{\pi}{6}\).

Applying the chain rule ensures our derivatives are calculated accurately for these layered functions. This is especially critical in calculus problem-solving, where errors in initial steps can lead to incorrect results further down the line.

Rate of Change

The rate of change gives us a powerful tool in calculus by telling us how rapidly things are shifting. This is especially informative in problems involving functions where something changes over time or space.

In our water depth example, calculating the rate of change tells us exactly how the depth is changing at a certain point in time. Using the derivative \(D'(t)\), we assessed the rate at which the depth of water changes as time moves from any given hour. Specifically, at 6 a.m., substituting \(t = 6\) into our derivative expression gives us the rate at that instant.

So, why is this important? This shows how tides affect water levels as they rise and fall, enabling someone at the dock to predict depth behavior at precise moments. Comprehending rates of change can turn abstract mathematical concepts into tangible predictions for real-world applications.

Trigonometric Functions

Trigonometric functions cover a wide range of applications, especially when dealing with periodic phenomena like tides. These functions, including sine, cosine, and tangent, are central to modeling cyclical changes, such as those in our initial exercise about water depth.

Simplifying a trigonometric function isn't always easy. Key elements in trigonometric manipulations include knowing how to differentiate and understand these functions' behavior. In this problem:

• The sine function models the water depth over time as it naturally oscillates.
• The derivative involves the cosine function, showcasing how these functions are interrelated.

Understanding trigonometric functions and their properties can help solve calculus problems and predict outcomes in fields such as engineering, physics, and beyond.

Calculus Problem-Solving

Problem-solving with calculus requires a steady approach and understanding of multiple interconnected principles. Solutions often include several steps: set-up, differentiation, and evaluation.

Take our water depth example:

• First, define the function and understand its components, here applying how \(D(t)\) reflects trigonometric behavior.
• Next, differentiate using the correct rules, such as the chain rule, and derive the expression for the rate of change.
• Finally, substitute any given values to solve for the specific moment in question, like finding when the clock strikes 6 a.m.

By following these steps, and using skills like differentiating and applying real-world context, calculus becomes powerful and practical. Good problem-solving skills transform theoretical equations into predictable, useful results.