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Chapter 3: Problem 389

The question concern the water level in Ocean City, New Jersey, in January, which can be approximated by \(w(t)=1.9+2.9 \cos \left(\frac{\pi}{6} t\right), \quad\) where \(t\) is measured in hours after midnight, and the height is measured in feet. Find \(w^{\prime}(3)\). What is the physical meaning of this value?

Short Answer

Expert verified
\(w'(3) \approx -1.52\): The water level is decreasing at 3 AM.

Step by step solution

01

Understand the Function

The given function is \(w(t) = 1.9 + 2.9 \cos \left(\frac{\pi}{6} t\right)\). This function represents the water level in feet at a time \(t\), where \(t\) is hours after midnight. Our task is to find \(w'(3)\), the derivative at \(t = 3\).
02

Find the Derivative

We first need to compute the derivative \(w'(t)\) of the function \(w(t) = 1.9 + 2.9 \cos \left(\frac{\pi}{6} t\right)\). Using the derivative formula for \(\cos(x)\), we know that \(\frac{d}{dt}\cos(x) = -\sin(x)\) and by the chain rule, \(\frac{d}{dt}\cos(ax) = -a\sin(ax)\). Therefore, \(w'(t) = 2.9 \cdot -\left(\frac{\pi}{6}\right) \sin \left(\frac{\pi}{6} t\right) = -\frac{2.9\pi}{6} \sin \left(\frac{\pi}{6} t\right)\).
03

Evaluate the Derivative at t = 3

Substitute \(t = 3\) into the derivative \(w'(t)\). We have \(w'(3) = -\frac{2.9\pi}{6} \sin \left(\frac{\pi}{6} \times 3\right)\). Simplify the argument of the sine function: \(\frac{\pi}{6} \times 3 = \frac{\pi}{2}\). Thus, \(w'(3) = -\frac{2.9\pi}{6} \cdot \sin\left(\frac{\pi}{2}\right)\).
04

Simplify the Expression

Recall that \(\sin\left(\frac{\pi}{2}\right) = 1\). Therefore, \(w'(3) = -\frac{2.9\pi}{6} \times 1 = -\frac{2.9\pi}{6}\). Compute the value: \(\frac{2.9\pi}{6} \approx 1.520665\), hence \(w'(3) \approx -1.520665\).
05

Interpret the Physical Meaning

The value of \(w'(3)\) represents the rate of change of the water level at 3 hours after midnight. Since \(w'(3)\) is negative, it indicates that the water level is decreasing at that time.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Functions
Trigonometric functions like sine and cosine are essential in modeling periodic phenomena. The function given, \[ w(t) = 1.9 + 2.9 \cos\left(\frac{\pi}{6} t\right) \], is a cosine function that models the water level over time. The cosine function is periodic, meaning it repeats itself over regular intervals. This is perfect for depicting something like waves or tides, which have a regular rising and falling pattern.

Understanding the nature of trigonometric functions:
  • The amplitude, which in this context is 2.9, defines the height of the wave from its mean position.
  • The phase shift and frequency factor, like \(\frac{\pi}{6}\), alters how quickly the cycles repeat.
  • Cosine and sine are fundamentally similar, shifted by a phase difference, where cosine starts at a maximum value.
When examining a function like this, it's important to recognize how these elements work together to capture natural rhythms.
Chain Rule
The chain rule is a fundamental technique in calculus used to differentiate composite functions. In the exercise, we see it applied to the function \[ w(t) = 1.9 + 2.9 \cos\left(\frac{\pi}{6} t\right) \].

Here's a breakdown of how the chain rule works:
  • First, identify the outer and inner functions. Here, \( \cos(x) \) is the outer function, and \( \frac{\pi}{6}t \) is the inner function.
  • The derivative of the cosine function is \(-\sin(x)\).
  • By the chain rule, the derivative of a composite function \( \cos(u) \) with respect to \( t \) is \(-u' \sin(u)\).
In this application, the inner function \( u = \frac{\pi}{6}t \) has the derivative \( u' = \frac{\pi}{6} \), which is then multiplied by \(-\sin(u)\). The chain rule efficiently handles differentiation where a function is nested within another function.
Rate of Change
The concept of rate of change is crucial in understanding how a quantity varies over time. In our problem, the derivative \( w'(t) \) represents the rate at which the water level changes. By calculating \[ w'(3) = -1.520665, \]we find out how fast and in which direction the water level is changing at 3 hours after midnight.

Important aspects of rate of change:
  • A positive value implies an increasing quantity, whereas a negative value indicates a decrease.
  • The magnitude shows how rapidly the change occurs at that specific instant.
  • In the context of this problem, a negative derivative denotes that the water level is dropping at a rate of about 1.52 feet per hour.
Understanding rates of change provides insights into dynamic processes, such as tidal movements, making it an invaluable tool in many fields.
Differentiation
Differentiation is the process through which we determine the rate of change of a function. This technique is central to calculus and has broad applications in science and engineering. In the context of the exercise, differentiation allows us to examine the water level fluctuations over a day.

Key points about differentiation:
  • It is derived from the limits and measures how a function changes as its variable changes.
  • Calculating derivatives such as \( w'(t) \) helps predict the behavior of a function at any given point.
  • The process involves applying rules such as the chain rule to find the derivative of composite functions.
By differentiating the given function, we could assess vital characteristics like when the water level is rising or falling and at what rate. Differentiation is a powerful tool that simplifies understanding complex, dynamic systems.

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