Problem 39 - Thomas Calculus in SI Units

Question

Find the function with the given derivative whose graph passes through the point \(P\). $$r^{\prime}(\theta)=8-\csc ^{2} \theta, \quad P\left(\frac{\pi}{4}, 0\right)$$

Step-by-Step Solution

Answer

The function is \( r(\theta) = 8\theta + \cot \theta - 2\pi - 1 \).

Step 1: Identify the Problem

We need to find the function \( r(\theta) \) whose derivative is given by \( r'(\theta) = 8 - \csc^2 \theta \) and that passes through the point \( P \left( \frac{\pi}{4}, 0 \right) \).

Step 2: Integrate to Find the Function

Since the derivative of the function is \( r'(\theta) = 8 - \csc^2 \theta \), we need to find \( r(\theta) \) by integrating this derivative. The integral of 8 with respect to \( \theta \) is \( 8\theta \). The integral of \( -\csc^2 \theta \) with respect to \( \theta \) is \( \cot \theta \) since the derivative of \( \cot \theta \) is \( -\csc^2 \theta \). Thus,\[r(\theta) = \int (8 - \csc^2 \theta) \, d\theta = 8\theta + \cot \theta + C\]where \( C \) is the constant of integration.

Step 3: Use the Initial Condition to Find the Constant

We use the point \( P(\frac{\pi}{4}, 0) \) to find the constant \( C \). Substitute \( \theta = \frac{\pi}{4} \) and \( r(\theta) = 0 \) into the equation:\[0 = 8 \left( \frac{\pi}{4} \right) + \cot \left( \frac{\pi}{4} \right) + C\]Knowing that \( \cot \left( \frac{\pi}{4} \right) = 1 \), we have:\[0 = 2\pi + 1 + C\]Solving for \( C \), we get \( C = -2\pi - 1 \).

Step 4: Write the Final Function

Now that we have \( C \), the function \( r(\theta) \) can be written as:\[r(\theta) = 8\theta + \cot \theta - 2\pi - 1\]This is the function whose derivative is \( 8 - \csc^2 \theta \) and passes through the given point.

Key Concepts

Trigonometric Integration

Trigonometric integration involves finding the integral of functions involving trigonometric expressions. It's a key skill in calculus because many real-world phenomena are modeled using sine, cosine, tangent, and their related functions.

In this exercise, we deal with the derivative \( r'(\theta) = 8 - \csc^2 \theta \). To find the original function, we integrate this expression. Let's break down the process:

• **Integrate the constant term**: The integral of a constant, like 8, over a variable, \( \theta \), is straight forward. It becomes \( 8\theta \).
• **Integrate the trigonometric term**: For \( -\csc^2 \theta \), it helps to recall its relationship to the derivatives of trigonometric functions. \( \csc^2 \theta \) is the derivative of \( \cot \theta \). Therefore, integrating \( -\csc^2 \theta \) gives us \( \cot \theta \).

Putting these together, we find the integral to be:\[ r(\theta) = 8\theta + \cot \theta + C \]This step-by-step integration process reveals how trigonometric functions work within calculus, demonstrating how derivatives and integrals are inverse operations.

Constant of Integration

When integrating a function in calculus, an arbitrary constant, often denoted as \( C \), is included with the integral. This is known as the constant of integration. It acknowledges that there are infinitely many antiderivatives (original functions) that differ only by a constant.

In our integration result, \( r(\theta) = 8\theta + \cot \theta + C \), \( C \) appears because any constant would disappear when differentiating to get back to \( r'(\theta) \).

• This means that without additional information, such as an initial condition, we can't pinpoint the exact function, only a family of functions corresponding to different values of \( C \).
• Understanding the role of this constant reminds us that integrals undo differentiation, recovering the general form of the original function.

The constant of integration is resolved only when specific values or conditions are known, as shown in the next part.

Initial Condition Problem

An initial condition problem is characterized by solving for a specific constant in an integrated function. You use given conditions that often include a point through which the graph of the function passes.

In our case, we know the function passes through \( P\left( \frac{\pi}{4}, 0 \right) \). This initial condition allows us to solve for the constant \( C \).

• Using \( r\left( \frac{\pi}{4} \right) = 0 \), we plug \( \theta = \frac{\pi}{4} \) into the integrated function \( r(\theta) = 8\theta + \cot \theta + C \).
• This gives us the equation \( 0 = 8\left( \frac{\pi}{4} \right) + \cot\left( \frac{\pi}{4} \right) + C \).
• Knowing \( \cot\left( \frac{\pi}{4} \right) = 1 \), we solve for \( C \) to be \( -2\pi - 1 \).

Thus, the initial condition pinpoints the specific antiderivative, giving us the unique function fitting both the derived expression and the given point. This shows how initial conditions ensure integration results tie back to real-world constraints or specific values.