Problem 35 - Thomas Calculus

Question

\(r=f(\boldsymbol{\theta})\) vs. \(\boldsymbol{r}=2 \boldsymbol{f}(\boldsymbol{\theta}) \quad\) Can anything be said about the relative lengths of the curves \(r=f(\theta), \alpha \leq \theta \leq \beta,\) and \(r=2 f(\theta)\) \(\alpha \leq \theta \leq \beta ?\) Give reasons for your answer.

Step-by-Step Solution

Answer

The length of the curve \( r=2f(\theta) \) is twice that of \( r=f(\theta) \).

Step 1: Understand the Problem

We are given two polar curves: \( r = f(\theta) \) and \( r = 2f(\theta) \). The problem asks whether we can infer anything about the relative lengths of these curves over the interval \( \alpha \leq \theta \leq \beta \).

Step 2: Formula for Arc Length of Polar Curves

The length of a polar curve \( r = g(\theta) \) over \( \alpha \leq \theta \leq \beta \) is given by the integral \[ L = \int_{\alpha}^{\beta} \sqrt{\left( \frac{dr}{d\theta} \right)^2 + r^2} \, d\theta \].

Step 3: Set Up Integrals for Both Curves

For the first curve \( r = f(\theta) \), the arc length \( L_1 \) is \[ L_1 = \int_{\alpha}^{\beta} \sqrt{\left( \frac{d}{d\theta}(f(\theta)) \right)^2 + f(\theta)^2} \, d\theta \]. Similarly, for the second curve \( r = 2f(\theta) \), the arc length \( L_2 \) is \[ L_2 = \int_{\alpha}^{\beta} \sqrt{\left( \frac{d}{d\theta}(2f(\theta)) \right)^2 + (2f(\theta))^2} \, d\theta \].

Step 4: Differentiate the Curves

For \( r = f(\theta) \), we have \( \frac{dr}{d\theta} = f'(\theta) \). For \( r = 2f(\theta) \), we have \( \frac{dr}{d\theta} = 2f'(\theta) \).

Step 5: Simplify Integrals

Substitute \( \frac{dr}{d\theta} \) into each integral: - For \( L_1: \) \[ L_1 = \int_{\alpha}^{\beta} \sqrt{f'(\theta)^2 + f(\theta)^2} \, d\theta \] - For \( L_2: \) \[ L_2 = \int_{\alpha}^{\beta} \sqrt{(2f'(\theta))^2 + (2f(\theta))^2} \, d\theta \] \[ = \int_{\alpha}^{\beta} \sqrt{4f'(\theta)^2 + 4f(\theta)^2} \, d\theta \] \[ = \int_{\alpha}^{\beta} 2\sqrt{f'(\theta)^2 + f(\theta)^2} \, d\theta = 2L_1 \].

Step 6: Conclusion

From the calculations, we can see that \( L_2 = 2L_1 \), meaning the length of the second curve is twice that of the first curve over the interval \( \alpha \leq \theta \leq \beta \).

Key Concepts

Polar Coordinates

Polar coordinates offer a unique way to describe a point in a plane, using a distance from a reference point and an angle from a reference direction. This system differs from the Cartesian coordinates, which are based on x and y axes to pinpoint a location. In polar coordinates, a point is defined as \((r, \theta)\), where:

• \(r\) is the radial distance from the origin, which is the point \( (0,0) \) in Cartesian coordinates.
• \(\theta\) is the angle measured in radians from the positive x-axis.

Polar coordinates are especially useful in problems involving circular or rotational motion, as they simplify the mathematics by focusing on radial distances and angular positions.
In our exercise, two curves are plotted using polar coordinates: \(r = f(\theta)\) and \(r = 2f(\theta)\). These represent how the radial distance from the origin changes with the angle \(\theta\), allowing for a neat visualization of complex curves or shapes on the plane.

Integration

Integration is a fundamental concept in calculus, serving as the mechanism to accumulate quantities. In the context of polar curves, we use integration to determine the arc length of a curve defined by \(r = f(\theta)\). The formula for arc length in polar coordinates requires the integration of a square root expression, specifically:
\[ L = \int_{\alpha}^{\beta} \sqrt{\left( \frac{dr}{d\theta} \right)^2 + r^2} \, d\theta \]
where \(\alpha\) and \(\beta\) are the bounds of angle \(\theta\). This integral sums up the infinitesimal lengths of segments along the curve between these angular limits.
For instance, the arc length \(L_1\) of the curve \(r = f(\theta)\) was calculated using:
\[ L_1 = \int_{\alpha}^{\beta} \sqrt{f'(\theta)^2 + f(\theta)^2} \, d\theta \]
Integration allows us to handle such calculations efficiently, proving particularly important when variables change continuously, which is the essence of the curves defined in polar forms.

Differentiation

Differentiation is another cornerstone of calculus, crucial for understanding how functions change. When working with polar coordinates, differentiation helps in determining how the radial distance \(r\) varies with respect to the angle \(\theta\). This becomes essential when evaluating the derivative \( \frac{dr}{d\theta} \).
The exercise examines two polar curves: \( r = f(\theta) \) and \( r = 2f(\theta) \). Differentiating these functions yields:

• For \( r = f(\theta) \), the derivative is \( \frac{dr}{d\theta} = f'(\theta) \).
• For \( r = 2f(\theta) \), the derivative is \( \frac{dr}{d\theta} = 2f'(\theta) \).

These derivatives measure how sharply the radial distance from the origin changes with the angle. In the context of arc length calculation, they play a pivotal role in forming the integrand, highlighting how rates of change govern the behavior and properties of curves within polar coordinates.