Problem 33 - Thomas Calculus

Question

The length of the curve \(r=f(\boldsymbol{\theta}), \boldsymbol{\alpha} \leq \boldsymbol{\theta} \leq \boldsymbol{\beta}\) Assuming that the necessary derivatives are continuous, show how the substitutions $$x=f(\theta) \cos \theta, \quad y=f(\theta) \sin \theta$$ (Equations 2 in the text) transform $$L=\int_{\alpha}^{\beta} \sqrt{\left(\frac{d x}{d \theta}\right)^{2}+\left(\frac{d y}{d \theta}\right)^{2}} d \theta$$ into $$L=\int_{\alpha}^{\beta} \sqrt{r^{2}+\left(\frac{d r}{d \theta}\right)^{2}} d \theta$$

Step-by-Step Solution

Answer

We substitute derivatives and simplify to transform the integral as given.

Step 1: Differentiate x with respect to θ

Given \( x = f(\theta) \cos \theta \), differentiate with respect to \( \theta \). Use the product rule: \( \frac{d}{d\theta} [u \cdot v] = u'v + uv' \). Let \( u = f(\theta) \) and \( v = \cos \theta \). Hence, \[ \frac{d x}{d \theta} = f'(\theta) \cos \theta - f(\theta) \sin \theta. \]

Step 2: Differentiate y with respect to θ

Given \( y = f(\theta) \sin \theta \), differentiate with respect to \( \theta \) using the product rule. Let \( u = f(\theta) \) and \( v = \sin \theta \). Thus, \[ \frac{d y}{d \theta} = f'(\theta) \sin \theta + f(\theta) \cos \theta. \]

Step 3: Substitute derivatives into the length formula

Substitute \( \frac{d x}{d \theta} \) and \( \frac{d y}{d \theta} \) into the formula for \( L \): \[ L = \int_{\alpha}^{\beta} \sqrt{\left( f'(\theta) \cos \theta - f(\theta) \sin \theta \right)^2 + \left( f'(\theta) \sin \theta + f(\theta) \cos \theta \right)^2} \, d\theta. \]

Step 4: Simplify the expression

Simplify the expression inside the square root: \( \left( f'(\theta) \cos \theta - f(\theta) \sin \theta \right)^2 + \left( f'(\theta) \sin \theta + f(\theta) \cos \theta \right)^2 \). This becomes: \[ (f'(\theta))^2 (\cos^2 \theta + \sin^2 \theta) + f(\theta)^2 (\cos^2 \theta + \sin^2 \theta) = (f'(\theta))^2 + f(\theta)^2. \] Use the identity \( \cos^2 \theta + \sin^2 \theta = 1 \).

Step 5: Final Length Expression

Recognize that \( f(\theta) = r \) and \( f'(\theta) = \frac{dr}{d\theta} \). Thus, the expression simplifies to \( \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \). Hence, \[ L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta. \] This completes the transformation.

Key Concepts

Polar Coordinates

Polar coordinates provide a unique way of describing a point in a plane using a distance and an angle. Unlike Cartesian coordinates that rely on horizontal and vertical distances (x and y), polar coordinates use the radius, denoted by \( r \), and the angle, denoted by \( \theta \). This is particularly useful for curves like spirals and circles, which are naturally oriented around a central point.

In polar coordinates, a point is represented as \((r, \theta)\). The radius \( r \) is the distance from the origin to the point, and \( \theta \) is the angle between the positive x-axis and the line connecting the origin to the point. Polar coordinates are widely used in scenarios where symmetry and rotations are involved.

To convert polar coordinates to Cartesian coordinates (i.e., \( x \) and \( y \)), the transformations used are:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)

These transformations allow us to switch between the two coordinate systems as required by different contexts or problems.

Derivatives

In mathematics, a derivative represents the rate of change of a function with respect to a variable. When dealing with functions in polar coordinates, such as \( x = f(\theta) \cos \theta \) and \( y = f(\theta) \sin \theta \), we use derivatives to measure how these functions change as \( \theta \) changes.

For example, to find the derivative of \( x \) with respect to \( \theta \), we apply the product rule:

• \( \frac{d}{d\theta} [u \cdot v] = u'v + uv' \)

This rule is essential for differentiating products of functions, where one function is \( u = f(\theta) \), and the other is trigonometric.

The derivative calculations are core when transforming the length of the curve from polar back into Cartesian representation. Specifically, we need the derivatives \( \frac{d x}{d \theta} \) and \( \frac{d y}{d \theta} \) to be substituted into the integral formula for arc length.

Integration

Integration is used to calculate quantities like area, volume, and, in this context, the length of the curve. By integrating a function over an interval, we accumulate values to find a total quantity.

For the arc length of polar curves, the integral formula is given by:

• \( L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} d\theta \)

Here, \( \alpha \) and \( \beta \) are the limits of integration, representing the range of \( \theta \) over which we want to measure the curve.

Integration allows us to sum up very small segments of the curve to find the total length. Understanding the integration process involves recognizing how each infinitesimal element contributes to the whole, which is crucial when dealing with continuously changing curves.

Curve Transformation

Curve transformation refers to the process of changing a curve’s representation from one form to another, such as from Cartesian coordinates to polar coordinates, or vice versa. Transformations can greatly simplify problem-solving by utilizing the most convenient form.

In our context, we started with a curve expressed in Cartesian coordinates:\[(x, y) = (f(\theta) \cos \theta, f(\theta) \sin \theta).\]
Through differentiation and substitution, we were able to transform it into a polar form defined by \( r = f(\theta) \). The key transformation involved recognizing the length of the curve as an integral involving these new parameters.

This ability to transform encourages flexibility in mathematical problem-solving, allowing us to approach a problem from different angles depending on which representation simplifies our work. Each form has its own advantages; the choice depends on the symmetry and nature of the problem.