Problem 29 - Thomas Calculus

Question

The length of the curve \(r=f(\theta), \alpha \leq \theta \leq \beta\) 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

Substituting derivatives and simplifying leads to matching the given curve length formula in terms of \(r\).

Step 1: Find Derivatives for x and y

To start, use the given substitutions \(x=f(\theta)\cos\theta\) and \(y=f(\theta)\sin\theta\). Find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\) using the product rule: \(\frac{dx}{d\theta} = \frac{d}{d\theta}(f(\theta) \cos \theta) = \frac{df}{d\theta}\cos\theta - f(\theta)\sin\theta\) and \(\frac{dy}{d\theta} = \frac{d}{d\theta}(f(\theta) \sin \theta) = \frac{df}{d\theta}\sin\theta + f(\theta)\cos\theta\).

Step 2: Substitute Derivatives into Integral

Substitute the derivatives from Step 1 into the expression for \(L\): \(L = \int_\alpha^\beta \sqrt{(\frac{df}{d\theta}\cos\theta - f(\theta)\sin\theta)^2 + (\frac{df}{d\theta}\sin\theta + f(\theta)\cos\theta)^2} d\theta\).

Step 3: Simplify the Expression

Expand the squared expressions inside the integral. After simplifying by using trigonometric identities like \(\sin^2\theta + \cos^2\theta = 1\), the expression reduces to \(\sqrt{(\frac{df}{d\theta})^2 + f(\theta)^2}\).

Step 4: Express in Terms of r

Substitute back \(r = f(\theta)\) into the simplified expression: \(L = \int_\alpha^\beta \sqrt{r^2 + (\frac{dr}{d\theta})^2}d\theta\), which matches the transformed formula for \(L\).

Key Concepts

Parametric Equations

Parametric equations are a powerful tool in calculus and geometry that allow us to describe a curve in terms of a parameter, often denoted as \( t \) or \( \theta \). Instead of expressing \( y \) directly as a function of \( x \), parametric equations define both \( x \) and \( y \) as functions of a third variable, the parameter. For example, you might have \( x = f(t) \) and \( y = g(t) \).These equations are especially useful for describing curves that are difficult or impossible to represent with a single function. For example:

• Circular and elliptical paths, such as orbits
• Complex geometric shapes like spirals
• Physical trajectories involving changing speeds and directions

Using parametric equations provides a more complete view of a curve, allowing us to analyze each component's behavior individually.
When dealing with parametric equations, it's important to understand how to find derivatives and calculate integrals based on these expressions. In cases like this, the product rule and chain rule often come into play for differentiation. Similarly, to find the length of a curve described parametrically, integrating the speed (or the magnitude of the velocity vector) over a specified interval is central.

Polar Coordinates

Polar coordinates provide an alternative way of representing points in a plane using a radius and an angle, rather than the traditional \( x \) and \( y \) Cartesian coordinates. In polar form, a point is written as \( (r, \theta) \), where \( r \) is the distance from the origin and \( \theta \) is the angle measured from the positive x-axis.This system is particularly handy for plotting curves that are more naturally expressed in terms of angles and distances, like spirals, circles, and rose curves. Converting between polar and Cartesian coordinates involves trigonometric transformations:

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

The polar coordinate system can make some types of analysis more straightforward, such as when integrating over circular regions or finding the length of curves defined by polar equations.
In calculus, polar coordinates are often used to simplify the integration process for curves defined by \( r = f(\theta) \). By treating \( \theta \) as the parameter, they allow for simpler differentiation and integration, especially when finding curve properties like area and arc length.

Curve Length

Finding the length of a curve is a common problem in calculus. For a curve defined by parametric equations \( x(t) \) and \( y(t) \), the formula for arc length \( L \) over an interval \( [a, b] \) is derived from the Pythagorean theorem and is given by:\[ L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]This integral measures the total distance traveled along the curve by combining the horizontal and vertical components of small segments into a single expression.When working with curves in polar coordinates, we adjust this formula to account for the radial nature of the system. For a polar curve \( r = f(\theta) \), the length is:\[ L = \int_\alpha^\beta \sqrt{r^2 + \left( \frac{dr}{d\theta} \right)^2} \, d\theta \]This result is derived by substituting the derivatives of the Cartesian coordinates into the general formula for curve length, then simplifying using trigonometric identities.
Understanding how to find the curve length is essential for solving a variety of practical problems, such as determining the material length needed to construct a physical representation of the curve or analyzing the motion along a path. By mastering these techniques, students can better appreciate the interconnectedness of geometry and calculus.