Question

Find the length of the entire spiral \(r=e^{-a \theta},\) for \(\theta \geq 0\) and \(a>0\).

Short answer

Answer: The length of the entire spiral is given by \(L = \frac{\sqrt{1 + a^2}}{a}\).

Step-by-step solution

Find the derivative of r with respect to θ

First, we need to find the derivative of \(r\) with respect to \(\theta\). Given the polar equation \(r = e^{-a \theta}\), we have: $$\frac{dr}{d\theta} = -ae^{-a \theta}$$

Substitute r and dr/dθ into the arc length formula

Now we will substitute \(r = e^{-a \theta}\) and \(\frac{dr}{d\theta} = -ae^{-a \theta}\) into the arc length formula: $$L = \int_0^\infty \sqrt{\left(e^{-a \theta}\right)^2 + \left(-ae^{-a \theta}\right)^2} d\theta$$

Simplify the integrand

Simplify the expression inside the square root: $$L = \int_0^\infty \sqrt{e^{-2a \theta} + a^2e^{-2a \theta}} d\theta$$ Factor out \(e^{-2a\theta}\): $$L = \int_0^\infty \sqrt{e^{-2a \theta}(1 + a^2)} d\theta$$ Take out the constant factor from the square root: $$L = \int_0^\infty \sqrt{1 + a^2} \cdot e^{-a \theta} d\theta$$

Calculate the integral

Now we calculate the integral: $$L = \sqrt{1 + a^2} \int_0^\infty e^{-a \theta} d\theta$$ This is an exponential integral, which can be solved using integration by part or by looking up in a table of integrals. The result is: $$L = \sqrt{1 + a^2} \left[-\frac{1}{a}e^{-a \theta}\right]_0^\infty$$

Evaluate the integral

Now we evaluate the integral at the limits: $$L = \sqrt{1 + a^2} \left[\left(-\frac{1}{a}e^{-a \infty}\right) - \left(-\frac{1}{a}e^{-a \cdot 0}\right)\right]$$ Simplifying the terms: $$L = \sqrt{1 + a^2} \left[0 - \left(-\frac{1}{a}\right)\right]$$ $$L = \frac{\sqrt{1 + a^2}}{a}$$

Present the final result

The length of the entire spiral for the given polar equation \(r = e^{-a \theta}\), where \(\theta \geq 0\) and \(a > 0\), is: $$L = \frac{\sqrt{1 + a^2}}{a}$$

Key concepts

Polar Coordinates

Polar coordinates are an alternative way to represent points on a plane, different from the familiar Cartesian (x, y) coordinates. Instead of using horizontal and vertical distances from a reference point, polar coordinates use:

• The distance from the origin, represented by \(r\).
• The angle \(\theta\) from the positive x-axis.

In the context of the spiral given by \(r = e^{-a\theta}\), \(r\) changes as \(\theta\) increases. This creates a path that winds around the origin, known as a spiral, that gets closer to the origin as \(\theta\) becomes larger. This type of representation is handy for describing curves and situations involving circular-like paths, like spirals and cycles.
Polar coordinates can be converted to Cartesian coordinates with equations:

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

Converting between these systems often helps simplify complex geometry and calculations.

Exponential Functions

Exponential functions are mathematical functions of the form \(f(x) = a^x\), where \(a\) is a constant and \(x\) is the variable. In our spiral equation, we have an exponential function \(r = e^{-a\theta}\), where \(e\) is the base of natural logarithms, approximately equal to 2.71828. The significance of exponential functions includes:

• Rapid growth (if the exponent is positive).
• Rapid decay (if the exponent is negative).

For \(r = e^{-a\theta}\), as \(\theta\) increases, the negative exponent causes the function to decay quickly towards zero. Exponential decay describes how the spiral gets progressively smaller and tighter as it winds toward the origin. Exponential functions play a crucial role in compound growth, radioactive decay, and learning curves, so they are essential to understand in various fields of science and engineering.

Integration Techniques

Integration is a fundamental concept in calculus used to find areas under curves and solve differential equations. When calculating arc lengths, like for the spiral \(r = e^{-a\theta}\), integration is crucial to determine the length along the curved path. The arc length formula in polar coordinates is:

• \(L = \int_a^b \sqrt{\left(\frac{dr}{d\theta}\right)^2 + r^2} \,d\theta\).

For spirals, substitution and simplification steps are used. First, the derivative \(\frac{dr}{d\theta}\) is determined and substituted into the formula. The complexity of these integrals often requires more advanced techniques like:

• Integration by parts.
• Simplifying the integrand using algebraic techniques.
• Using known integral tables for exponential and trigonometric functions.

Practicing these techniques efficiently solves complex calculus problems. Integration, especially with exponential decay, often involves setting limits at infinity, as seen in the arc length of the spiral, ensuring proper evaluation of the integral.