Question

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

Short answer

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

Step-by-step solution

Identify the polar function and its derivative

The polar function is given by \(r = e^{-a\theta}\). To find the arc length, we will need to find the derivative of the function with respect to \(\theta\). The derivative is: $$\frac{dr}{d\theta} = -ae^{-a\theta}$$

Set up the arc length formula

Now we will set up the arc length formula, which is given by \(L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}d\theta\). Our integral will be from \(\theta = 0\) to \(\theta = \infty\), since we want to find the length of the entire spiral: $$L = \int_{0}^{\infty} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}d\theta$$

Substitute the polar function and its derivative

Now we will substitute the polar function and its derivative into the integral: $$L = \int_{0}^{\infty} \sqrt{(e^{-a\theta})^2 + (-ae^{-a\theta})^2}d\theta$$

Simplify the integrand

We can simplify the integrand in the integral by combining the two terms inside the square root: $$L = \int_{0}^{\infty} \sqrt{e^{-2a\theta} + a^2e^{-2a\theta}}d\theta$$ Factor out \(e^{-2a\theta}\) from inside the square root: $$L = \int_{0}^{\infty} \sqrt{e^{-2a\theta}(1 + a^2)}d\theta$$ Take the constant (\(1+a^2\)) outside the square root since it doesn't depend on \(\theta\): $$L = \sqrt{1+a^2}\int_{0}^{\infty} e^{-a\theta}d\theta$$

Solve the integral

Since we now have a simple exponential function inside the integral, we can easily integrate it: $$L = \sqrt{1+a^2}\left[-\frac{1}{a}e^{-a\theta}\right]_{0}^{\infty}$$

Evaluate the integral

Now we will evaluate the integral at the limits of integration: $$L = \sqrt{1+a^2}\left(-\frac{1}{a}e^{-a\infty} - -\frac{1}{a}e^{-a(0)}\right)$$ Since the exponential function approaches 0 as its argument approaches infinity, \(e^{-a\infty} = 0\): $$L = \sqrt{1+a^2}\left(\frac{1}{a}\right)$$

Final Answer

The arc length of the entire spiral \(r = e^{-a\theta}\) is: $$ L = \frac{\sqrt{1+a^2}}{a} $$

Key concepts

Polar Coordinates

Polar coordinates are a unique way of describing points on a plane using a combination of distance and angle. Instead of using traditional Cartesian coordinates like \((x, y)\), polar coordinates express points as \((r, \theta)\), where:

• \(r\) is the radial distance from the origin to the point.
• \(\theta\) is the angular position relative to a reference direction (usually the positive x-axis).

In polar systems, curves can be more elegantly represented using polar equations. For example, a spiral can be defined using equations like \(r = e^{-a\theta}\). This specific type of curve expands or contracts as \(\theta\) changes, which is very different from the linear or parabolic shapes often seen in Cartesian coordinates. In polar coordinates, understanding these relationships is key for calculating properties like arc length.

Integration Techniques

Integration techniques involve finding the area under a curve or solving problems involving continuous sums. To calculate the arc length of a polar curve, we use a specific integration formula: \[ L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}d\theta \]This formula allows us to sum up small segments of the polar curve to find the total arc length.
Here's a step-by-step breakdown:- Identify the polar function and its derivative, \(\frac{dr}{d\theta}\).- Plug these values into the arc length formula.The integral bounds from \(\theta = 0\) to \(\theta = \infty\) describe the entire extent of the spiral, capturing its full length by considering each infinitesimal piece. Integration transforms this idea into a summation that evaluates to the length of the curve. This demonstrates how calculus can be applied to solve complex geometry problems involving curves.

Exponential Functions

Exponential functions are mathematical expressions where the variable is in the exponent. An example in the given problem is \(r = e^{-a\theta}\). These functions are known for their rapid changes:

• \(e^{-a\theta}\) decreases as \(\theta\) increases if \(a > 0\).
• They exhibit continuous growth or decay depending on the sign and value of the exponent.

In polar curves like spirals, this exponential nature controls how tightly the spiral wraps as \(\theta\) changes. Integrals involving exponential functions often simplify due to their predictable behavior. This makes them easier to handle compared to polynomials or other complex equations. Understanding these properties helps when evaluating integrals or solving for values over infinite intervals, as seen in the arc length of the spiral problem, where the function simplifies and integrates elegantly.