Question

Find the length of the logarithmic spiral \(r=e^{\theta / 2}\) from \(\theta=0\) to \(\theta=2 \pi\)

Short answer

The length of the logarithmic spiral is \( \sqrt{\frac{5}} \times (e^{\pi} - 1) \).

Step-by-step solution

Understanding the Logarithmic Spiral

A logarithmic spiral is defined by the polar equation \( r = e^{\theta/2} \). In this equation, \( r \) is the distance from the origin to a point on the spiral at a given angle \( \theta \) in radians.

Length Formula for Polar Curves

The formula to calculate the arc length of a curve given in polar coordinates \( r=f(\theta) \) between two angles \( \theta=a \) and \( \theta=b \) is: \[ L = \int_{a}^{b} \sqrt{\left( \frac{dr}{d\theta} \right)^2 + r^2}\, d\theta \] Here, \( f(\theta) = e^{\theta/2} \), so we need to find the derivative \( \frac{dr}{d\theta} \).

Compute the Derivative \( \frac{dr}{d\theta} \)

Differentiate the function \( r = e^{\theta/2} \) with respect to \( \theta \):\[ \frac{dr}{d\theta} = \frac{1}{2}e^{\theta/2} \]We use the chain rule here, identifying the inner function as \( u = \theta/2 \), so the derivative \( \frac{du}{d\theta} = 1/2 \).

Substitute into Length Formula

Substitute \( r = e^{\theta/2} \) and \( \frac{dr}{d\theta} = \frac{1}{2} e^{\theta/2} \) into the length formula:\[ L = \int_{0}^{2\pi} \sqrt{\left(\frac{1}{2} e^{\theta/2}\right)^2 + (e^{\theta/2})^2}\, d\theta \] This simplifies to:\[ L = \int_{0}^{2\pi} \sqrt{\frac{1}{4}e^{\theta} + e^{\theta}}\, d\theta \] \[ = \int_{0}^{2\pi} \sqrt{\frac{5}{4}e^{\theta}}\, d\theta \] \[ = \sqrt{\frac{5}{4}} \int_{0}^{2\pi} e^{\theta/2}\, d\theta \]

Solve the Integral

Evaluate the integral:\[ \int_{0}^{2\pi} e^{\theta/2}\, d\theta \] Use substitution: let \( u = \theta/2 \) so \( d\theta = 2\,du \) and changes limits of integration from \( \theta=0 \) to \( u = 0 \) and \( \theta=2\pi \) to \( u=\pi \):\[ = 2 \int_{0}^{\pi} e^{u}\, du \] \[ = 2 \left[ e^{u}\right]_{0}^{\pi} \] \[ = 2(e^{\pi} - e^{0}) \] \[ = 2(e^{\pi} - 1) \]

Final Calculation

Substitute the evaluated integral result back: \[ L = \sqrt{\frac{5}{4}} \times 2(e^{\pi} - 1) \] \[ = \sqrt{\frac{5}{4}} \times 2e^{\pi} - \sqrt{\frac{5}{4}} \times 2 \] \[ = \sqrt{\frac{5}} \times (e^{\pi} - 1) \]

Conclusion: Length of the Spiral

The total length of the spiral from \( \theta = 0 \) to \( \theta = 2\pi \) is \( \sqrt{\frac{5}} \times (e^{\pi} - 1) \).

Key concepts

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system that represents points on a plane based on a distance and an angle. Unlike the standard Cartesian coordinate system which uses an (x, y) format, polar coordinates use \( (r, \theta) \). Here, \( r \) is the radial distance from the origin to the point, and \( \theta \) is the angle from the positive x-axis in radians.

Understanding polar coordinates is essential when dealing with curves like spirals, circles, and other shapes that are naturally defined by rotation and radial distance rather than horizontal and vertical alignment. In the case of the logarithmic spiral, our formula is \( r = e^{\theta/2} \). This means that for any angle \( \theta \), the distance from the origin increases exponentially.

• \( r \) increases as \( \theta \) increases, showing the nature of the spiral as it "unwinds".
• \( \theta \) represents the angle that aligns with the typical counterclockwise increase from the positive x-axis.

Such descriptions help visualize how the spiral grows as \( \theta \) changes. It also highlights the dynamic nature of polar coordinates in mapping complex paths and curves.

Arc Length

Arc length, in polar coordinates, refers to the distance along a curve from one point to another. For a polar curve defined by \( r = f(\theta) \), the arc length \( L \) between two angles \( \theta = a \) and \( \theta = b \) is given by the formula:
\[ L = \int_{a}^{b} \sqrt{\left( \frac{dr}{d\theta} \right)^2 + r^2} \, d\theta \]

In this exercise, we use the given expression \( r = e^{\theta/2} \) to calculate the arc length of the logarithmic spiral. The integral calculates the sum of tiny distances along the curve, ensuring that even the smallest twist or turn of the spiral is accounted for.

• This approach carefully navigates the radial growth and the angular motion.
• It is particularly useful in finding lengths of curves that are not straight or simple geometric figures.

The combination of differentiation and integration is powerful in resolving such distances in polar coordinates. These calculations rely on both the changing radius as you move around the angle and how steeply that radius changes with \( \theta \).

Differentiation

Differentiation in calculus is the process of finding the rate at which a function is changing at any given point. For the logarithmic spiral \( r = e^{\theta/2} \), differentiation helps determine how rapidly \( r \) changes as \( \theta \) changes.

By computing \( \frac{dr}{d\theta} \), we identify this rate of change:
\[ \frac{dr}{d\theta} = \frac{1}{2}e^{\theta/2} \]

This derivative shows the slope of the tangent to the curve at any point \( \theta \). This is extremely useful:

• It gives insight into the spiral's growth rate.
• Helps understand the nature of the curve's expansion.

When substituting into the arc length formula, this differential term combines with the radial distance \( r \) to account for both linear and radial curvatures, securing an accurate arc length computation. Differentiation here ensures we are considering the precise dynamics at every infinitesimal segment of the spiral.

Integral Calculus

Integral calculus plays a key role in solving the length of the logarithmic spiral. Through integration, we calculate the total arc length by summing up infinitesimally small segments of the curve.

The integral required for solving the spiral problem is:
\[ \int_{0}^{2\pi} e^{\theta/2} \, d\theta \]

This involves substitution, a method used to simplify the complex expressions before integrating:
\[ u = \frac{\theta}{2}, \quad d\theta = 2 \, du \]
The integral becomes easier to manage:

• It translates the problem to a more straightforward form.
• Allows us to substitute and solve efficiently.

The outcome provides the cumulative length from starting to ending angle in its simplest form. Integral calculus, therefore, assembles these tiny lengths into a comprehensible whole, showcasing its indispensable utility in measuring complex geometric figures like spirals.