Problem 20 - Thomas Calculus

Question

Find the lengths of the curves. The spiral \(r=e^{\theta} / \sqrt{2}, \quad 0 \leq \theta \leq \pi\)

Step-by-Step Solution

Answer

The length of the curve is \( e^{\pi} - 1 \).

Step 1: Recall the formula for the length of a polar curve

The length of a polar curve \( r = f(\theta) \) from \( \theta = a \) to \( \theta = b \) can be calculated using the formula: \[ L = \int_{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \] In our case, \( a = 0 \), \( b = \pi \), and \( r = \frac{e^{\theta}}{\sqrt{2}} \).

Step 2: Compute the derivative \( \frac{dr}{d\theta} \)

Given \( r = \frac{e^{\theta}}{\sqrt{2}} \), compute the derivative \( \frac{dr}{d\theta} \): \[ \frac{dr}{d\theta} = \frac{1}{\sqrt{2}} \frac{d}{d\theta}(e^{\theta}) = \frac{e^{\theta}}{\sqrt{2}} \] As \( \frac{d}{d\theta}(e^{\theta}) = e^{\theta} \).

Step 3: Substitute the expressions into the arc length formula

We substitute \( \frac{dr}{d\theta} = \frac{e^{\theta}}{\sqrt{2}} \) and \( r = \frac{e^{\theta}}{\sqrt{2}} \) into the arc length formula: \[ L = \int_{0}^{\pi} \sqrt{ \left(\frac{e^{\theta}}{\sqrt{2}}\right)^2 + \left(\frac{e^{\theta}}{\sqrt{2}}\right)^2 } \, d\theta \] This simplifies to: \[ L = \int_{0}^{\pi} \sqrt{ 2 \left(\frac{e^{\theta}}{\sqrt{2}}\right)^2 } \, d\theta \] which further simplifies to: \[ L = \int_{0}^{\pi} \sqrt{ \frac{e^{2\theta}}{2} + \frac{e^{2\theta}}{2} } \, d\theta \]

Step 4: Simplify within the integral

Combine the terms under the square root: \[ L = \int_{0}^{\pi} \sqrt{ \frac{2e^{2\theta}}{2} } \, d\theta = \int_{0}^{\pi} \sqrt{ e^{2\theta} } \, d\theta = \int_{0}^{\pi} e^{\theta} \, d\theta \]

Step 5: Evaluate the integral

Now integrate: \[ L = \int_{0}^{\pi} e^{\theta} \, d\theta \] This evaluates to: \[ L = \left[ e^{\theta} \right]_{0}^{\pi} = e^{\pi} - e^{0} = e^{\pi} - 1 \] The length of the curve is \( e^{\pi} - 1 \).

Key Concepts

Polar Curves

Polar curves are a way to represent mathematical curves on a plane using polar coordinates. In polar coordinates, each point is defined by an angle and a radius, denoted as \( r \) and \( \theta \), respectively.
This means instead of using rectangular coordinates \((x, y)\), we use \((r, \theta)\) such that \(x = r \cos \theta\) and \(y = r \sin \theta\).

When dealing with polar curves, it's important to become familiar with the concept of \(r\) being a function that depends on \(\theta\).
For example, in the given exercise, the spiral curve is defined by \( r = \frac{e^{\theta}}{\sqrt{2}} \). This equation means that as \(\theta\) changes, \(r\) changes according to an exponential function.

Polar curves can illustrate shapes like spirals and circles that are difficult to express in Cartesian coordinates. They offer a very natural and insightful way to study curves in calculus, especially when dealing with problems involving rotating movements or periodic patterns. When visualizing them, think of the way fireworks expand outwards from a central point.

Arc Length

Arc length is a measure of the distance along a curve. For a polar curve, the process to find this involves integrating a specific formula over a defined interval of \(\theta\).

In our given task, we used the formula for the arc length of a polar curve:
\[ L = \int_{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \]
where \(a\) and \(b\) are the boundaries of \(\theta\) over which you want to calculate the arc length.
In this case, \(a = 0\) and \(b = \pi\) for the curve described.

This formula accounts for changes in both the radius \(r\) and the angle \(\theta\), offering a comprehensive way to track movement along a polar curve.
Through integration, it sums up infinitesimally small "pieces" of the curve's distance altogether, giving us the total length. Think of it like measuring the length of a path with a very tiny ruler, where the summation of all those tiny measurements gives you the complete arc length.

Integral Calculus

Integral calculus is the branch of mathematics that deals with accumulation of quantities and the areas under and between curves.
In integral calculus, we primarily focus on integrals, which helps us find areas, volumes, central points, and many useful things, particularly in physics and engineering.

In the case of curve length calculation, integral calculus allows us to sum an infinite number of infinitely small lines to find the whole length. This can be seen in our example by integrating the function \(e^{\theta}\) from \(0\) to \(\pi\).
The computation involves evaluating an integral of the form \( \int_{0}^{\pi} e^{\theta} \, d\theta \).

Knowing how to work with and solve integrals is essential for successful calculus problem-solving. The fundamental theorem of calculus connects differentiation with integration, providing powerful tools for evaluating integrals that represent real-world phenomena. Whether you are working with polar curves or any other kind of mathematical function, integral calculus is indispensable in understanding and quantifying change.