Problem 22 - Thomas Calculus

Question

Find the lengths of the curves in Exercises \(21-28 .\) The spiral \(r=e^{\theta} / \sqrt{2}, \quad 0 \leq \theta \leq \pi\)

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We need to find the arc length of the spiral given in polar coordinates as \( r = \frac{e^{\theta}}{\sqrt{2}} \) for \( 0 \leq \theta \leq \pi \). The arc length \( L \) of a polar curve from \( \theta = a \) to \( \theta = b \) is given by the integral formula:\[L = \int_a^b \sqrt{\left(\frac{dr}{d\theta}\right)^2 + r^2} \ d\theta\]

Step 2: Derive the Formula for Arc Length

First, we determine \( \frac{dr}{d\theta} \) by differentiating \( r = \frac{e^{\theta}}{\sqrt{2}} \) with respect to \( \theta \). This yields:\[\frac{dr}{d\theta} = \frac{e^{\theta}}{\sqrt{2}}\]The formula for arc length becomes:\[L = \int_0^\pi \sqrt{\left( \frac{e^{\theta}}{\sqrt{2}} \right)^2 + \left( \frac{e^{\theta}}{\sqrt{2}} \right)^2} \ d\theta\]

Step 3: Simplify the Expression Under the Integral

Simplify the expression under the square root:\[\sqrt{\left( \frac{e^{\theta}}{\sqrt{2}} \right)^2 + \left( \frac{e^{\theta}}{\sqrt{2}} \right)^2} = \sqrt{2 \cdot \left( \frac{e^{\theta}}{\sqrt{2}} \right)^2} = \sqrt{\frac{2 \cdot e^{2\theta}}{2}} = \sqrt{e^{2\theta}} = e^{\theta}\]Thus, the integral simplifies to:\[L = \int_0^\pi e^{\theta} \ d\theta\]

Step 4: Evaluate the Integral

To find the value of the integral, evaluate:\[\int_0^\pi e^{\theta} \ d\theta = \left[ e^{\theta} \right]_0^\pi = e^{\pi} - e^0 = e^{\pi} - 1\]

Step 5: Conclusion

The length of the curve of the spiral over the given interval is:\[\boxed{e^{\pi} - 1}\]

Key Concepts

Differentiation in Polar Coordinates

Differentiation in polar coordinates involves finding the derivative of a function that is expressed in terms of the polar coordinate system, rather than the more common Cartesian coordinates. In polar coordinates, a point is represented by \((r, \theta)\), where \(r\) is the radial distance from the origin and \(\theta\) is the angle from the positive x-axis. Understanding how to differentiate in this setting is crucial when dealing with curves defined in a polar format.

In this exercise, we begin with the function \( r = \frac{e^{\theta}}{\sqrt{2}} \). To find \( \frac{dr}{d\theta} \), the derivative of \( r \) with respect to \( \theta \), we apply the basic rules of differentiation as follows:

• Recognize \(r\) is a function of \(\theta\) involving an exponential component.
• Differentiate \( e^{\theta} \) to get \( \frac{e^{\theta}}{\sqrt{2}} \) since \( \frac{1}{\sqrt{2}}\) is a constant multiplier.

This gives us the same form for the derivative as the original function, \( \frac{dr}{d\theta} = \frac{e^{\theta}}{\sqrt{2}} \). This derivative is then used in the computation of the arc length of the curve.

Exponential Functions

Exponential functions are a fundamental component of mathematics and appear frequently in calculus and mathematical modeling. The general form of an exponential function is \( f(x) = a e^{bx} \), where \(e\) is the base of the natural logarithm, approximately 2.718.

The function \( r = \frac{e^{\theta}}{\sqrt{2}} \) is an exponential function of \( \theta \). Here we work with:

• Base \( e \), representing the continuous growth rate, crucial for various applications.
• The expression \( e^{\theta} \) meaning the output grows exponentially as \( \theta \) increases.

Exponential functions are characterized by their constant relative growth rates, meaning the rate of growth is proportional to the current value. This is evident under transformations such as differentiation or integration, maintaining their form which simplifies solving the arc length in polar coordinates.

Integral Calculus

Integral calculus is essential for computing areas, volumes, and lengths, particularly arc lengths of curves. To find the arc length of a curve in polar coordinates, integral calculus comes into play. The length of a polar curve from \( \theta = a \) to \( \theta = b \) can be found using the formula:
\[ L = \int_a^b \sqrt{\left(\frac{dr}{d\theta}\right)^2 + r^2} \ d\theta \]

With our function \( r = \frac{e^{\theta}}{\sqrt{2}} \), this becomes:

• First, substitute \( \frac{dr}{d\theta} = \frac{e^{\theta}}{\sqrt{2}} \) in the formula, resulting in the expression \( \sqrt{2 \cdot \left( \frac{e^{\theta}}{\sqrt{2}} \right)^2} = e^{\theta} \).
• This simplification allows us to directly integrate \( e^{\theta} \) from 0 to \( \pi \).

Evaluating this integral involves understanding how the antiderivative of \(e^{\theta}\), which is itself, applies within limits from 0 to \(\pi \). Thus, the computation
\( \int_0^\pi e^{\theta} \ d\theta = \left[ e^{\theta} \right]_0^\pi = e^{\pi} - 1 \),illustrates how integral calculus allows us to determine the total arc length of the curve.