Problem 219 - Calculus Volume 3

Question

Find the length of the curve over the given interval. \(r=e^{3 \theta}\) on the interval \(0 \leq \theta \leq 2\)

Step-by-Step Solution

Answer

The length of the curve is \( \frac{\sqrt{10}}{3} \times (e^6 - 1) \).

Step 1: Formula for Arc Length in Polar Coordinates

To find the length of a curve given in polar coordinates, we use the formula \( L = \int_{\alpha}^{\beta} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \), where \( r = f(\theta) \) and \( \alpha \) to \( \beta \) is the interval over which we are integrating. Here, \( r = e^{3\theta} \) and our interval is \( 0 \leq \theta \leq 2 \).

Step 2: Differentiate r with Respect to θ

Find the derivative of \( r = e^{3\theta} \) with respect to \( \theta \). This is \( \frac{dr}{d\theta} = 3e^{3\theta} \), using the chain rule, because the derivative of \( e^{u} \) is \( e^{u} \cdot \frac{du}{d\theta} \).

Step 3: Substitute into the Arc Length Formula

Substitute \( r = e^{3\theta} \) and \( \frac{dr}{d\theta} = 3e^{3\theta} \) into the arc length formula. The integrand becomes \( \sqrt{ (3e^{3\theta})^2 + (e^{3\theta})^2 } \).

Step 4: Simplify the Integrand

Inside the square root, simplify \( (3e^{3\theta})^2 + (e^{3\theta})^2 = 9e^{6\theta} + e^{6\theta} = 10e^{6\theta} \). Thus, the integrand becomes \( \sqrt{10} e^{3\theta} \).

Step 5: Integrate Over the Interval

The integral becomes \( L = \sqrt{10} \int_{0}^{2} e^{3\theta} \, d\theta \). Integrate using the formula for the integral of an exponential function: \( \int e^{ax} \, dx = \frac{1}{a} e^{ax} + C \). Here, \( a = 3 \), so we get \( \frac{e^{3\theta}}{3} \) after integration.

Step 6: Evaluate the Definite Integral

Evaluate \( \sqrt{10} \left[ \frac{e^{3\theta}}{3} \right]_{0}^{2} \). This becomes \( \sqrt{10} \left[ \frac{e^{6}}{3} - \frac{e^{0}}{3} \right] \), which simplifies to \( \sqrt{10} \times \frac{e^{6} - 1}{3} \).

Key Concepts

Arc Length in Polar Coordinates

When we talk about calculating the arc length of a curve in polar coordinates, we're really delving into the beauty of integrating a unique form. Unlike Cartesian coordinates, polar coordinates define points using the radius and angle \( (r, \theta) \). To find the arc length \( L \) of a curve \( r = f(\theta) \), you use the formula:

• \( L = \int_{\alpha}^{\beta} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \)

Here, \( dr/d\theta \) is the derivative of the radius with respect to \( \theta \). This is logical since we’re tracing the path of the curve as \( \theta \) changes over our given interval \([\alpha, \beta] \). By applying this formula to the curve \( r = e^{3\theta} \) over the interval \( 0 \leq \theta \leq 2 \), we find the distance, or arc length, between the two angular positions.

Differentiation of Polar Functions

Differentiation in polar coordinates is a bit different from the Cartesian system. For our exercise, we needed to determine the rate at which the radius \( r \) changes with \( \theta \). We began with the function \( r = e^{3\theta} \). By applying the chain rule, which is vital in calculus, we find the derivative:

• \( \frac{dr}{d\theta} = 3e^{3\theta} \)

This derivative tells us how rapidly \( r \) is increasing as \( \theta \) moves from \( 0 \) to \( 2 \). The chain rule is key here, as it allows us to handle composite functions beautifully, with the function \( e^{3\theta} \) effectively acting as such due to its exponential nature.

Integral Calculus

Integral calculus is essential for finding the arc length in our example. After simplifying inside the radical, we substitute the differentiated function into our integral for arc length:

• \( L = \sqrt{10} \int_{0}^{2} e^{3\theta} \, d\theta \)

The process involves evaluating the exponentiated function \( e^{3\theta} \) across the interval from \( 0 \) to \( 2 \). Integration, especially with exponential functions, allows us to sum up infinitesimally small changes over this interval. Utilizing the fundamental theorem of calculus, we simplify our work:

• \( \frac{e^{3\theta}}{3} \bigg|_{0}^{2} \) gives us the evaluated form of the integral.

Thus, we compute how much the curve "grows" from the start to the end of the interval.

Exponential Functions

Exponential functions are a central topic in mathematics due to their unique property of constant relative growth. In our exercise, we examined the function \( r = e^{3\theta} \). Here, the base \( e \) is Euler's number, an irrational number approximately equal to 2.718. Functions of this form grow rapidly, especially with increasing powers, as displayed in \( e^{3\theta} \).

• They retain the form \( y = a^{x} \), where the derivative's form has a constant ratio to the original function.
• With differentiation, the rate of growth is again exponential, yet scaled by the inner derivative, which we saw as 3 in \( 3e^{3\theta} \).

Exponential functions, therefore, offer powerful models for continuous growth processes, a key concept in both math and nature.