Problem 217 - Calculus Volume 3

Question

Find a definite integral that represents the arc length. \(r=e^{\theta}\) on the interval \(0 \leq \theta \leq 1\)

Step-by-Step Solution

Answer

The arc length is \(L = \sqrt{2} (e - 1)\).

Step 1: Formula for Arc Length in Polar Coordinates

The formula to find the arc length \(L\) of a curve described by \(r=f(\theta)\) from \(\theta=a\) to \(\theta=b\) in polar coordinates is given by \[L = \int_{a}^{b} \sqrt{\left(\frac{dr}{d\theta}\right)^2 + r^2} \, d\theta.\]

Step 2: Compute Derivative of the Polar Function

Given \(r = e^\theta\), we find the derivative \(\frac{dr}{d\theta}\). Derivative of \(e^\theta\) with respect to \(\theta\) is itself, so \(\frac{dr}{d\theta} = e^\theta.\)

Step 3: Substitute Values into Arc Length Formula

Substitute \(r = e^\theta\) and \(\frac{dr}{d\theta} = e^\theta\) into the arc length formula. This gives us:\[L = \int_{0}^{1} \sqrt{(e^\theta)^2 + (e^\theta)^2} \, d\theta.\]

Step 4: Simplify the Integrand

Simplify the expression under the square root:\[(e^\theta)^2 + (e^\theta)^2 = 2(e^\theta)^2.\]So, the integrand becomes:\[\sqrt{2(e^\theta)^2} = \sqrt{2} e^\theta.\]

Step 5: Evaluate the Definite Integral

Now, evaluate the integral:\[L = \sqrt{2} \int_{0}^{1} e^\theta \, d\theta.\]The integral of \(e^\theta\) is itself. Therefore, we find:\[L = \sqrt{2} [e^\theta]_0^1.\]

Step 6: Compute the Integral's Value

Evaluate \([e^\theta]_0^1\):\[[e^\theta]_0^1 = e^1 - e^0 = e - 1.\]

Step 7: Find the Arc Length

Substitute the evaluated bounds into the arc length expression:\[L = \sqrt{2} (e - 1).\]

Key Concepts

Polar Coordinates

In the world of mathematics, polar coordinates are often used to represent points on a plane. While Cartesian coordinates use a grid of x and y axes, polar coordinates use a different system. Here, each point is defined by a distance from the origin (often noted as "r") and an angle (noted as "θ") from a fixed direction, usually the positive x-axis. This concept is useful for dealing with circular and spiral shapes, where angles and radii provide a clearer understanding of position. Imagine a spiral going outward from a center point; describing this with polar coordinates simplifies calculations and graphing. In our exercise, the polar coordinate system helps represent the exponential spiral described by the equation \(r = e^\theta\) on the interval \(0 \leq \theta \leq 1\). Knowing polar coordinates is essential for solving various integrals involving spirals and circular shapes. Polar equations like \(r = f(\theta)\) make it easier to integrate values over given angular intervals by focusing on radius changes.

Definite Integral

The concept of a definite integral in calculus relates to calculating the accumulation of quantities, like areas under curves or in our case, lengths of arcs. When we mention a definite integral, it's key to remember we are working with specific boundaries. In our example, \(\int_{a}^{b}\) signifies that we're calculating the arc length from \(\theta = 0\) to \(\theta = 1\). This provides us a concrete result rather than a general formula, hence being termed as definite. With definite integrals, the limits of integration (here \(0\) and \(1\)) are critical, as they guide the region over which the function is evaluated. This encapsulates the idea of finite boundaries where the integral calculation starts and stops, providing a complete and specific measure of the spiral path's length over this range of angles.

Integral Calculus

Integral calculus is one of the main branches of calculus, alongside differential calculus. It's all about the idea of summing up small parts to form a whole, otherwise known as integration. In our specific task, integral calculus is utilized to find the total arc length of the curve \(r = e^\theta\), a procedure that involves integrating over given limits. Here, the process involves the function and its derivative. As seen in Steps 1 to 3 of our solution, integral calculus builds upon differentiating the function to set up the integral for its entire curve's length. Integration involves inputting calculated expressions and solving them within specific boundaries. It’s an essential tool for linking the concept of smooth continuous growth, summing up infinitely many infinitesimally small elements, and finding impactful macroscopic results like our arc length.

Arc Length Formula

The arc length formula offers a way to determine the length of a curve, especially in different coordinate systems such as polar coordinates. In polar coordinates, the arc length formula is given as: \[L = \int_{a}^{b} \sqrt{\left(\frac{dr}{d\theta}\right)^2 + r^2} \, d\theta.\] This formula is key to calculating the length of curves described by \(r=f(\theta)\). Understanding the formula requires knowing how to compute and substitute the derivative \(\frac{dr}{d\theta}\) and the radius \(r\) itself, as shown in our exercise. In our case:

• Substitute \(r = e^\theta\) and \(\frac{dr}{d\theta} = e^\theta\) into the formula.
• Simplify the integrand to \(\sqrt{2} e^\theta\).
• Evaluate the resulting definite integral.

Through simplification and evaluation, the arc length formula not only emphasizes the importance of integral calculus but also demonstrates how functions evolve over specified bounds.