Chapter 8: Problem 23
Find the arc length of the curve \(r=e^{\theta}\) from \(\theta=\pi / 2\) to \(\theta=\pi\)
Short Answer
Expert verified
The arc length is \( \sqrt{2}(e^{\pi} - e^{\pi/2}) \).
Step by step solution
01
Understanding the Arc Length Formula for Polar Curves
The formula to find the arc length of a curve given in polar coordinates is \( L = \int_{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \), where \( r = f(\theta) \). For the given problem, \( r = e^{\theta} \). This formula will be used to calculate the arc length from \( \theta = \pi/2 \) to \( \theta = \pi \).
02
Calculate the Derivative of r with Respect to θ
Calculate \( \frac{dr}{d\theta} \). Since \( r = e^{\theta} \), differentiate with respect to \( \theta \) to find \( \frac{dr}{d\theta} = e^{\theta} \).
03
Set Up the Integral for Arc Length
Substitute \( r = e^{\theta} \) and \( \frac{dr}{d\theta} = e^{\theta} \) into the arc length formula. This gives: \[ L = \int_{\pi/2}^{\pi} \sqrt{(e^{\theta})^2 + (e^{\theta})^2} \, d\theta = \int_{\pi/2}^{\pi} \sqrt{2(e^{\theta})^2} \, d\theta = \int_{\pi/2}^{\pi} \sqrt{2} e^{\theta} \, d\theta \].
04
Solve the Integral
Factor out \( \sqrt{2} \) from the integral to simplify calculation: \[ L = \sqrt{2} \int_{\pi/2}^{\pi} e^{\theta} \, d\theta \]. Compute the integral: \[ \int e^{\theta} \, d\theta = e^{\theta} \]. Evaluate the definite integral from \( \pi/2 \) to \( \pi \): \[ L = \sqrt{2} \left[ e^{\theta} \right]_{\pi/2}^{\pi} = \sqrt{2}(e^{\pi} - e^{\pi/2}) \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Polar Coordinates
When studying curves, polar coordinates provide a unique way to express points on a plane using a radius and an angle. It differs from the Cartesian coordinate system we commonly use, which relies on x and y coordinates. In polar coordinates, each point is defined by:
- The radius (\(r\)): the distance from the origin (center) to the point.
- The angle (\(\theta\)): the direction from the origin to the point, measured in radians from the positive x-axis.
Derivative Calculation in Polar Coordinates
Calculating the derivative is a crucial step in many mathematical problems, including finding arc lengths. When a curve is given in polar form, like \(r = e^{\theta}\), we need to calculate the derivative of \(r\) with respect to \(\theta\) to apply it in the arc length formula.
The derivative \(\frac{dr}{d\theta}\) represents how the radius changes as the angle \(\theta\) changes. For \(r = e^{\theta}\), the derivative is quite simple:
The derivative \(\frac{dr}{d\theta}\) represents how the radius changes as the angle \(\theta\) changes. For \(r = e^{\theta}\), the derivative is quite simple:
- \(\frac{dr}{d\theta} = e^{\theta}\)
Integrating to Find the Definite Integral
In calculus, a definite integral calculates the area under a curve or other related measurements. For arc length determination of a polar curve, the integral formula often includes the square root of sums of squares of derivatives.
Using the formula \(L = \int_{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta\), we substitute the values obtained:
Using the formula \(L = \int_{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta\), we substitute the values obtained:
- \(\frac{dr}{d\theta} = e^{\theta}\)
- \(r = e^{\theta}\)
Understanding Exponential Functions in Polar Coordinates
Exponential functions are pervasive in mathematics due to their unique properties and growth patterns. In this context, \(r = e^{\theta}\) is an exponential function that generates a spiral as \(\theta\) increases. Let's explore some key characteristics:
- Exponential Growth: As the angle \(\theta\) increases, the radius \(r\) grows exponentially. This means that with each increase of \(\theta\) by 1, \(r\) multiplies by about 2.718 (Euler's number).
- Continuous and Smooth: The exponential function is continuous and infinitely differentiable, making it smooth and without abrupt changes, an ideal candidate for calculating accurate arc lengths.