Problem 47 - Calculus Early Transcendentals

Question

\(45-48\) Find the exact length of the polar curve. $$ r=\theta^{2}, \quad 0 \leqslant \theta \leqslant 2 \pi $$

Step-by-Step Solution

Answer

The length of the polar curve is \( \frac{1}{3} (8(\pi^2 + 1)^{3/2} - 8) \).

Step 1: Understand the Problem

We need to find the length of the polar curve given by the equation \( r = \theta^2 \) for \( 0 \leq \theta \leq 2\pi \). To do this, we'll use the formula for the length of a polar curve.

Step 2: Recall the Formula for Length

The length \( L \) of a polar curve given by \( r = f(\theta) \) from \( \theta = a \) to \( \theta = b \) is derived from: \[ L = \int_{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \] This covers the interval \( [a, b] \) where \( r \) is in terms of \( \theta \).

Step 3: Compute the Derivative \( \frac{dr}{d\theta} \)

Given \( r = \theta^2 \), compute \( \frac{dr}{d\theta} \):\[ \frac{dr}{d\theta} = 2\theta \]

Step 4: Substitute into Length Formula

Substitute \( r = \theta^2 \) and \( \frac{dr}{d\theta} = 2\theta \) into the length formula:\[ L = \int_{0}^{2\pi} \sqrt{(2\theta)^2 + (\theta^2)^2} \, d\theta \]Simplify under the square root:\[ L = \int_{0}^{2\pi} \sqrt{ 4\theta^2 + \theta^4 } \, d\theta \]

Step 5: Simplify the Integral Expression

Simplify the integral:\[ L = \int_{0}^{2\pi} \sqrt{ \theta^4 + 4\theta^2 } \, d\theta \]Factor \( \theta^2 \) from the expression under the square root:\[ L = \int_{0}^{2\pi} \sqrt{ \theta^2(\theta^2 + 4) } \, d\theta \]Now rewrite it as:\[ L = \int_{0}^{2\pi} \theta \sqrt{ \theta^2 + 4 } \, d\theta \]

Step 6: Solve the Integral

To solve \( \int_{0}^{2\pi} \theta \sqrt{ \theta^2 + 4 } \, d\theta \), use substitution:Let \( u = \theta^2 + 4 \), so \( du = 2\theta \, d\theta \).Therefore, \( \theta \, d\theta = \frac{1}{2} du \).Change limits:- When \( \theta = 0 \), \( u = 4 \).- When \( \theta = 2\pi \), \( u = (2\pi)^2 + 4 = 4\pi^2 + 4 \).This converts our integral into:\[ L = \frac{1}{2} \int_{4}^{4\pi^2 + 4} \sqrt{u} \, du \]

Step 7: Evaluate the Integral

Now solve the integral:\[ \frac{1}{2} \int \sqrt{u} \, du = \frac{1}{2} \cdot \frac{2}{3}u^{3/2} = \frac{1}{3} u^{3/2} \]Evaluate from 4 to \( 4\pi^2 + 4 \):\[ L = \frac{1}{3} \left( (4\pi^2 + 4)^{3/2} - 4^{3/2} \right) \]

Step 8: Final Calculation

Calculate:- \( 4\pi^2 + 4 \) evaluates to \( 4(\pi^2 + 1) \).- \( (4\pi^2 + 4)^{3/2} = (4(\pi^2 + 1))^{3/2} = 8(\pi^2 + 1)^{3/2} \).- \( 4^{3/2} = 8 \).Thus, the length of the curve is:\[ L = \frac{1}{3} (8(\pi^2 + 1)^{3/2} - 8) \]

Key Concepts

Calculus

Calculus is a branch of mathematics that studies how things change and how quantities are related to each other. It is essential for understanding motion, growth, and various physical phenomena. Calculus consists mainly of two parts:

• Differential Calculus: This part is concerned with the concept of derivatives, which represent rates of change. It's like looking at the steepness of a hill at any point.
• Integral Calculus: This part deals with integrals, which are essentially accumulations of quantities. Think of it as the total amount of something, like collecting raindrops over time.

These concepts help solve various real-world problems, ranging from physics and engineering to economics.

Polar Coordinates

Polar coordinates offer an alternative way to describe the position of a point on a plane. Instead of using traditional Cartesian coordinates (\(x, y\)), this system uses:

• Radius (\(r\)): The distance from a fixed point called the pole.
• Angle (\(\theta\)): The direction measured from a fixed line called the polar axis.

This is especially useful when dealing with curves and shapes that are naturally circular or spiral-like. For example, in our exercise, the curve is expressed by \(r = \theta^2\), which describes a spiral when plotted.

Integral Calculus

Integral calculus involves finding the accumulated value or the total sum of quantities over a range. It's often used to determine areas, volumes, and curve lengths. In our exercise, the task was to find the length of a polar curve, which requires setting up an integral based on the curve's equation. The general formula for the length \(L\) of a polar curve \(r = f(\theta)\) is:\[ L = \int{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \]This involves integrating over the angle \(\theta\) to accumulate the curve's length from one point to another, considering both the radial distance and the angular change.

Derivatives

Derivatives express how a function changes as its input changes. They are the backbone of differential calculus. In polar coordinates, the derivative \(\frac{dr}{d\theta}\) represents the rate at which the radius \(r\) changes with the angle \(\theta\). For our example curve \(r = \theta^2\), the derivative is calculated as:\[ \frac{dr}{d\theta} = 2\theta \]This helps find the necessary data to compute the integral for the curve's length. Understanding derivatives is crucial for solving problems involving change, like motion, slope, and curvature of graphs.