Problem 57 - Essential Calculus

Question

The curve represented by the equation \(r=a \theta,\) where \(a\) is a constant, is called the spiral of Archimedes. (a) Use a graphing utility to graph \(r=\theta,\) where \(\theta \geq 0\). What happens to the graph of \(r=a \theta\) as \(a\) increases? What happens if \(\theta \leq 0 ?\) (b) Determine the points on the spiral \(r=a \theta(a>0, \theta \geq 0)\) where the curve crosses the polar axis. (c) Find the length of \(r=\theta\) over the interval \(0 \leq \theta \leq 2 \pi\). (d) Find the area under the curve \(r=\theta\) for \(0 \leq \theta \leq 2 \pi\).

Step-by-Step Solution

Answer

(a) The spiral gets wider as 'a' increases, for \(\theta \leq 0\) the graph is a reflection in the polar axis. (b) The curve crosses the polar axis at the origin. (c) Length = \(5\pi\). (d) Area = \(2\pi^3\).

Step 1: Graph Interpretation for varying 'a' and \(\theta < 0\)

A graphing tool or software is used to plot \(r=\theta\). It is observed that as the value of \(a\) increases, the spiral gets wider - meaning it unfurls at a quicker rate. If \(\theta \leq 0\), the graph is a mirror image of \(r=a \theta\), (\(\theta \geq 0\)), reflected about the polar axis.

Step 2: Identifying crossing points on the polar axis

This curve crosses the polar or x-axis when \(r\) equals 0. Solving \(0 = a \theta\), since \(a > 0\) it implies that \(\theta = 0\). Therefore, the curve crosses the polar axis at the origin.

Step 3: Arc Length Calculation

The length of a curve given by \(r=f(\theta)\) on the interval \(\alpha \leq \theta \leq \beta\) is given by the integral \(\int_{\alpha}^\beta \sqrt {r^2 + (dr/d\theta)^2} d\theta\). In this case, we need to calculate the arc length of \(r=\theta\) for \(\theta\) ranging from 0 to \(2\pi\). After calculating, we find that the length is \(5\pi\).

Step 4: Area Calculation

The area under a curve in polar coordinates given by \(r=f(\theta)\) for \(\alpha \leq \theta \leq \beta\) is \(A = 1/2 \int_{\alpha}^\beta (f(\theta))^2 d\theta\). When we apply this, the area under the curve \(r=\theta\) for \(0 \leq \theta \leq 2\pi\), we get \(2\pi^3\).

Key Concepts

Polar Coordinates

Imagine you're at the center of a circular track, and you want to describe the location of a runner on that track. You could use the runner's distance from you and the direction they're facing to explain exactly where they are. That's exactly what polar coordinates do!

In mathematics, polar coordinates are a two-dimensional coordinate system where each point on a plane is determined by a distance from a reference point and an angle from a reference direction. The reference point is known as the pole, typically represented as the origin (0,0) in the Cartesian coordinate system, and the reference direction is usually the positive x-axis.

A point in polar coordinates is expressed as a pair \( (r, \theta) \), where \( r \) is the radius or the distance from the pole, and \( \theta \) is the angle measured in radians from the reference direction (counter-clockwise). For example, the polar coordinates \((5, \pi/4)\) tell us that the point is 5 units away from the pole at an angle of \(\frac{\pi}{4}\) radians (or 45 degrees).

Using this system, the shape known as the Spiral of Archimedes is described by the equation \( r = a \theta \), where \( a \) is a constant that affects how tightly or loosely the spiral curves as it moves away from the pole.

Arc Length of a Curve

Have you ever wondered how long a roller coaster track is or the distance of one lap around a curvy racetrack? What you're thinking about is the arc length of a curve.

The arc length is the measure of the distance along a curved line. In polar coordinates, this gets a bit trickier than just measuring a straight line. To find the arc length of a curve represented by a polar equation \( r = f(\theta) \) between two angles \( \alpha \) and \( \beta \), we use the integral formula:\[ L = \int_{\alpha}^{\beta} \sqrt{\left(r(\theta)\right)^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta \]

Let's break this down: We're integrating (adding up) the tiny bits of length along the curve, where each bit is a diagonal like the hypotenuse of a right-angle triangle. The terms under the square root represent the square of the vertical and horizontal components of that tiny length. So, we're taking those 'baby steps' along the curve from one angle to another, adding those bits all together until we've measured the entire curve.

In the spiral of Archimedes example, the arc length from \( \theta = 0 \) to \( \theta = 2\pi \) would be the total

Area in Polar Coordinates

If you're trying to figure out how much paint you need to cover a circular floor, you're thinking about the area. When it comes to curves described by polar coordinates, we can still calculate areas, but they're made up of many slices of a circular pie.

To find the area enclosed by a polar curve from \( r = f(\theta) \) between angles \( \alpha \) and \( \beta \) the formula we use is a bit more intuitive. Think about it like this: you're cutting up the whole area into tiny triangles with a very small base and height that reaches out to the curve, and then adding up all their areas to get the total area. This idea gives us the formula:\[ A = \frac{1}{2} \int_{\alpha}^{\beta} \left[f(\theta)\right]^2 d\theta \]

Here, \( \frac{1}{2} f(\theta)^2 \) essentially represents the area of each tiny triangle, and we integrate to sum all these up across our interval from \( \alpha \) to \( \beta \). It's as if you drew radii out from the pole to the curve making these slices (triangles), and then flattened them to paint that circular floor without missing a spot.

For our Spiral of Archimedes, the area under the curve \( r = \theta \) for the interval \( 0 \leq \theta \leq 2\pi \) was calculated using this formula, and it turned out to be \( 2\pi^3 \), that’s square units of whatever system you’re using (like square meters or square feet).