Question

Plot the spiral, $$r=\frac{1}{\theta}$$ Show thät it has an asymptote parallel to the prime vector. Suggestion. Consider the distance of a point \(P\) of the curve from the prime direction, and find the limit of this distance when \(\theta\) approaches 0 . Determine the angle at which the radius vector corresponding to \(\theta=\pi / 2\) meets this curve.

Short answer

Answer: The angle between the radius vector at \(\theta=\frac{\pi}{2}\) and the curve \(r=\frac{1}{\theta}\) is given by \(\phi = \arctan{(\frac{T(\frac{\pi}{2})}{r(\frac{\pi}{2})})}\), where \(T(\frac{\pi}{2}) = (1 + (\frac{-4}{\pi^{2}})^{2})^{\frac{1}{2}}\), and \(r(\frac{\pi}{2}) = \frac{2}{\pi}\).

Step-by-step solution

Plot the curve \(r=\frac{1}{\theta}\) in polar coordinates.

To plot the curve, take various angles \(\theta\), and calculate the corresponding values of \(r\). For example, when \(\theta = \pi, \,\frac{\pi}{2}, \, \frac{\pi}{4}, ...\), we can find the corresponding r values. Then, plot these in polar coordinates.

Consider the distance of a point P on the curve from the prime direction.

The polar axis can be thought of as a straight line going through the origin at an angle of 0. The polar coordinate equation of this line is \(r = a\sec\theta\), where \(a>0\) is a constant. To find the distance between any point \(P(r, \theta)\) in the curve \(r = \frac{1}{\theta}\) and the direction of the polar axis, we project P onto this line and find the length of the perpendicular segment.

Find the limit of the distance when \(\theta\) approaches 0.

Using the formula \(d = r \sin(\alpha-\theta)\) for the distance between a point \(P(r, \theta)\) and a line \(r=a\sec\theta\), where \(\alpha\) is the angle of this line relative to the polar axis, and calculating the limit as \(\theta\) approaches 0, $$\lim_{\theta \to 0} d = \lim_{\theta \to 0} \frac{\sin(\alpha-\theta)}{\theta}$$ If \(\alpha = 0\), i.e., the line is parallel to the prime vector, the formula simplifies to $$\lim_{\theta \to 0} \frac{1}{\theta} \sin(-\theta) = \lim_{\theta \to 0} \frac{-\sin(\theta)}{\theta}$$ Since the curve approaches the polar axis but does not cross it, and the distance approaches 0, the asymptote is parallel to the prime vector.

Determine the angle between the radius vector (for \(\theta=\frac{\pi}{2}\)) and the curve.

We can calculate the tangent vector to the curve at a given point to find the angle \(\phi\) between the radius vector (when \(\theta=\frac{\pi}{2}\) )of a curve and its tangent vector. We can write \(r\) as a function of \(\theta\), \(r(\theta) = \frac{1}{\theta}\). The derivative of \(r\) with respect to \(\theta\) is \(\frac{dr}{d\theta} = -\frac{1}{\theta^2}\). At \(\theta =\frac{\pi}{2}\) , \(\frac{dr}{d\theta} = \frac{-4}{\pi^2}\) . In polar coordinates, the tangent vector is given by \(T(\theta) = (1 + (\frac{dr}{d\theta})^2)^{1/2}\), therefore tangent vector at \(\theta = \frac{\pi}{2}\) is \(T(\frac{\pi}{2}) = (1 + (\frac{-4}{\pi^{2}})^{2})^{\frac{1}{2}}\). Thus the angle between the radius vector and the tangent is given by \(\phi = \arctan{(\frac{T(\frac{\pi}{2})}{r(\frac{\pi}{2})})}\) Finally, calculate \(\phi\) and find the angle between the radius vector and the curve, when \(\theta = \frac{\pi}{2}\).

Key concepts

Asymptote Analysis

When we talk about asymptotes in polar coordinates, we're looking for lines or curves that a given polar curve approaches as the angle \theta or radial distance r increases or decreases without bound. To find this for the curve \( r = \frac{1}{\theta} \), we consider the limit of the distance between a point on the curve and a line parallel to the prime vector, which is the polar axis, as \theta approaches zero.

The prime vector is treated as the baseline direction, often referred to as the pole, and in this case, where \( \alpha = 0 \) for the polar axis, we simplify the limit of the distance to \( \lim_{\theta \to 0} \frac{-\sin(\theta)}{\theta} \). As \theta gets closer to 0, the radial distance \( r \) grows without limit, but because the \( \sin(\theta) \) in the numerator also approaches 0, this shows that the distance to the line \( r=0 \), which is the polar axis, also gets closer to 0. This confirms that the polar axis is indeed an asymptote to the curve.

Polar Curves Plotting

Plotting polar curves requires a different approach than Cartesian curves since we pair every angle \theta with a radius r to describe locations in the plane. For the curve \( r=\frac{1}{\theta} \), we begin by choosing a series of values for \theta and computing the corresponding r values.

As \theta decreases, r increases, creating a spiral that moves outwards from the pole as \theta approaches 0. Points such as \( (\frac{1}{\pi}, \pi) \), \( (\frac{1}{\frac{\pi}{2}}, \frac{\pi}{2}) \), and \( (\frac{1}{\frac{\pi}{4}}, \frac{\pi}{4}) \) are plotted on polar graph paper or with computer software to visualize the spiral pattern this function depicts. It's essential to note how the curve behaves near the pole and the direction it moves as \theta changes.

Limits in Polar Coordinates

Studying limits in polar coordinates can be somewhat different from the Cartesian system. The limit process often involves considering how close the curve comes to the pole (the origin in polar coordinates) or an asymptote as \theta varies.

For the curve \( r = \frac{1}{\theta} \), we explore the limit as \theta goes to 0 to understand the curve's behavior near the pole. As previously mentioned, applying the limit to the distance function gives us a measure of how the curve behaves with respect to the polar axis. Since \( \lim_{\theta \to 0} \frac{-\sin(\theta)}{\theta} = -1 \) (applying L'Hôpital's Rule if necessary), we deduce that the radius grows indefinitely while the curve never quite touches the polar axis, strengthening our understanding of the curve's relationship with its asymptote.

Tangent to Polar Curve

In polar coordinates, the tangent to a curve at a point is not merely the derivative at that point as in Cartesian coordinates. Instead, we use the slope of the tangent line in terms of \( \frac{dr}{d\theta} \), which describes how quickly the radius changes as \theta changes.

For our spiral, we found \( \frac{dr}{d\theta} = -\frac{1}{\theta^2} \). At \( \theta = \frac{\pi}{2} \), this gives us \( \frac{dr}{d\theta} = \frac{-4}{\pi^2} \). We consider the length of the tangent vector \( T(\theta) \) and use it to find the angle \( \phi \) between the tangent and the radius vector. The resultant \( \phi = \arctan{(\frac{T(\frac{\pi}{2})}{r(\frac{\pi}{2})})} \) gives the precise angle at which the radius vector and the tangent to the curve intersect when \( \theta = \frac{\pi}{2} \), thus characterizing the local behavior of the curve at this particular point.