Tangent Lines at the Pole

Why the Pole Requires a Directional Test

A polar curve reaches the pole whenever its radius is zero. At that point the polar angle does not specify a direction from the pole to the point itself: every angle describes the same origin. To find a tangent line, therefore, examine the direction from the pole to nearby points on the curve.

This method matters especially when the radius changes sign or when the usual formula for slope gives an indeterminate expression. A negative radius reverses the direction of the position vector, but it does not change the line through the pole that the vector determines.

Definition: For a polar curve \(r=r(\theta)\), a parameter value \(\theta_0\) reaches the pole if \(r(\theta_0)=0\). A tangent line at the pole associated with \(\theta_0\) is the limiting unoriented direction of the position vectors of nearby curve points, provided that non-pole points approach the pole through that parameter value.

Position Vectors and Pole Parameters

Using Cartesian coordinates, the position vector of a polar curve is

\boldsymbol{\gamma}(\theta) = \bigl(r(\theta)\cos\theta,\;r(\theta)\sin\theta\bigr) = r(\theta)(\cos\theta,\sin\theta).

The pole parameters are the solutions of \(r(\theta)=0\) in the parameter interval being considered. At any such value \(\theta_0\), the curve point is \((0,0)\). For nearby values with \(r(\theta)\ne 0\), the position vector points along the radial direction at angle \(\theta\), or in the opposite direction if \(r(\theta)<0\).

A tangent line is unoriented: a direction and its reversal describe the same line. Thus the relevant limiting direction is the line spanned by the position vector, not an oriented arrow that must point consistently one way. The line associated with \(\theta_0\), when the limit exists, is

L_{\theta_0} = \left\{ t(\cos\theta_0,\sin\theta_0):t\in\mathbb{R} \right\}.

This line passes through the pole. Its slope is \(\tan\theta_0\) when \(\cos\theta_0\ne 0\); if \(\cos\theta_0=0\), it is the vertical line \(x=0\). The line direction depends on \(\theta_0\) modulo \(\pi\): angles differing by \(\pi\) give the same line.

Why the Limiting Direction Gives the Tangent

Here is a direct justification using secants from the pole. Write \(\mathbf{e}_{\theta}=(\cos\theta,\sin\theta)\). For a nearby parameter value where \(r(\theta)\ne 0\), the unit position vector is

\frac{\boldsymbol{\gamma}(\theta)} {\|\boldsymbol{\gamma}(\theta)\|} = \frac{r(\theta)\mathbf{e}_{\theta}}{|r(\theta)|} = \operatorname{sgn}(r(\theta))\mathbf{e}_{\theta}.

The sign may change, so the oriented unit vectors may point in opposite directions on opposite sides of \(\theta_0\). But each vector spans the same line as \(\mathbf{e}_{\theta}\), because multiplying a nonzero vector by \(-1\) does not change its span. As \(\theta\to\theta_0\), the vector \(\mathbf{e}_{\theta}\) tends to \(\mathbf{e}_{\theta_0}\), since sine and cosine are continuous. Consequently, the secant lines tend to the line spanned by \(\mathbf{e}_{\theta_0}\). This proves the formula for \(L_{\theta_0}\).

The argument requires nearby parameter values that give points other than the pole. If the curve remains at the pole throughout a whole parameter interval, there are no nearby secants from that interval to use. Also, if several distinct parameter values reach the pole, apply the test at each one: different values can give different tangent lines.

Theorem: If \(r(\theta_0)=0\) and non-pole curve points approach the pole as \(\theta\to\theta_0\), the limiting tangent line associated with that parameter value is spanned by \((\cos\theta_0,\sin\theta_0)\). A change in the sign of \(r\) does not alter this unoriented line.

Worked Examples

Worked Example: Two Pole Parameters with the Same Tangent

Find the tangent lines at the pole for \(r=2\sin\theta\) at the pole parameters in \([0,2\pi)\).

First solve for the pole parameters:

2\sin\theta=0 \quad\Longrightarrow\quad \sin\theta=0 \quad\Longrightarrow\quad \theta=0,\ \pi \qquad (0\leq\theta<2\pi).

The Cartesian coordinates are \(x=2\sin\theta\cos\theta\) and \(y=2\sin^2\theta\). At nearby parameter values where \(x\ne 0\), their ratio is

\frac{y}{x} = \frac{2\sin^2\theta}{2\sin\theta\cos\theta} = \tan\theta.

At \(\theta_0=0\), \(\tan\theta\to 0\), so the limiting line has slope \(0\), namely \(y=0\). At \(\theta_0=\pi\), \(\tan\theta\to 0\) again, so the tangent is also \(y=0\). The line formula gives the same result: \(\theta_0=0\) spans \((1,0)\), while \(\theta_0=\pi\) spans \((-1,0)\); these vectors determine the same line. Substitution into the parameterization confirms that both parameter values give \((x,y)=(0,0)\).

Worked Example: A Radius That Vanishes to Higher Order

Find the tangent at the pole for \(r=1+\cos\theta\) when \(\theta=\pi\).

Since \(r(\pi)=1+\cos\pi=0\), this parameter reaches the pole. Put \(\delta=\theta-\pi\), so that \(\delta\to 0\) as \(\theta\to\pi\). Then

r=1+\cos(\pi+\delta)=1-\cos\delta.

Using \(\cos(\pi+\delta)=-\cos\delta\) and \(\sin(\pi+\delta)=-\sin\delta\), the coordinates near the pole are

x=-(1-\cos\delta)\cos\delta, \qquad y=-(1-\cos\delta)\sin\delta.

For sufficiently small nonzero \(\delta\), the common factor \(1-\cos\delta\) is nonzero and \(\cos\delta\ne 0\). Therefore

\frac{y}{x} = \frac{-(1-\cos\delta)\sin\delta} {-(1-\cos\delta)\cos\delta} = \tan\delta \longrightarrow 0.

The limiting tangent line is \(y=0\), the horizontal axis. As a check using the direction formula, \((\cos\pi,\sin\pi)=(-1,0)\), which spans exactly the horizontal axis. The fact that the radius is especially small near \(\pi\) does not change the limiting line.

Worked Example: Distinct Tangents from Distinct Pole Parameters

For \(r=\sin 2\theta\), find the tangent lines at the pole for \(0\leq\theta<\pi\).

The radius vanishes when

\sin 2\theta=0 \quad\Longrightarrow\quad 2\theta=k\pi \quad\Longrightarrow\quad \theta=\frac{k\pi}{2}.

In the specified interval, the pole parameters are \(0\) and \(\pi/2\). The Cartesian coordinates are \(x=(\sin 2\theta)\cos\theta\) and \(y=(\sin 2\theta)\sin\theta\). Near \(\theta=0\), canceling the common nonzero factor gives

\frac{y}{x} = \frac{(\sin 2\theta)\sin\theta} {(\sin 2\theta)\cos\theta} = \tan\theta \longrightarrow 0.

Thus the tangent at the pole associated with \(\theta=0\) is \(y=0\). Near \(\theta=\pi/2\), use the reciprocal ratio instead:

\frac{x}{y} = \frac{(\sin 2\theta)\cos\theta} {(\sin 2\theta)\sin\theta} = \cot\theta \longrightarrow 0.

Thus the tangent associated with \(\theta=\pi/2\) is \(x=0\). These lines are distinct, as also follows from the direction vectors \((1,0)\) and \((0,1)\). Substituting either parameter into \(r=\sin2\theta\) gives \(r=0\), confirming that both branches pass through the pole.

Connection with the Derivative

The limiting-position-vector method is useful whether or not the derivative vector immediately identifies the tangent. To see the relationship, differentiate the parameterization, using \(\mathbf{e}_{\theta}=(\cos\theta,\sin\theta)\) and \(\mathbf{e}'_{\theta}=(-\sin\theta,\cos\theta)\):

\boldsymbol{\gamma}'(\theta) = r'(\theta)\mathbf{e}_{\theta} + r(\theta)\mathbf{e}'_{\theta}.

At a pole parameter \(\theta_0\), the second term vanishes because \(r(\theta_0)=0\), leaving \(\boldsymbol{\gamma}'(\theta_0)=r'(\theta_0)\mathbf{e}_{\theta_0}\). If \(r'(\theta_0)\ne 0\), this derivative is nonzero and points along the same tangent line found from the limiting secants. If the derivative is zero, that fact alone does not determine whether a limiting tangent line exists; the secant-direction test directly examines the nearby curve points.

Common Pitfall: Treating an Indeterminate Slope as No Tangent

For \(r=1+\cos\theta\) at \(\theta=\pi\), the polar coordinate derivatives are \(x'=r'\cos\theta-r\sin\theta\) and \(y'=r'\sin\theta+r\cos\theta\). Since \(r(\pi)=0\) and \(r'(\pi)=-\sin\pi=0\), both derivatives vanish. Substituting into \(dy/dx=y'/x'\) produces

\left.\frac{dy}{dx}\right|_{\theta=\pi} = \frac{0}{0}.

The incorrect conclusion is that there is no tangent. The calculation only shows that this direct derivative quotient is indeterminate at that parameter. The nearby secant calculation in the worked example gives \(y/x=\tan(\theta-\pi)\to 0\), so the limiting tangent is the horizontal line \(y=0\).

Key takeaway: At the pole, solve \(r(\theta)=0\), then use the parameter angle to determine the limiting line. A zero radius, a sign change, or an indeterminate derivative quotient does not by itself rule out a tangent.

Using the Method Reliably

For each pole parameter, keep the calculation focused on that branch of the curve. Find all relevant zeros of the radius in the parameter interval, and then determine the line direction associated with each zero. If two angles differ by \(\pi\), they give the same line; if their difference is not a multiple of \(\pi\), they give different lines. This distinction is important when a polar curve passes through the pole more than once.

Find pole parameters.
Solve \(r(\theta)=0\) on the interval of interest.

Examine each parameter separately.
Nearby non-pole position vectors determine the branch direction at that visit to the pole.

Take the unoriented limit.
The tangent line is spanned by \((\cos\theta_0,\sin\theta_0)\); opposite vector signs describe the same line.

Compare the resulting lines.
Angles equal modulo \(\pi\) give the same tangent; other directions give distinct tangents.

The central advantage of this approach is that it uses the geometry of the curve near the pole rather than relying on a slope formula that may be undefined there. It also keeps the parameter value visible, so separate visits to the pole are not accidentally treated as one branch.

Check Your Understanding

For each question, identify the relevant pole parameter or parameters before deciding on a tangent line.

For \(r=3\cos\theta\), find the tangent line associated with \(\theta=\pi/2\).
For \(r=\sin\theta\), determine the tangent line associated with \(\theta=\pi\).
Why does a change in the sign of \(r\) not necessarily change the tangent line at the pole?
For \(r=\sin 2\theta\), why do the pole parameters \(0\) and \(\pi/2\) give different tangent lines?
What does a \(0/0\) result in the polar slope formula establish, and what does it fail to establish?

What You'll Learn