Horizontal and Vertical Tangencies in Polar Form

Why tangencies require a derivative check

A polar curve is described by a radius \(r\) that depends on an angle \(\theta\). Its Cartesian coordinates are \(x=r\cos\theta\) and \(y=r\sin\theta\), so changing \(\theta\) traces a parametrized curve. To locate horizontal or vertical tangencies, we examine how each coordinate changes with the parameter.

A zero in one coordinate derivative gives a candidate, not automatically a tangency. We must also check that the other derivative is nonzero. If both derivatives vanish, the parameter test alone does not determine the tangent direction.

Key idea: At a parameter \(\theta_0\), a regular polar parametrization has a horizontal tangent when \(\frac{dy}{d\theta}=0\) and \(\frac{dx}{d\theta}\ne 0\). It has a vertical tangent when \(\frac{dx}{d\theta}=0\) and \(\frac{dy}{d\theta}\ne 0\).

Coordinate derivatives for a polar curve

For a polar equation \(r=f(\theta)\), its Cartesian parametrization is

x(\theta)=r(\theta)\cos\theta, \qquad y(\theta)=r(\theta)\sin\theta.

Applying the product rule to each coordinate gives

\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta-r\sin\theta, \qquad \frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta+r\cos\theta.

These are derivatives with respect to the parameter \(\theta\), not with respect to \(x\). Where \(\frac{dx}{d\theta}\ne 0\), the chain rule gives the curve's slope as \(\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}\). The tangency tests follow from this ratio, together with a check that it is valid.

Definition: A parameter \(\theta_0\) is a regular point of the parametrization if \(\bigl(\frac{dx}{d\theta},\frac{dy}{d\theta}\bigr)\ne(0,0)\) there. At such a point, a horizontal tangent requires \(\frac{dy}{d\theta}=0\) and \(\frac{dx}{d\theta}\ne0\); a vertical tangent requires \(\frac{dx}{d\theta}=0\) and \(\frac{dy}{d\theta}\ne0\).

Why the tests work

The velocity vector of the parametrized curve is \(\bigl(\frac{dx}{d\theta},\frac{dy}{d\theta}\bigr)\). At a regular point, this vector gives the direction of the tangent. A nonzero horizontal vector has zero vertical component; a nonzero vertical vector has zero horizontal component. This explains both the zero-derivative condition and the need for the other derivative to be nonzero.

The slope test gives the same result when \(\frac{dx}{d\theta}\ne0\). Specifically, if \(\frac{dy}{d\theta}=0\), then \(\frac{dy}{dx}=0\), so the tangent is horizontal. For a vertical tangent, \(\frac{dx}{d\theta}=0\) while \(\frac{dy}{d\theta}\ne0\); the tangent direction is vertical, although the slope ratio is not defined there.

Differentiate both coordinates.
Use the product rule on \(x=r\cos\theta\) and \(y=r\sin\theta\).

Solve for candidates.
Set \(\frac{dy}{d\theta}=0\) for horizontal candidates and \(\frac{dx}{d\theta}=0\) for vertical candidates.

Check the other derivative.
Keep a candidate as a horizontal or vertical tangency only when the other derivative is nonzero.

Report parameters in the stated interval.
Use the interval to avoid counting repeated parameter values as separate traces unless the task calls for them.

Worked examples

Worked Example: Tangencies for \(r=2+\cos\theta\)

Find all horizontal and vertical tangencies for \(0\le\theta<2\pi\).

First, \(r'= -\sin\theta\). The coordinate derivatives are

\begin{aligned} x'&=r'\cos\theta-r\sin\theta\\ &=-\sin\theta\cos\theta-(2+\cos\theta)\sin\theta\\ &=-2\sin\theta(1+\cos\theta),\\[4pt] y'&=r'\sin\theta+r\cos\theta\\ &=-\sin^2\theta+(2+\cos\theta)\cos\theta\\ &=2\cos^2\theta+2\cos\theta-1. \end{aligned}

For vertical candidates, solve \(x'=0\). The factors give \(\sin\theta=0\) or \(\cos\theta=-1\). The latter is already included among the solutions to \(\sin\theta=0\), so the candidates in the interval are \(\theta=0\) and \(\theta=\pi\). At \(\theta=0\), \((x',y')=(0,3)\). At \(\theta=\pi\), \((x',y')=(0,-1)\). Thus both are vertical tangencies.

For horizontal candidates, set \(y'=0\) and write \(c=\cos\theta\):

2c^2+2c-1=0, \qquad c=\frac{-1\pm\sqrt{3}}{2}.

The value \(\frac{-1-\sqrt{3}}{2}\) is less than \(-1\), so it cannot be a cosine. The other value \(a=\frac{-1+\sqrt{3}}{2}\) lies between \(-1\) and \(1\), giving two parameters in the interval, one with positive sine and one with negative sine. At either one, \(y'=0\), while \(x'=-2\sin\theta(1+a)\ne0\): here \(\sin\theta\ne0\) and \(1+a=\frac{1+\sqrt{3}}{2}\ne0\). Both are horizontal tangencies.

Check: substituting \(c=a\) into the horizontal derivative gives \(2a^2+2a-1=0\), exactly the equation used to find \(a\). The vertical candidates have the nonzero \(y'\) values computed above, so none has both derivatives zero.

Worked Example: Tangencies for \(r=2+\sin\theta\)

Find the horizontal and vertical tangencies for \(0\le\theta<2\pi\).

Since \(r'=\cos\theta\), differentiating gives

\begin{aligned} x'&=\cos^2\theta-(2+\sin\theta)\sin\theta =1-2\sin^2\theta-2\sin\theta,\\ y'&=\sin\theta\cos\theta+(2+\sin\theta)\cos\theta =2\cos\theta(1+\sin\theta). \end{aligned}

For horizontal tangencies, \(y'=0\) requires \(\cos\theta=0\) or \(\sin\theta=-1\). The latter is already one of the \(\cos\theta=0\) candidates. Thus the candidate parameters are \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\). At \(\frac{\pi}{2}\), \(\sin\theta=1\), so \(x'=1-2-2=-3\ne0\). At \(\frac{3\pi}{2}\), \(\sin\theta=-1\), so \(x'=1-2+2=1\ne0\). Both are horizontal tangencies.

For vertical tangencies, solve \(x'=0\). Set \(s=\sin\theta\):

1-2s^2-2s=0 \quad\Longleftrightarrow\quad 2s^2+2s-1=0 \quad\Longleftrightarrow\quad s=\frac{-1\pm\sqrt{3}}{2}.

The root \(\frac{-1-\sqrt{3}}{2}\) is less than \(-1\), so it is not a sine value. The root \(b=\frac{-1+\sqrt{3}}{2}\) lies in \((-1,1)\), giving two parameters in the interval. At either, \(x'=0\) by the equation solved, and \(y'=2\cos\theta(1+b)\ne0\), since \(\cos\theta\ne0\) and \(1+b=\frac{1+\sqrt{3}}{2}\ne0\). Both are vertical tangencies.

Check: the two horizontal parameters give \(x'=-3\) and \(x'=1\), respectively, confirming regular horizontal directions. At the vertical candidates, the defining equation makes \(x'=0\), and the stated nonzero factors ensure \(y'\ne0\).

Worked Example: A candidate where both derivatives vanish

For the cardioid \(r=1+\cos\theta\), check the parameter \(\theta=\pi\). Here \(r=1+\cos\pi=0\) and \(r'=-\sin\pi=0\). The coordinate derivatives are

x'=r'\cos\theta-r\sin\theta=0, \qquad y'=r'\sin\theta+r\cos\theta=0.

Thus the velocity vector is \((0,0)\), and neither the horizontal nor the vertical test applies. This calculation does not classify the tangent at that parameter; it identifies a case requiring a separate analysis. The important check is that \(x'=0\) alone is not enough to declare a vertical tangent.

Check: substituting \(r=0\), \(r'=0\), \(\sin\pi=0\), and \(\cos\pi=-1\) into each derivative gives zero, as claimed.

A common mistake: keeping every zero as a tangency

A frequent error is to solve \(x'=0\) and label every resulting parameter a vertical tangency, without checking \(y'\). The cardioid calculation above shows why this fails: at \(\theta=\pi\), both \(x'\) and \(y'\) are zero. The tangent direction cannot be read from a zero velocity vector.

The correction is to treat each equation as producing candidates, then apply the nonzero condition. For a horizontal candidate, evaluate \(x'\); for a vertical candidate, evaluate \(y'\). If the other derivative vanishes too, report the parameter as inconclusive under these tests rather than assigning it a horizontal or vertical tangent.

Why parameter intervals matter

Polar equations can trace the same geometric point at different parameter values. The tangency test is performed at parameter values, while a question may ask for tangencies over a particular interval or for distinct points on the curve. Always state the interval you use, solve within it, and be alert to endpoints that represent the same trace when a full period is included.

This method applies to many polar curves, including the standard families studied in earlier material on polar graphs. It turns a geometric question into two derivative equations and a careful check. Its main limitation is equally important: when both coordinate derivatives vanish, further analysis is needed rather than a routine horizontal-versus-vertical classification.

Check Your Understanding

For each question, state the derivative condition and explain what additional check is required.

What two conditions identify a horizontal tangent at a regular parameter?
What two conditions identify a vertical tangent at a regular parameter?
Why does \(\frac{dx}{d\theta}=0\) by itself not prove that a tangent is vertical?
For \(r=2+\cos\theta\), what equation in \(c=\cos\theta\) produces horizontal candidates?
What does it mean for both coordinate derivatives to vanish at a candidate parameter?