Why a Polar Slope Quotient Can Be Misleading
A formula for the slope of a polar curve can become indeterminate at a parameter value, even though the curve itself has a well-defined tangent. In particular, when both Cartesian derivatives vanish, direct substitution into \(dy/dx\) gives \(0/0\). Taking a limit may recover the tangent direction, but the limit alone does not tell us whether the curve has a cusp or simply passes through a stationary parameter.
To distinguish these possibilities, examine nearby parameter values on both sides of the singular value. A cusp has two local branches that approach the same tangent direction from the same side. At a stationary passage, the branches continue through the point in opposite directions along the tangent line. Both situations can produce a removable singularity in the slope quotient.
When Do Both Derivatives Vanish?
Let a polar curve be parametrized by \(r=r(\theta)\), with Cartesian coordinates \(x=r\cos\theta\) and \(y=r\sin\theta\). Differentiating with respect to \(\theta\) gives the velocity components. The two unit vectors \((\cos\theta,\sin\theta)\) and \((-\sin\theta,\cos\theta)\) are perpendicular, so the squared speed is the sum of two squares:
Consequently, both derivatives vanish at \(\theta=\theta_0\) exactly when \(r(\theta_0)=0\) and \(r'(\theta_0)=0\). The curve is at the pole, and the radius has a zero of at least second order, provided \(r\) is sufficiently differentiable.
Read the Local Shape from the Radius
Set \(u=\theta-\theta_0\). Suppose the radius has a zero of finite order \(m\geq 2\), meaning that, near \(u=0\),
Here \(O(u^k)\) denotes terms whose size is bounded by a constant times \(|u|^k\) for sufficiently small \(u\). Rotate the coordinate axes so that the first axis points in the radial direction at \(\theta_0\), and call the resulting coordinates \(U\) and \(V\). The rotation does not change whether branches lie along the same or opposite directions of a tangent line. In these coordinates,
Since \(\cos u=1+O(u^2)\) and \(\sin u=u+O(u^3)\), the leading terms are
The first coordinate therefore determines whether the two sides of the parameter approach the same or opposite sides of the tangent line. If \(m\) is even, \(U\) has the same sign for small positive and negative \(u\), while \(V\) changes sign. The two branches approach the same tangent ray: the point has cusp-type behavior. When \(m=2\), this is the usual ordinary cusp. If \(m\) is odd, \(U\) changes sign, so the branches approach from opposite directions along the tangent line. This is a stationary passage, not a cusp.
Worked Example: A Second-Order Zero Produces a Cusp
Worked Example: Analyze \(r=\theta^2\) at the Pole
Consider \(r(\theta)=\theta^2\) near \(\theta=0\). The curve is \(x=\theta^2\cos\theta\), \(y=\theta^2\sin\theta\), and \(r(0)=r'(0)=0\), so the parameter is singular. Differentiate:
For \(\theta\ne 0\), cancel the common factor \(\theta\) in the slope quotient. Its limit is
The limiting tangent is horizontal. But the limit does not show whether the point is a cusp. To check the branches, observe that \(x=\theta^2\cos\theta>0\) for small nonzero \(\theta\), while \(y=\theta^2\sin\theta\) is negative for \(\theta<0\) and positive for \(\theta>0\). Also \(y/x=\tan\theta\to 0\) from both sides. Both branches approach the pole along the positive horizontal ray, from opposite sides of that ray. The second-order zero therefore gives a cusp-type point (in this case, an ordinary cusp).
Check the derivative calculation by substituting \(\theta=0\) into the factored derivatives: both are \(0\), as required for a singular parameter. The slope limit is consistent with the horizontal tangent, but the branch check is what identifies the cusp.
Worked Example: An Odd-Order Zero Gives a Stationary Passage
Worked Example: Analyze \(r=\theta^3\) at the Pole
Now let \(r(\theta)=\theta^3\). Again \(r(0)=r'(0)=0\). The coordinate derivatives are
For \(\theta\ne 0\), cancel \(\theta^2\) in the quotient. Then
The slope formula has a removable singularity, and its limiting tangent is horizontal. To classify the point, check whether the curve passes through to the opposite side. The horizontal coordinate derivative is \(x'=\theta^2(3\cos\theta-\theta\sin\theta)\). The factor in parentheses is positive near zero because its value at zero is \(3\) and it is continuous. Thus \(x\) increases through the pole: it is negative just before \(\theta=0\) and positive just after.
For additional confirmation, \(x\sim\theta^3\) and \(y\sim\theta^4\), so \(y/x\to 0\), as expected for a horizontal tangent. The odd-order zero gives a stationary passage, not a cusp: the parameter velocity vanishes at the pole, but the curve continues from one side of the tangent line to the other.
Finally, substituting \(\theta=0\) into the factored derivatives gives \(x'(0)=y'(0)=0\). This verifies that the parameter is stationary even though the geometric curve has a tangent and passes through the point.
Worked Example: A Higher-Order Even Zero
Worked Example: Analyze \(r=\theta^4\) at the Pole
Consider \(r(\theta)=\theta^4\), whose zero at the origin has order four. Differentiation gives
For nonzero \(\theta\), cancel the common factor \(\theta^3\). The slope limit is
Since \(x=\theta^4\cos\theta>0\) for small nonzero \(\theta\), both sides lie to the right of the pole. Meanwhile \(y=\theta^4\sin\theta\) changes sign with \(\theta\). Thus both branches approach along the positive horizontal tangent ray: the point has higher-order cusp-type behavior. The order-four zero predicts this because it is even. Substitution into the factored derivatives confirms that both derivatives vanish at \(\theta=0\); the slope quotient's finite limit does not remove the cusp.
Why the Order Test Works
The classification follows from how the leading powers behave on the two sides of \(u=0\). In the rotated coordinates, the leading terms are \(U\sim a u^m\) and \(V\sim a u^{m+1}\). The sign of \(u^m\) is unchanged when \(u\) is replaced by \(-u\) if \(m\) is even, and reversed if \(m\) is odd. The power \(u^{m+1}\) has the opposite parity, so \(V\)'s sign reverses exactly when \(U\)'s does not. This yields the same-ray cusp-type pattern for even \(m\) and the opposite-direction passage for odd \(m\).
The tangent direction is radial at \(\theta_0\): as \(u\to0\), the position vector points in direction \((\cos\theta_0,\sin\theta_0)\) or its opposite. For a nonvertical tangent, its slope is \(\tan\theta_0\). If \(\cos\theta_0=0\), the tangent is vertical, and it is more useful to examine \(dx/dy\) or the tangent direction directly than to expect a finite \(dy/dx\) limit.
A Common Mistake: Treating a Slope Limit as a Cusp Test
A common error is to conclude that a point is not a cusp merely because \(dy/dx\) has a finite limit. For \(r=\theta^2\), one might cancel the common factor and compute a horizontal limiting slope, then call the point a smooth passage. That conclusion ignores the branches: here \(x>0\) on both sides of zero, so both branches approach along the same tangent ray. It is a cusp-type point.
The correction is to use the slope limit only to identify a candidate tangent direction. Then examine the signs of suitable local coordinates on both sides of the parameter. Same tangent ray indicates cusp-type behavior; opposite directions along the tangent line indicate a stationary passage. A finite slope quotient answers a tangent question, not by itself a branch-classification question.
Practical Analysis
Use \(x'=r'\cos\theta-r\sin\theta\) and \(y'=r'\sin\theta+r\cos\theta\). Both vanish exactly when \(r=r'=0\).
Write a local expansion \(r(\theta_0+u)=a u^m+O(u^{m+1})\), if such a finite-order expansion is available.
An even \(m\) gives cusp-type behavior; an odd \(m\) gives a stationary passage. Confirm by checking the signs of nearby coordinates.
Factor the common vanishing terms from \(dy/dx\), then take the limit. Check for a vertical tangent if the denominator still tends to zero.
These steps matter because a singular parameter and a singular point of the geometric curve are not interchangeable ideas. A parametrization may stop momentarily while tracing a curve that has a clear tangent and continues through the point. Conversely, a cusp can have a well-defined limiting tangent even though its two branches meet with cusp-type geometry. For accurate analysis, use both the derivative quotient and the local branch behavior.
Check Your Understanding
Use nearby parameter values as well as derivative limits when answering.
- For a polar parametrization, what conditions on \(r(\theta_0)\) and \(r'(\theta_0)\) make both Cartesian derivatives vanish?
- If \(r(\theta_0+u)=a u^5+O(u^6)\), is the local behavior cusp-type or a stationary passage? Explain using the parity of the order.
- Why does a finite limit of \(dy/dx\) fail to distinguish a cusp from a stationary passage?
- For \(r=\theta^4\), which side of the horizontal tangent line do both local branches approach?
- What limitation should be kept in mind when applying the finite-order zero test?
