When the Derivative Pair Vanishes
A parametrized plane curve is described by coordinate functions \(x=x(t)\) and \(y=y(t)\), so its point at parameter \(t\) is \((x(t),y(t))\). At a parameter value \(t_0\), the derivative pair is \((x'(t_0),y'(t_0))\). If this pair is not \((0,0)\), the parameter value is regular. If both derivatives vanish, the parameter value is singular.
A singular parameter value needs further investigation. It does not automatically produce a cusp: the parametrization may simply slow down at a point where the curve itself is smooth. To tell these possibilities apart, examine where nearby parameter values lie and the directions in which the curve approaches and leaves the point.
One-Sided Directions from Nearby Coordinates
Let \(h\) be a small nonzero change in the parameter. The displacement from the point \(P\) is
When this displacement is nonzero, divide it by its length to obtain a unit direction from \(P\) toward the nearby curve point. Taking the limit as \(h\) approaches zero from the positive or negative side gives the one-sided tangent directions, when those limits exist. The two directions can reveal whether the curve approaches along opposite rays of a line or along the same ray.
For common polynomial parametrizations, the first nonzero terms in the local coordinate expansions give a useful guide. If the first nonzero vector term in the displacement is \(\mathbf{c}h^k\), where \(\mathbf{c}\ne\mathbf{0}\), then the unit displacement approaches the direction of \(\mathbf{c}\) from the right. From the left, it approaches the same direction if \(k\) is even, and the opposite direction if \(k\) is odd. This conclusion follows because \(h^k\) has the same sign on both sides when \(k\) is even and opposite signs when \(k\) is odd.
This leading-term test is a guide, not a complete classification by itself. The actual nearby coordinates matter: two sides may lie on the same geometric branch, or they may form two distinct branches meeting at a cusp.
Worked Example: A Semicubical Cusp
Worked Example: A Semicubical Cusp
Consider \(x=t^2\), \(y=t^3\), near \(t_0=0\). First differentiate the coordinate functions:
The parameter value is singular, and the point is \(P=(0,0)\). The nearby displacement is \((t^2,t^3)\). Its length is \(\sqrt{t^4+t^6}=t^2\sqrt{1+t^2}\), since \(t^2\ge 0\). Thus the unit displacement is
Both sides approach along the same ray, pointing to the right. The nearby coordinates confirm that there are two distinct branches: when \(t>0\), \(y>0\), and when \(t<0\), \(y<0\), while \(x=t^2>0\) on both sides. Eliminating the parameter gives \(y^2=t^6=(t^2)^3=x^3\), with \(x\ge0\). This is the standard semicubical cusp at the origin.
As a further check away from \(t=0\), the parametric slope is \(dy/dx=y'(t)/x'(t)=3t/2\). It tends to \(0\) from either side, agreeing with the horizontal tangent line suggested by the direction limits. The slope formula itself cannot be evaluated at \(t=0\), because both derivatives there are zero.
A Singular Parametrization Can Trace a Smooth Curve
The derivative pair identifies a problem with the parametrization at a value; by itself, it does not determine the shape of the curve. A smooth curve can have a singular parameter value if the parameter does not move through its points regularly.
Worked Example: A Smooth Parabola with a Singular Parameter
Consider \(x=t^3\), \(y=t^6\), again at \(t_0=0\). The derivatives and point are
Since \(x=t^3\), the parameter takes both negative and positive values of \(x\) as \(t\) passes through zero. Also, \(y=t^6=(t^3)^2=x^2\). The curve trace is the smooth parabola \(y=x^2\), not a cusp.
To check the directions, the displacement is \((t^3,t^6)\), with length \(|t|^3\sqrt{1+t^6}\). Therefore
As \(t\to0^+\), this tends to \((1,0)\); as \(t\to0^-\), it tends to \((-1,0)\). The curve approaches and leaves along opposite rays of the horizontal tangent line, as expected for a smooth passage through the vertex. As a slope check, for \(t\ne0\), \(dy/dx=6t^5/(3t^2)=2t^3\), which tends to \(0\); the graph \(y=x^2\) also has slope \(2x=0\) at \(x=0\).
A Cusp with a Vertical Tangent
A cusp need not point horizontally. The same approach works when the dominant change is vertical. Compare both one-sided directions and verify that the nearby points really form two branches.
Worked Example: A Rotated Semicubical Cusp
Let \(x=t^3\), \(y=t^2\), with \(t_0=0\). Then
The displacement is \((t^3,t^2)\), and its length is \(t^2\sqrt{1+t^2}\). Its unit direction is
For \(t>0\), the nearby points have \(x>0\); for \(t<0\), they have \(x<0\). Thus two distinct branches approach the origin along the same upward ray. Eliminating the parameter gives \(x^2=t^6=(t^2)^3=y^3\), with \(y\ge0\). These coordinates describe a cusp with a vertical tangent line.
For \(t\ne0\), the slope is \(dy/dx=2t/(3t^2)=2/(3t)\), which is unbounded in magnitude as \(t\to0\). That is consistent with the vertical tangent line indicated by the one-sided direction vectors.
Why the Direction Test Works
The one-sided direction calculation is based on the displacement itself, not on a slope that may be undefined at a singular parameter value. Here is the leading-term argument for polynomial parametrizations, or for smooth parametrizations whose first nonzero Taylor terms have the stated form.
Write the displacement from \(P\) as \(\mathbf{c}h^k+\) terms of higher order, where \(k\) is the first order with a nonzero vector coefficient \(\mathbf{c}\).
Its length is \(|h|^k\|\mathbf{c}\|\) to leading order, so its unit direction is governed by \(\mathbf{c}h^k/|h|^k\).
If \(k\) is even, \(h^k/|h|^k=1\) on both sides, so the leading directions agree. If \(k\) is odd, the factor is \(1\) on the right and \(-1\) on the left, so the leading directions are opposite.
This explains why a same-ray limit is a warning sign for a cusp, while opposite rays often indicate a smooth passage. The rest of the coordinate behavior is essential: the examples show that same-ray directions produce a cusp when two distinct branches meet, whereas a singular parametrization can trace a smooth curve with opposite one-sided directions.
A Common Mistake
A frequent error is to calculate \(dy/dx=y'(t)/x'(t)\) at the singular parameter and treat the resulting \(0/0\) as a slope. The expression \(0/0\) is undefined; it gives no tangent information. It is also incorrect to conclude that every singular parameter gives a cusp.
Check Your Understanding
For each question, use both the derivative pair and the nearby coordinates rather than relying only on a slope formula.
- What condition makes a parameter value singular, and does that condition alone imply a cusp?
- For \(x=t^2,\ y=t^3\), what are the one-sided unit displacement limits at \(t=0\)?
- For \(x=t^3,\ y=t^6\), why is the curve trace smooth even though the parameter value is singular?
- What does it suggest geometrically when two one-sided tangent directions are opposite rays of the same line?
- Why is substituting a singular parameter directly into \(dy/dx=y'(t)/x'(t)\) not a valid way to find the tangent?
