Why Compare Tangent Directions?
A parametrized curve can pass through the same point at two different parameter values. At that point, the curve has two local branches, one associated with each parameter value. Comparing their tangent directions tells us whether the branches meet with distinct tangent lines or share a tangent line.
This comparison describes the branches to first order. It can establish a transverse crossing when the tangent lines are distinct. If the tangent lines coincide, however, the first derivatives alone do not tell us whether the branches touch, cross with matching tangents, or share part of the same path.
Repeated Points and Tangent Lines
Write a parametrized plane curve as \(\mathbf{r}(t)=(x(t),y(t))\). Suppose \(t_1\ne t_2\), and \(\mathbf{r}(t_1)=\mathbf{r}(t_2)=P\). The point \(P\) is a repeated point of the parametrization. If it is where distinct local branches meet, it is a self-intersection.
At a repeated point with two regular parameter values, the two tangent lines are determined by \(\mathbf{r}'(t_1)\) and \(\mathbf{r}'(t_2)\). A tangent line has no preferred orientation: a nonzero vector and its negative describe the same line. Thus the derivative vectors need only be scalar multiples for the tangent lines to coincide.
In particular, the determinant test applies only when both derivative vectors are nonzero. A zero derivative vector does not specify a tangent direction by itself; a singular parameter value requires further analysis.
Worked Example: A Transverse Crossing
Worked Example: A Transverse Crossing
Consider \(\mathbf{r}(t)=(t^2-1,\ t(t^2-1))\). Find the repeated point at \(t=1\) and \(t=-1\), then compare the tangent directions there.
Step 1: Check the point. At \(t=1\), the coordinates are \(x=1^2-1=0\) and \(y=1(1^2-1)=0\). At \(t=-1\), they are \(x=(-1)^2-1=0\) and \(y=(-1)((-1)^2-1)=0\). The two parameter values represent \(P=(0,0)\).
Step 2: Differentiate.
Step 3: Evaluate and test. The derivative vectors are \(\mathbf{r}'(1)=(2,2)\) and \(\mathbf{r}'(-1)=(-2,2)\). Both are nonzero, and $$ \det\bigl(\mathbf{r}'(1),\mathbf{r}'(-1)\bigr)=2(2)-2(-2)=8\ne 0. $$ So the tangent lines are distinct. Their slopes are \(2/2=1\) and \(2/(-2)=-1\), giving the lines \(y=x\) and \(y=-x\).
Check: Substituting \(t=1\) and \(t=-1\) into the original parametrization gives \((0,0)\) in both cases. Substituting each derivative vector into the line through the origin gives \(y=x\) and \(y=-x\), respectively. The branches meet at the same point with distinct tangent lines, so this is a transverse crossing.
Why Distinct Tangent Lines Give a Crossing
The determinant criterion has a geometric meaning. When two regular branches have distinct tangent lines at their common point, each branch can locally be written as a graph over a suitable common axis. The slopes of those graphs at the intersection differ. Their vertical difference is zero at the intersection and changes sign as one moves through it, so the branches cross there.
To justify this, use coordinates with a horizontal axis that is not parallel to either tangent line. In these coordinates, each branch is locally a graph, say \(y=f_1(x)\) and \(y=f_2(x)\), with \(f_1(x_0)=f_2(x_0)\) at the common point. Their tangent lines are distinct, so \(f_1'(x_0)\ne f_2'(x_0)\). Set \(g(x)=f_1(x)-f_2(x)\). Then \(g(x_0)=0\) and \(g'(x_0)\ne 0\). By the definition of derivative, \(g(x_0+h)=g'(x_0)h+o(h)\). For sufficiently small nonzero \(h\), this has opposite signs on the two sides of \(h=0\), and it is nonzero. Thus the branches switch sides at \(P\), with no other intersection sufficiently near \(P\).
Worked Example: Matching Tangent Lines, Opposite Orientations
Worked Example: Matching Tangent Lines, Opposite Orientations
Consider \(x=t^2-1\) and \(y=x+t x^3\). Compare the tangent directions at \(t=1\) and \(t=-1\).
Step 1: Locate the repeated point. At either parameter value, \(x=0\), so \(y=0+t(0)^3=0\). Both values represent \(P=(0,0)\).
Step 2: Differentiate carefully. Since \(x'=2t\), the product and chain rules give $$ y'=x'+\frac{d}{dt}(t x^3) =2t+x^3+3t x^2(2t) =2t+x^3+6t^2x^2. $$ At \(t=1\), \(x=0\), so \(\mathbf{r}'(1)=(2,2)\). At \(t=-1\), \(x=0\), so \(\mathbf{r}'(-1)=(-2,-2)\).
Step 3: Compare the vectors. Their determinant is $$ \det\bigl((2,2),(-2,-2)\bigr)=2(-2)-2(-2)=0. $$ Both vectors lie along the line \(y=x\). They point in opposite orientations because \((-2,-2)=-(2,2)\), but they give the same tangent line.
Check: The original parametrization gives \((0,0)\) at both \(t=1\) and \(t=-1\), and the differentiated coordinate functions give the stated vectors at those values. Their zero determinant confirms that the tangent lines coincide. This first-order test alone does not decide whether the branches cross or touch.
Matching Tangents Do Not Determine the Contact
When the determinant is zero, the branches have the same tangent line, but their behavior beyond that line can vary. Higher-order terms can make them stay on the same relative side of one another, or exchange sides. The next examples show both possibilities using the same horizontal coordinate \(x=t^2-1\).
Worked Example: Matching Tangents and a Higher-Order Crossing
For \(x=t^2-1\), let \(y=x+t x^3\), as in the previous example. Near \(x=0\), both parameter branches can be described using \(t=\sqrt{1+x}\) or \(t=-\sqrt{1+x}\), since \(t^2=1+x\). Their heights are $$ y_+(x)=x+\sqrt{1+x}\,x^3,\qquad y_-(x)=x-\sqrt{1+x}\,x^3. $$
Subtracting gives $$ y_+(x)-y_-(x)=2\sqrt{1+x}\,x^3. $$ For \(x\) close to zero, \(\sqrt{1+x}>0\). The difference is negative for \(x<0\) and positive for \(x>0\), so the branches exchange which one is higher as they pass through the common point. They cross, even though both have tangent line \(y=x\).
Check: At \(x=0\), both branch heights are zero. Their derivatives with respect to \(x\) at zero are both \(1\), because the extra terms contain \(x^3\). This agrees with the parameter-derivative calculation \((2,2)\) and \((-2,-2)\), whose slopes are both \(1\). Thus they share a tangent while crossing at higher order.
Worked Example: Matching Tangents Without a Crossing
Now keep \(x=t^2-1\), but take \(y=x+t x^2\). At \(t=1\) and \(t=-1\), the point is again \((0,0)\). Differentiating gives $$ x'=2t,\qquad y'=x'+x^2+2t x(2t)=2t+x^2+4t^2x. $$ Since \(x=0\) at both parameter values, the derivative vectors are \(\mathbf{r}'(1)=(2,2)\) and \(\mathbf{r}'(-1)=(-2,-2)\). Their determinant is \(2(-2)-2(-2)=0\), so both tangent lines are \(y=x\).
For a fixed \(x\) near zero, the branches correspond to \(t=\pm\sqrt{1+x}\). Their heights are \(y_+(x)=x+\sqrt{1+x}\,x^2\) and \(y_-(x)=x-\sqrt{1+x}\,x^2\). Therefore $$ y_+(x)-y_-(x)=2\sqrt{1+x}\,x^2>0 $$ for sufficiently small \(x\ne 0\). The same branch remains higher on both sides of the intersection; the branches do not exchange sides there.
Check: At \(x=0\), both heights are zero, and their difference has the same positive sign on either side. The derivative vectors confirm the shared tangent line, while the branch comparison shows that the branches do not cross locally.
A Common Mistake: Treating Opposite Vectors as Different Tangents
A common error is to conclude that \((2,2)\) and \((-2,-2)\) give different tangent directions because one vector points in the opposite direction. For example, comparing only their signs might lead to the incorrect conclusion that the branches have distinct tangents.
The correction is to compare the lines, not the orientations of the vectors. Here \((-2,-2)=-1(2,2)\), so the vectors are parallel. Equivalently, $$ \det\bigl((2,2),(-2,-2)\bigr)=2(-2)-2(-2)=0, $$ and both tangent lines are \(y=x\). Opposite orientations do not mean different tangent lines.
How to Analyze a Repeated Point
Identify distinct values \(t_1\) and \(t_2\) for which the parametrization gives the same point.
Compute both derivative vectors. If either is zero, the determinant test alone does not determine the tangent direction at that parameter value.
For two nonzero vectors, compute their determinant. A nonzero result means distinct tangent lines; zero means a shared tangent line.
Distinct tangent lines imply a transverse crossing. A shared tangent line records matching first-order directions, but does not by itself settle whether the branches cross or stay on the same side.
Why the Distinction Matters
A tangent direction is a first-order description of a curve near a point. It tells us the line that best describes the branch at that point, but it does not record all the branch’s nearby shape. The determinant test is therefore decisive when it is nonzero and limited when it is zero.
At a self-intersection, keep three ideas separate: the parameter values are distinct, the derivative vectors may have different orientations, and the tangent lines may still coincide. Comparing the vectors correctly identifies whether the branches meet transversely or share a tangent, while avoiding conclusions that require information beyond their first derivatives.
Check Your Understanding
Use the derivative vectors at the repeated point in each question. State what the comparison establishes, and do not infer more than it supports.
- If \(\mathbf{r}(t_1)=\mathbf{r}(t_2)\), with \(t_1\ne t_2\), and the two derivative vectors are \((3,1)\) and \((2,4)\), what does their determinant tell you?
- Why do the vectors \((1,-2)\) and \((-3,6)\) determine the same tangent line?
- What does a zero determinant establish when both derivative vectors are nonzero?
- Why can matching tangent lines alone not distinguish a tangential crossing from branches that stay on the same relative side?
- What additional caution is needed if one derivative vector at the repeated point is zero?