Definition 2.2.1.
A point where the tangent line to an algebraic plane curve is not defined is called a singularity.
Section 2.2 Singularities of Algebraic Plane Curves
Subsection 2.2.1 Defining Singularities
In Example 2.1.7, the algebraic plane curve had a bunch of self intersections. What would the tangent line (and slope) be at such an intersection? This is the first example of a place where defining the tangent become problematic in a way that is different from a vertical tangent. There is a name for these places.
Note one subtlety here: the a singularity is where the tangent line is not defined, not just where its slope is undefined. This means that points with vertical lines are not singularities; even though they don’t have a slope, they do have a well defined tangent line. Singularities are points where the whole notion of a tangent line breaks down. The classification of singularities of algebraic plane curve is a old and historically important mathematical problem. This is particular true for cubics (algebraic plane curves of degree 3); the study of cubics led to whole branches of 19th and 20th century mathematics.
Before I get to cubics, let me cover the only possible singulary in degree two. In degree one, there are no problems: the tangent line to a line is just the original line, so this is always defined, even for a vertical line. Likewise, almost all degree two curves have no singularities: circles, ellipses, parabolae and hyperbolae have no places where a tangent line cannot be undefined. There is only one kind of degree two curve which has a singularity: an intersection of two lines. An example, with the equation \(x^2 - y^2 = 0\text{,}\) is shown in Figure 2.2.2.
This strange degree two curve is just two lines. I can see this by factoring the equation: \(x^2 - y^2 = 0\) becomes \((x-y)(x+y) = 0\text{.}\) This is just the product of two linear equations: \(x-y=0\) and \(x+y=0\text{.}\) Those two lines are the lines through the origin with slopes \(1\) and \(-1\text{,}\) respectively. These are clearly the lines in Figure 2.2.2 as well. This degree two curve is just two lines put together. (You may have noticed that I haven’t called this a conic. It really isn’t, since it is just two lines. I’ll reserve the term ‘conic’ for degree two curves which are not the product of two lines.)
Now let me ask: what is the tangent line at the origin to the curve \(x^2 - y^2 = 0\text{?}\) This is a problem, since the line with slope \(1\) and the line with slope \(-1\) both seem to be tangent to parts of the curve. Therefore, the tangent line can’t be properly defined. The origin is a singularity of this curve. There is a name for this type of singularity.
Definition 2.2.3.
A singularity which is an intersection point of two or more pieces of an algebraic curve is called a node. The number of pieces at the intersection gives more precise names: two pieces is called a double point, three a triple point, and so on.
The curve \(x^2 - y^2=0\) has a double point node at the origin. In Example 2.1.7, the picture in Figure 2.1.8 looks like it has five different double point nodes. For an example of more than two intersecting pieces, consider the curve give by the following equation (shown in Figure 2.2.4.
\begin{align*}
\amp -x^4y + x^3 + 3x^3y + xy^3 - 3x^3 - 3x^2y - 3xy^2 - y^3 +
3x^2 \\
\amp + 4xy + 3y^2 - 2x - 3y + 1 = 0
\end{align*}
This curve has a quadruple point node at \((1,1)\text{,}\) since four different paths in the curve intersect at that point.
The theory of singularities of algebraic plane curves is a deep and complicated one. Nodes are just one among many types of singularities. There are several others which have names, and many other singularities which don’t have specific names but at classified by various catalogues and systems. Among those that have names, I want to define one more type of singularites. I’ll give a less technical, more descriptive definition here.
Definition 2.2.5.
A point on an algebraic plane curve which has a sharp corner (and is not the intersection of two or more pieces of the curve) is called a cusp.
A cusp can be seen in Figure 2.2.6 at the origin on the curve \(x^2 - y^3 = 0\text{.}\)
At a node, there were multiple possibilities for a tangent based on the various pieces of the curve passing through the point. For a cusp, the situation is even worse. At a sharp corner, there are infinitly many lines that could be drawn that touch the curve only once. There notion of a slope at a sharp corner make any sense at all.
Subsection 2.2.2 Investigating Singularities
I would like to use the implicit derivatives defined in Section 1.2 to undertand singularities of algebraic plane curves. In general, as I said before, this is a deep and complicated theory. Since I don’t have time to delve deeply into all of the various ways to use derivatives to understand singularities, I am going to specialize to a very particular class of curves. In this section, I’m only going to consider curves that have the following form, where \(p(x)\) is a polynomial of degree three or more.
\begin{equation*}
y^2 = p(x)
\end{equation*}
Even though this is specializing a great deal, it still allows for many interesting and important examples. The degree of the algebraic plane curve is still unbounded, since \(p(x)\) can have as high a degree as I want. If the degree of \(p(x)\) is three or four, the resulting curve is an elliptic curve (or, at least, one presentation of an elliptic curve). Elliptic curves are, themselves, an important branch of mathematics with many applications and connections. If the degree of \(p(x)\) is five or higher, the resulting curve is a hyperelliptic curve, another interesting and important type of object.
For the purposes of this course, curves of this form are accesible for several reasons. (I present the following statements without proof.)
- All singularities and vertical tangents to these curve will always be on the \(x\) axis.
- Vertical tangents and singularities are identified by places where the implicit derivative is not defined.
- At a point \((a,0)\) where the implicit derivative is not defined, the limit \(x \rightarrow a\) of the implicit derivative approaching this point will classify the behaviour of the point.
- The only possibilites for the behaviour in this case are vertical tangents, double point nodes, or cusps.
With these facts, I can build an algorithm to analyze the singularities of these special algebraic plane curves.
- I check the domain of possible \(x\) values that are allowed in the curve. This will have an impact on the directions of limits later in the process.
- I will take the implicit derivative.
- I will find the points \((a,0)\) where the implicit derivative is undefined.
- Using the equation of the orignal curve, I will replace \(y\) with \(\pm \sqrt{p(x)}\) in the expression for the implicit derivative. That gives an implicit derivative expressed entirely in \(x\text{.}\)
- I will take the limit \(x \rightarrow a\) of the implicit derivative.
- If this limit is \(\pm \infty\text{,}\) the slope of the tangent line is becoming steeper and steeper and I conclude there is vertical tangent.
- If this limit has two or more values (for example, because the \(\pm\) is still present in the final answer), then there are two or more possible tangent lines and I conclude this is a double point note.
- If this limit is zero, then I conclude there is a cusp.
Now I’ll work through a number of examples to show this algorithm in practice. In this examples, note that if I can factor \(p(x)\) (or if it comes already factored), the calculations and the limits can become a lot easier due to cancelation of some of the factors.
Before I start these examples, though, let me review a small bit of the rules of exponents. How do I simplify this expression>
\begin{equation*}
\frac{\sqrt{x-1}}{(x-1)^3}
\end{equation*}
I can do this by writing the square root as an exponent and then using the exponent laws.
\begin{equation*}
\frac{\sqrt{x-1}}{(x-1)^3} =
\frac{(x-1)^{\frac{1}{2}}}{(x-1)^3}
\end{equation*}
From this point, I can simplify as I wish. I might want everything in the numerator, or everything in the denominator. All of the following are possible result, depending on what kind of for you want.
\begin{equation*}
\frac{(x-1)^{\frac{1}{2}}}{(x-1)^3} =
\frac{(x-1)^{\frac{-5}}2}} = \frac{1}{(x-2)^{\frac{5}{2}}} =
\frac{1}{(x-2)^2 \sqrt{x-2}}
\end{equation*}
Throughout the examples and activity this week, I’ll use this simplification many times. I won’t necessary make all the exponential tricks explicit as I do the simplifications.
Example 2.2.7.
Consider the plane curve given by the equation \(y^2 = 4 -
3x^2 - x^3\text{.}\) This is a degree three curve, which factors as \(y^2 = (1-x)(2+x)^2\text{.}\) The domain is \(x \leq 1\text{,}\) since \(x \geq 1\) gives a negative on the right side, which is impossible with \(y^2\) on the left. I calculate the implicit derivative.
\begin{equation*}
2 y \frac{dy}{dx} = -6x - 3x^2 \implies \frac{dy}{dx} =
\frac{-6x-3x^2}{2y} = \frac{-3x(2+x)}{2y}
\end{equation*}
This is well defined except when \(y=0\) which, for this curve, happens at \((1,0)\) and \((-2,0)\text{.}\)
Now I solve for \(y\) to get \(y = \pm
\sqrt{(1-x)(2+x)^2} = \pm (2+x)\sqrt{1-x} \text{.}\) I use this to replace \(y\) in the implicit derivative.
\begin{equation*}
\frac{dy}{dx} = \pm \frac{-3x(2+x)}{2\sqrt{1-x} (2+x)}
\end{equation*}
Then I look at the limits as I approach the undefined points. When \(x=1\text{,}\) only the denominator goes to \(0\text{,}\) the numerator is finite. Therefore, the slope diverges to infinity and the curve will have a vertical tangent. Note that I can only take the limit approaching \(x=1\) from the left, due to the domain constraints, but the conclusion still holds.
However, at \(x=-2\text{,}\) both numerator and denominator have a \(x+2\) term, which cancel out. (The limit is an indeterminate form of type \(\frac{0}{0}\)). Evaluating the limit gives a slope of \(\pm \frac{6}{2\sqrt{3}} = \pm
\sqrt{3}\text{.}\) Since I get two values, this is a double point node. Figure 2.2.8 shows both the vertical tangent and the double point node that I have found with the limits of the implicit derivatives.
Example 2.2.9.
Consider the curve \(y^2 = \frac{x^3}{4}\text{.}\) This curve is only defined when \(x /geq 0\text{,}\) since negative \(x\) lead to a negative on the right side, which cannot happen with the square on the left. The implicit derivative is \(\frac{dy}{dx} = \frac{3x^2}{8y} =
\pm \frac{3x^2}{4\sqrt{x^3}}\text{.}\) This is undefined at the point \((0,0)\) on the curve.and the limit approaching the undefined point is \(0\text{.}\) Since I get a limit of \(0\text{,}\) I conclude that this should be a cusp. I can only approach \(0\) from the right due to the domain restrictions, but that doesn’t affect the conclusion. Figure 2.2.10 indeed shows this cusp.
