Week 2 Activity

Section 2.3 Week 2 Activity

Subsection 2.3.1 Singularities of Algebraic Plane Curves

Activity 2.3.1.

Find and classify the singularities of this algebraic plane curve.

\begin{equation*} y^2 = x^2 + x^4 \end{equation*}

Solution.

I can factor the right side of the curve as \(x^2(1+x^2)\text{.}\) There are no domain restrictions for this curve. I’ll use this factored form, since it will help to clearly understand the limits of the implicit derivative.

First I calculate the implicit derivative of the curve.

\begin{align*} \frac{d}{dx} y^2 \amp = \frac{d}{dx} x^2 (1+x^2) \\ 2y \frac{dy}{dx} \amp = 2x (1+x^2) + x^2 (2x) = 2x (1+x^2) + 2x^3 \\ \frac{dy}{dx} \amp = \frac{2x(1 + x^2) + 2x^3}{2y} = \frac{x(1+x^2) + x^3}{y} \end{align*}

Then I use the original curve to solve for \(y\) and replace \(y\) in the implicit derivative.

\begin{align*} y \amp = \pm \sqrt{x^2 (1+x^2)} = \pm x \sqrt{1+x^2} \\ \frac{dy}{dx} \amp = \pm \frac{x(1+x^2) + x^3}{x \sqrt{1+x^2}} \end{align*}

The singularities happen at the undefined points of the implicit derivative. These are the points where the denominator is zero. For this curve, the only potential singularity is at \(x=0\text{,}\) since \(1+x^2\) can never be zero.

For each of the undefined points, I calculate the limit of the implicit derivative approaching the undefined points. There is only one such point here.

\begin{align*} \lim_{x \rightarrow 0} \frac{dy}{dx} \amp = \lim_{x \rightarrow 0} \pm \frac{x(1+x^2) + x^3}{x \sqrt{1+x^2}}\\ \amp = \lim_{x \rightarrow 0} \pm \frac{(1+x^2) + x^2}{\sqrt{1+x^2}} = \pm \frac{1}{1} = \pm 1 \end{align*}

There are two possible values for the limit. I conclude that the singularity is a double point node. Figure 2.3.1 shows the curve and the double point node.

Figure 2.3.1.

Activity 2.3.2.

Find and classify the singularities of this algebraic plane curve.

\begin{equation*} y^2 = x^4 + x^3 \end{equation*}

Solution.

For this curve, I can factor the right side to get \(x^3(x+1)\text{.}\) This curve is not defined when \(-1 \lt x \lt 0\text{,}\) since the right side is negative in that range. Then I calculate the implicit derivative of the curve.

\begin{align*} \frac{d}{dx} y^2 \amp = \frac{d}{dx} x^3 (x+1)\\ 2y \frac{dy}{dx} \amp = 3x^2 (x+1) + x^3 \\ \frac{dy}{dx} \amp = \frac{3x^2 (x+1) + x^3}{2y} \end{align*}

Then I use the original curve to solve for \(y\) and replace \(y\) in the implicit derivative.

\begin{align*} y \amp = \pm \sqrt{x^3 (x+1)} = \pm x \sqrt{x(x+1)} \\ \frac{dy}{dx} \amp = \pm \frac{3x^2 (x+1) + x^3}{2x\sqrt{x(x+1)}} \end{align*}

The singularities happen at the undefined points of the implicit derivative. These are the points where the denominator is zero. For this curve, there are potential singularities at \(x=0\) and \(x=-1\text{.}\)

For each of the undefined points, I calculate the limit of the implicit derivative approaching the undefined points. I’ll start with \(x=0\text{.}\) I only approach from the right, due to the domain restrictions.

\begin{align*} \lim_{x \rightarrow 0^+} \frac{dy}{dx} \amp = \lim_{x \rightarrow 0^+} \pm \frac{3x^2 (x+1) + x^3}{2x\sqrt{x(x+1)}} \\ \amp = \lim_{x \rightarrow 0^+} \pm \left[ \frac{3x^2 (x+1)}{2x\sqrt{x(x+1)}} + \frac{x^3}{2x\sqrt{x(x+1)}} \right] \\ \amp = \lim_{x \rightarrow 0^+} \pm \left[ \frac{3\sqrt{x(x+1)}}{2} + \frac{x}{2} \sqrt{ \frac{x}{x+1}} \right] \\ \amp = \pm \left[ 0 + 0 \right] = 0 \end{align*}

The limit is zero. I conclude that there is a cusp at \(x=0\text{.}\) (There are some tricky subtleties with the square root calculations in this question. In the second piece of the second last step of the previous calculations, I wrote \(\sqrt{\frac{x}{x+1}}\text{.}\) Fitting with how I’ve written other square root, I might have written \(\frac{\sqrt{x}}{\sqrt{x+1}}\text{.}\) However, for \(x \lt -1\text{,}\) both of these square roots are undefined. As written, the square root of the fraction is defined, since the ratio is a negative divided by a negative, which is positive.)

Now I move on to \(x=-1\text{.}\) I only approach from the left, due to the domain restrictions.

\begin{align*} \lim_{x \rightarrow -1^-} \frac{dy}{dx} \amp = \lim_{x \rightarrow -1^-} \pm \frac{3x^2 (x+1) + x^3}{2x\sqrt{x(x+1)}} \\ \amp = \lim_{x \rightarrow -1^-} \pm \frac{3x^2 (x+1)}{2x \sqrt{x(x+1)}} + \frac{x^3}{2x\sqrt{x(x+1)}} \\ \amp = \lim_{x \rightarrow -1^-} \pm \frac{3x^2 (x+1)}{2x \sqrt{x(x+1)}} + \lim_{x \rightarrow -1} \frac{x^3}{2x\sqrt{x(x+1)}} \\ \amp = \lim_{x \rightarrow -1^-} \pm \frac{3\sqrt{x(x+1)}}{2} + \lim_{x \rightarrow -1^-} \frac{x^3}{2x\sqrt{x(x+1)}} \end{align*}

The first limit evaluates to \(0\text{.}\) The second limit is more complicated. I can’t cancel off all the term I want to cancel off, since that would lead to square roots of negative terms. However, I can understand this limit in the form it currently have. The numerator approaches \(-1\) and the denominator approaches \(0\text{.}\) Therefore, the limit must approach \(\pm \infty\text{.}\) I conclude that there is a vertical tangent at \(x=-1\text{.}\) Figure 2.3.2 shows the curve with both its cusp and its vertical tangent. It also shows the lack of points between \(x=-1\) and \(x=0\text{.}\)

Figure 2.3.2.

Activity 2.3.3.

Find and classify the singularities of this algebraic plane curve.

\begin{equation*} y^2 = (x-3)^2(x-4)^2 \end{equation*}

Solution.

The curve is already presented in factored form. There are no domain restrictions, since the right side is always possitive for any \(x\) value. I calculate the implicit derivative of the curve.

\begin{align*} \frac{d}{dx} y^2 \amp = \frac{d}{dx} (x-3)^2 (x-4)^2 \\ 2y \frac{dy}{dx} \amp = 2(x-3)(x-4)^2 + 2(x-3)^2(x-4)\\ \frac{dy}{dx} \amp = \frac{2(x-3)(x-4)^2 + 2(x-3)^2(x-4)}{2y} \end{align*}

Then I use the original curve to solve for \(y\) and replace \(y\) in the implicit derivative.

\begin{align*} y \amp = \pm \sqrt{(x-3)^2(x-4)^2} = \pm (x-3)(x-4)\\ \frac{dy}{dx} \amp = \pm \frac{(x-3)(x-4)^2 + (x-3)^2(x-4)}{(x-3)(x-4)} \end{align*}

The singularities happen at the undefined points of the implicit derivative. These are the points where the denominator is zero. For this curve, the possible singularities are at \(x=3\) and \(x=4\text{.}\)

For each of the undefined points, I calculate the limit of the implicit derivative approaching the undefined points. I start with \(x=3\text{.}\)

\begin{align*} \lim_{x \rightarrow 3} \frac{dy}{dx} \amp = \lim_{x \rightarrow 3} \pm \frac{(x-3)(x-4)^2 + (x-3)^2(x-4)}{(x-3)(x-4)} \\ \amp = \lim_{x \rightarrow 3} \pm ((x-4) + (x-3) = \pm 1 \end{align*}

There are two limit values, so this is a double point node.

\begin{align*} \lim_{x \rightarrow 4} \frac{dy}{dx} \amp = \lim_{x \rightarrow 4} \pm \frac{(x-3)(x-4)^2 + (x-3)^2(x-4)}{(x-3)(x-4)} \\ \amp = \lim_{x \rightarrow 4} \pm ((x-4) + (x-3) = \pm 1 \end{align*}

Again, there are two limit values, so this is a double point node. Figure 2.3.3 shows the curve and both the double point nodes.

Figure 2.3.3.

Activity 2.3.4.

Find and classify the singularities of this algebraic plane curve.

\begin{equation*} y^2 = x(x-1)(x-2) \end{equation*}

Solution.

The curve is already factored. There are domain restrictions, since the right side cannot be negative. The right side is positive when \(0 \leq x \leq 1\) and \(x \geq 2\text{,}\) so those are the valid \(x\) coordinates for points on the curve. I calculate the implicit derivative of the curve.

\begin{align*} \frac{d}{dx} y^2 \amp = \frac{d}{dx} x(x-1)(x-2)\\ 2y \frac{dy}{dx} \amp = (x-1)(x-2) + x(x-2) + x(x-1) \\ \frac{dy}{dx} \amp = \frac{(x-1)(x-2) + x(x-2) + x(x-1)}{2y} \end{align*}

Then I use the original curve to solve for \(y\) and replace \(y\) in the implicit derivative.

\begin{align*} y \amp = \pm \sqrt{x(x-1)(x-2)} \\ \frac{dy}{dx} \amp = \pm \frac{(x-1)(x-2) + x(x-2) + x(x-1)}{2\sqrt{x(x-1)(x-2)}} \end{align*}

For each of the undefined points, I calculate the limit of the implicit derivative approaching the undefined points. I’ll start with \(x=0\text{.}\) Due to the domain restrictions, I only approach from the right.

\begin{align*} \amp \lim_{x \rightarrow 0^+} \frac{dy}{dx} = \frac{1}{2} \lim_{x \rightarrow 0^+} \pm \frac{(x-1)(x-2) + x(x-2) + x(x-1)}{\sqrt{x(x-1)(x-2)}}\\ \amp = \frac{1}{2} \lim_{x \rightarrow 0^+} \pm \left[ \frac{(x-1)(x-2)}{\sqrt{x(x-1)(x-2)}} + \frac{x(x-2)}{\sqrt{x(x-1)(x-2)}} + \frac{x(x-1)}{\sqrt{x(x-1)(x-2)}} \right] \\ \amp = \frac{1}{2} \lim_{x \rightarrow 0^+} \pm \left[ \frac{\sqrt{(x-1)(x-2)}}{\sqrt{x}} + \sqrt{\frac{x(x-2)}{x-1}} + \sqrt{\frac{x(x-1)}{x-2}} \right]\\ \amp = \pm \frac{1}{2} \left[ \lim_{x \rightarrow 0^+} \frac{\sqrt{(x-1)(x-2)}}{\sqrt{x}} + \lim_{x \rightarrow 0^+} \sqrt{\frac{x(x-2)}{x-1}} + \lim_{x \rightarrow 0^+} \sqrt{\frac{x(x-1)}{x-2}} \right] \end{align*}

The second and third limits are zero. In the first limit, the numerator approaches \(2\) and the denominator approaches \(0\text{,}\) so I conclude the limit is \(\pm \infty\) and that there is a vertical tangent at \(x=0\text{.}\) I’ll next look at \(x=1\text{.}\) Due to the domain restrictions, I only approach from the left.

\begin{align*} \lim_{x \rightarrow 1^-} \frac{dy}{dx} \amp = \frac{1}{2} \lim_{x \rightarrow 1^-} \pm \frac{(x-1)(x-2) + x(x-2) + x(x-1)}{\sqrt{x(x-1)(x-2)}}\\ \amp = \pm \frac{1}{2} \left[ \lim_{x \rightarrow 1^-} \frac{\sqrt{(x-1)(x-2)}}{\sqrt{x}} + \lim_{x \rightarrow 1^-} \sqrt{\frac{x(x-2)}{x-1}} + \lim_{x \rightarrow 1^-} \sqrt{\frac{x(x-1)}{x-2}} \right] \end{align*}

The first and third limits are zero. In the second limit, the numerator approaches \(-1\) and the denominator approaches \(0\text{,}\) so I conclude the limit is \(\pm \infty\) and that there is a vertical tangent at \(x=0\text{.}\) The last point to consider is \(x=2\text{.}\) Due to the domain restrictions, I only approach from the right.

\begin{align*} \lim_{x \rightarrow 2^+} \frac{dy}{dx} \amp = \frac{1}{2} \lim_{x \rightarrow 2^+} \pm \frac{(x-1)(x-2) + x(x-2) + x(x-1)}{\sqrt{x(x-1)(x-2)}}\\ \amp = \pm \frac{1}{2} \left[ \lim_{x \rightarrow 2^+} \frac{\sqrt{(x-1)(x-2)}}{\sqrt{x}} + \lim_{x \rightarrow 2^+} \frac{\sqrt{x(x-2)}}{\sqrt{x-1}} + \lim_{x \rightarrow 2^+} \frac{\sqrt{x(x-1)}}{\sqrt{x-2}} \right] \end{align*}

The first and second limits are zero. In the third limit, the numerator approaches \(2\) and the denominator approaches \(0\text{,}\) so I conclude the limit is \(\pm \infty\) and that there is a vertical tangent at \(x=2\text{.}\) Figure 2.3.4 shows the curve and all three the vertical tangents. The curve has no singularities. (This question also has subtleties in how the square roots are written. You might notice in the first two calculation, the square root need to capture the whole fraction as one, while in the last calculation, the squares can be separated in the numerator and denominator. This is all done to make the square root well-defined, so that there is no risk of taking the square root of a negative number in any of these limits.)

Figure 2.3.4.

Subsection 2.3.2 Conceptual Review Questions

What are the ways that an implicit derivative can be undefined?
What is a singularity?

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