How to validate if the horizontal and vertical tangent lines exist for implicit functions?
Given an implicit function $x^2 + y^3 - 15xy = 0$, I needed to calculate the horizontal tangent line and the vertical tangent line. When we calculate the derivative from this function we get:$$\frac{dy}{dx}= \frac{15y-2x}{3y^2-15x}$$ So mine approach of calculating the vertical tangent line was to set the denominator to 0 or $dx$ that approaches 0. Since $dx$ will approach 0, $\frac{dy}{dx}$ will become $\pm\infty$. That means that we need to solve $3y^2 - 15x = 0$. By solving for $x$, we obtain $x = \frac{y^2}{5}$. Substitute this back to the function we have $(\frac{y^2}{5})^2 + y^3 - 15(\frac{y^2}{5})y = 0$. Simplifying and solving for $y$, we find the solutions $y = 0\vee y = 50$. For the corresponding x-values, we get $x = 0$ and $x = 500$ respectively (although $x = 250$ was also a solution, but this wasn't valid because plugging it back into $\frac{dy}{dx}$ will result $\frac{1}{15}$ which is not $\pm\infty$).
But I have seen that using $\frac{dx}{dy}=0$ also works. This means that in this case we should differentiate with respect to $y$ instead of $x$. Differentiating the function with respect to $y$ gives: $$\frac{d}{dy} \left( x^2 + y^3 - 15xy \right) = 0$$ and this results in: $$\frac{dx}{dy} = \frac{15x - 3y^2}{2x - 15y}$$ And of course by solving $\frac{dx}{dy}=0$, we will eventually end up solving the same equation but only with an additional restriction where $2x - 15y\neq0$. $$$$
So my question for this is:
Why is using the $\frac{dx}{dy}=0$ method considered better? If I remember correctly I've heard that my calculus teacher said that using $\frac{dx}{dy}$ method is more formal and rigorous then doing $dx = 0$. Why is this the case? Are there any specific examples where this distinction becomes particularly relevant? $$$$
Then there is another problem.
Let me first explain how I calculated the horizontal tangent line. Since we want to know where the horizontal tangent line is, we simply solve $\frac{dy}{dx}=0$ where $3y^2-15x\neq0$. This will result $x=\frac{15y}{2}$. Substitute this back to the the original function, we obtain the equation $(\frac{15y}{2})^2 + y^3 - 15(\frac{15y}{2})y = 0$. Simplifying and solving this equation yields $y = 0\vee y = \frac{225}{4}$. Substituting these y-values into the original function returns the corresponding x-values: $x = 0$ and $x = \frac{3375}{8}$ respectively. However, since there was a restriction $3y^2-15x\neq0$, the solution $y = 0$ does not hold. Therefore, at $y = 0$ there is no horizontal tangent line.
Since we didn't verify this when we calculated the vertical tangent line it should also be checked. The solutions for the vertical tangent line were $y = 0$ and $y = 500$. In this case the solution $y = 0$ also does not hold, since we had a restriction $2x - 15y\neq0$. That should mean that based on the calculations, there is only 1 horizontal and 1 vertical tangent line. BUT if we graph the function we can see that there is actually 3 tangent lines (2 vertical and 1 horizontal)!
And more surprisingly, the extra vertical tangent line is at the point y = 0!
Which makes it very confusing and suspicious, where one would think that there might be some mistakes in my calculations. But even after looking at the steps I did, I didn't see any mistakes or problems that might be the case.
This leads to my second question:
Why or how can this happen? Why is there actually a valid vertical tangent? Whereas there is no horizontal tangent?
1 answer
Answering your questions a bit out of order, I'll start with the "non-rigorousness" of talking about $dx$ by itself. While this seems to be commonly poorly explained, the derivative (of a function from and to reals) is an operation that takes functions to functions. Let's consider the typical high school definition of the derivative: $$\frac{df}{dx}(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$
This takes a function $f$ and a number $x$, and gives you a new number $\lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$. There is no $dx$ piece you can take off. To this end, Newton's $\dot f$ notation is more honest and coherent. To rationalize, "setting $dx=0$", you could have just as well said you were setting $\frac{dx}{dy} = 0$. Indeed, that's exactly what you do in the following part. "Intuitively", assuming $b\neq 0$ so $a/b$ is well-defined, then $a/b = 0$ if and only if $a = 0$. In Leibniz' notation, we have $dx/dy = \frac{1}{dy/dx}$ or $\frac{dy}{dx}\frac{dx}{dy} = 1$. This is actually hiding a fairly important and non-trivial result known as the inverse function theorem. Informally, the inverse function theorem states that if a function has a non-zero derivative in some sufficiently small open interval around some point, then that function has an inverse on that interval (whose derivative is necessarily the reciprocal of the original function). This is why we can talk about $x$ as a function of $y$ and take its derivative.
Ultimately, I think Leibniz notation is extremely misleading and that it doesn't scale well to higher dimensions. There's more I could say here, but I won't.
For your other questions, the first question you should ask yourself is "what is the definition of a tangent line?"
Intuitively speaking, there are (at least) two obvious ways to think about this for a curve.
One approach would be to parameterize the curve, i.e. think of it as a trajectory in "time". Given a well-behaved parameterization, our "velocity" along the trajectory, i.e. the derivative of the trajectory, will be a vector tangent to the curve. One property we want for the trajectory to be well-behaved is that its derivative is never zero. Thus, this gives us a non-zero tangent vector, and thus tangent line, at every point of the parameter space. If two points in the parameter space, i.e. in "time", map to the same point on the curve, then we could potentially have two (or more) tangent vectors at that point of the curve. Indeed, in the self-intersecting case you provide, this is what would happen.
The big downside of this approach is that we need a suitable parameterization. So another intuitive approach is to view the curve as a level curve or contour of a real function on the plane, $\varphi$. In your case, $\varphi(x,y) = x^2 + y^3 - 15xy$ and the curve is the level curve determined by $\varphi(x,y) = 0$. Intuitively, a tangent vector at a point would be a direction that keeps us on the curve, i.e. a direction in which $\varphi$ does not change. More formally, we could say it is the vectors $\mathbf v$ such that the directional derivative of $\varphi$ in the direction $\mathbf v$ (at a given point $(x,y)$) is $0$. Using the vector derivative, we can write the "derivative in the $\mathbf v$ direction" operator as $\mathbf v \cdot \nabla$, and thus we're asking for $\mathbf v \cdot \nabla\varphi = 0$. $\nabla \varphi$ is also the gradient of $\varphi$, and, in this case, we have that $(\mathbf v \cdot \nabla)\varphi = \mathbf v \cdot (\nabla \varphi)$. $\mathbf v \cdot (\nabla \varphi) = 0$ is the statement that $\mathbf v$ is in the orthogonal complement of the gradient of $\varphi$. The orthogonal complement to a vector in 2D is a line.
In this case, it is indeed the tangent line to the point on the curve. Intuitively, the gradient tells us the direction of greatest increase and its negative would be the direction of greatest decrease. We want the direction that makes no change. If we move in a direction with a non-zero projection onto the gradient, then we will increase or decrease the value of $\varphi$ and move off the curve. Thus any tangent vector must have zero projection onto the gradient, which is another way of saying it's in the orthogonal complement.
In some sense, what you're doing is this latter approach albeit you're hamstrung by the need to have $y$ be a function of $x$ or vice versa which adds complexity. The gradient of $\varphi$ is $(2x - 15y, 3y^2 - 15x)$. The vertical tangent lines will occur where the second component is zero (remember we want the orthogonal complement to this vector), and the horizontal tangent lines will occur where the first component is zero. Of course, there's nothing special about vertical and horizontal. We can compute the tangent line at any point on the curve by computing the orthogonal complement of this vector at that point. This is supposing that the gradient isn't zero which is what you're running into at the self-intersection at $(0,0)$.
It's questionable whether you'd want to consider any line a tangent line at a self-intersection like this, which is part of why I said you should think about what your definition of "tangent line" even is. Nevertheless, in this case one solution would be to take the limit as we approach this point. This doesn't help by itself as the limit is still zero. Continuity requires that the magnitude continuously decreases to zero to get a zero vector. But we only care about the direction, so we could instead take the limit of the normalized gradient, i.e. $\nabla\varphi / |\nabla \varphi|$. This leads to a discontinuity at $(0,0)$. In this case, this means the limit we get depends on how we approach it. For your purposes, you can then just take the limit with the $x$-component constant to test for vertical tangent lines, and again with the $y$-component constant to test for horizontal tangent lines. For this example, taking the limit with $x=0$ as $y$ approaches $0$ will lead to a normalized gradient with a non-zero $x$-component and zero $y$-component, and thus a vertical tangent. If we take the limit with $y=0$ as $x$ approaches $0$, we end up with a normalized gradient where neither component is zero, so neither a horizontal nor vertical tangent line.
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