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?
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