A logo has been created for a company. The logo can be modelled by the following function:
$$x^2+y^3-5xy = 0$$
Here is a graph on Desmos for visual aid:
This logo has been printed on paper and needs to be cut out. For simplicity, the cut will happen along specific tangents to the logo, where they are only horizontal or vertical. Find the equation of these tangents that need to be cut along in Quadrant 1.
My attempt:
- To begin, I noted that horizontal gradients are $0$ and vertical gradients are infinite. I found the derivative by implicitly differentiating with respect to $x$ and the product rule of the term $-5xy$
So, we have:
- $\frac{d}{dx}(x^2 + y^3 -5xy) = 0$
- $2x + 3y^2(\frac{dy}{dx}) + y(-5) + (-5x)(\frac{d}{dx}(y))$
Then by factoring the derivative on one side:
- $(3y^2 - 5x)(\frac{dy}{dx}) = 5y - 2x$
- $\frac{dy}{dx} = \frac{5y-2x}{3y^2-5x}$
Ok so here is where I am stuck. I was originally hoping that one of the terms would cancel out, such as the $y$, so it would be easier to solve for $x$ by equating the derivative to $0$ or infinity(by a limit), but it did not turn out that way. So, I am looking for some suggestions or directions that will help me. Thanks!
From your implicit derivative, $ \ \frac{dy}{dx} \ = \ \frac{5y-2x}{3y^2-5x} \ \ , \ $ you have found that horizontal tangents occur where the numerator is zero, or where $ \ 5y \ = \ 2x \ \ , $ and vertical tangents occur where the denominator is zero, or where $ \ 3y^2 \ = \ 5x \ \ . $ This is, of course, provided that numerator and denominator are not both zero at a particular point; we will consider this in a moment.
So you have presumably already determined that a horizontal tangent occurs for $$ x^2 \ + \ \left(\frac25 \ x \right)^3 \ - \ 5x·\left(\frac25 \ x \right) \ \ = \ \ \frac{8}{125} \ x^3 \ - \ x^2 \ \ = \ \ x^2·\left(\frac{8}{125} \ x \ - \ 1 \right) \ \ = \ \ 0 $$ $$ \Rightarrow \ \ x \ = \ 0 \ \ , \ \ y \ \ = \ \ \frac25 · 0 \ \ \ ; \ \ \ x \ = \ \frac{125}{8} \ \ = \ \ 15.625 \ \ , \ \ y \ \ = \ \ \frac25 · \frac{125}{8} \ \ = \ \ \frac{25}{4} \ \ = \ \ 6.25 \ \ . $$
The latter point is marked on the curve in the graph below by a blue dot, where the line $ \ 5y \ = \ 2x \ $ intersects the curve. We'll say more about $ \ (0 \ , \ 0) \ $ shortly.
For vertical tangents, we have $$ \left(\frac35 \ y^2 \right)^2 \ + \ y^3 \ - \ 5·\left(\frac35 \ y^2 \right)·y \ \ = \ \ \frac{9}{25} \ y^4 \ - \ 2y^3 \ \ = \ \ y^3·\left(\frac{9}{25} \ y \ - \ 2 \right) \ \ = \ \ 0 $$ $$ \Rightarrow \ \ y \ = \ 0 \ \ , \ \ x \ \ = \ \ \frac35 · 0^2 \ \ \ ; \ \ \ y \ = \ \frac{50}{9} \ \ \approx \ \ 5.56 \ \ , \ \ x \ \ = \ \ \frac35 · \left(\frac{50}{9} \right)^2 \ \ = \ \ \frac{500}{27} \ \ \approx \ \ 18.52 \ \ . $$
The horizontal parabola $ \ 3y^2 \ = \ 5x \ $ is overlain in green on our curve, with the intersection indicated with a green dot. Again, the origin turns up as a solution, which suggests something peculiar is happening there (and is the main reason for my writing an answer).
The problem at the origin is that our implicit derivative $ \ \frac{5y-2x}{3y^2-5x} \ $ is indeterminate there; this is why we are supposedly finding horizontal and vertical tangents at that point on the curve. Is any of that correct? Viewing the graph suggest that there is a tangent line with a finite, non-zero slope there, so how are we to account for that?
The way we will eventually learn to work with this in calculus is to "parameterize" the curve by introducing a variable which "traces out" the curve as we change its value. We will then find that there are two values of the parameter at the origin, each one with a different "slope" for the tangent line. The indeterminacy of the derivative you obtained is often a sign of a self-intersection of the curve at that point; the parameterization offers a way to investigate the curve's behavior there.
But what are we to do at this stage of our studies? One method would be to "extract" the explicit function curves that are assembled in the given curve. Fortunately, the curve equation is not overly complicated and can be "solved" by the quadratic formula:
$$ x^2 \ - \ 5y·x \ + \ y^3 \ \ = \ \ 0 \ \ \rightarrow \ \ x \ \ = \ \ \frac{5y}{2} \ \pm \ \frac{\sqrt{25y^2 \ - \ 4y^3}}{2} \ \ , $$
showing us that our curve is comprised of two curves with $ \ x \ $ as a function of $ \ y \ \ . $ In the graph below, we see that these "explicit function" curves are separated by the line $ \ x \ = \ \frac52 \ y \ \ , $ with the green portion representing $ \ x \ \ = \ \ \frac{5y}{2} \ + \ \frac{\sqrt{25y^2 \ - \ 4y^3}}{2} \ \ $ and the red part representing $ \ x \ \ = \ \ \frac{5y}{2} \ - \ \frac{\sqrt{25y^2 \ - \ 4y^3}}{2} \ \ . $ These present us with the derivatives $$ \frac{dx}{dy} \ \ = \ \ \frac52 \ \pm \ \frac{50y \ - \ 12y^2}{4·\sqrt{25y^2 \ - \ 4y^3}} \ \ . $$ We find that there are two "directional limits" as $ \ y \ \rightarrow \ 0 \ \ , \ \frac{dx}{dy} \ = \ \frac52 \ \pm \ \frac52 \ \ . $ So there is a tangent line with finite slope $ \ y \ = \ \frac15 \ x \ $ at the origin, but also a "vertical tangent" there $ \ ( \ \frac{dx}{dy} \ = \ 0 \ \Rightarrow \ \frac{dy}{dx} \ $ undefined ).