I came across this example in an old book.



I have this question here. What is this point P, how is it defined? I did some calculations (implicit differentiation) and it seems to me it's the point where $x = \sqrt[3]{4}, y = \sqrt[6]{4}$. Why? Because it seems to me that's when the derivative $\frac{dy}{dx}$ is not defined. Is that correct?


1 Answer 1


Yes, that is the correct idea. At the point $P$, the derivative $dy/dx$ is undefined; or equivalently, the derivative $dx/dy = 0$. The function $f(x,y)$ satisfies reflection symmetry about $y = x$; i.e., $$f(y,x) = f(x,y)$$ for all $(x,y)$. Thus the reflection of $P$ about $y = x$ gives another point $P'$ whose derivative $dy/dx = 0$.

In any case, either point can be located via implicit differentiation. To locate $P$, we differentiate $f = 0$ implicitly with respect to $y$: $$0 = \frac{df}{dy} = \frac{d}{dy}\left[x^3 + y^3 - 3xy\right] = 3x^2 \frac{dx}{dy} + 3y^2 - 3y \frac{dx}{dy} - 3x,$$ hence $$\frac{dx}{dy} = \frac{x-y^2}{x^2-y}.$$ Then $P$ satisfies $dx/dy = 0$, i.e. $x = y^2$ under the condition $x^2 \ne y$ (since if the denominator also equals zero, then $dx/dy$ is indeterminate, not zero). So we know $P$ lies on a parabola. The intersection of this parabola with $f = 0$ is obtained by substitution: $$0 = f(x,y) = f(y^2,y) = (y^2)^3 + y^3 - 3(y^2)y = y^6 - 2y^3 = y^3(y^3 - 2),$$ hence $y \in \{0, 2^{1/3}\}$ and we have the candidate solutions $$(x,y) \in \{(0,0), (2^{2/3}, 2^{1/3})\}.$$ But as stated before, we also require $x^2 \ne y$, which excludes the candidate $(0,0)$. So the unique solution is $P = (2^{2/3}, 2^{1/3})$. We also have by symmetry $P' = (2^{1/3}, 2^{2/3})$, the unique nontrivial horizontal tangent to $f = 0$.

As a further exercise, can you locate the points $Q$, $Q'$ such that the tangent to $f = 0$ is parallel to the line $y = x$? What does this say about the "diagonal width" of the loop?


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