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Context

The context of the problem is the derivative doesn't exist at the point $(0,0)$ if you look at the curve as is in the $xy$ plane. It an intersection point. So by parametrizing the curve we can determine a suitable $t$ to determine appropriate slope given a particular $t$

Question

I'm definitely struggling with this one. The obvious $x=t$ doesn't work here as you can't write the equations as a function of $y$. I thought maybe a substitution may work but the lack of symmetry here feels like I can't actually parametrize it.

Math Attack
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Mando
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    What is the context of this problem? What parametrization techniques have you seen? – Greg Martin Jan 08 '24 at 17:12
  • The context of the problem is the derivative doesn't exist at the point (0,0) if you look at the curve as is in the xy plane. It an intersection point. So by parametrizing the curve we can determine a suitable "t" to determine appropriate slope given a particular "t". – Mando Jan 08 '24 at 17:15
  • I know of basic parametrization techniques using substitution, with the obvious being for any function, take x = t and y = f(t). But this is not a function, not a circle, so the cost and sint doesn't work. – Mando Jan 08 '24 at 17:31
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    @Mando I edited your question so that there is a little context (which is what you provided in the comments) – Math Attack Jan 08 '24 at 17:41
  • Thank you! I appreciate it. It's been a while since I posted here. – Mando Jan 08 '24 at 17:42

2 Answers2

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Using polar coordinate: $$x=\rho\cos(\theta)\qquad y=\rho\sin(\theta)$$

$$y^{2}-2x^{2}=x^{3}-2y^{3}$$ $$\rho^{2}\sin\left(\theta\right)^{2}-2\rho^{2}\cos\left(\theta\right)^{2}=\rho^{3}\cos\left(\theta\right)^{3}-2\rho^{3}\sin\left(\theta\right)^{3}$$ $$\sin\left(\theta\right)^{2}-2\cos\left(\theta\right)^{2}=\rho\cos\left(\theta\right)^{3}-2\rho\sin\left(\theta\right)^{3}$$ $$\rho=\frac{\sin\left(\theta\right)^{2}-2\cos\left(\theta\right)^{2}}{\cos\left(\theta\right)^{3}-2\sin\left(\theta\right)^{3}}\qquad \theta\in[0,\pi]$$

If $\theta=\arctan\left(\frac{t}{2}\right)$ you have: $$\left(\frac{\sin\left(\theta\right)^{2}-2\cos\left(\theta\right)^{2}}{\cos\left(\theta\right)^{3}-2\sin\left(\theta\right)^{3}}\cdot\cos\left(\theta\right),\frac{\sin\left(t\right)^{2}-2\cos\left(t\right)^{2}}{\cos\left(t\right)^{3}-2\sin\left(t\right)^{3}}\cdot\sin\left(\theta\right)\right)$$ $$\left(\frac{8-t^{2}}{t^{3}-4},\frac{t}{2}\cdot\frac{8-t^{2}}{t^{3}-4}\right)$$

The graph here for you

Math Attack
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You can also use the substitution $y=tx$ to get a rational parametrisation:

\begin{align*} x&=\frac{t^2-2}{1-2t^3}\\ y&=\frac{(t^2-2)t}{1-2t^3} \end{align*}

Note the origin $x=y=0$ corresponds to both $t=\sqrt{2}$ and $t=-\sqrt{2}$.

One can think of this as locating the intersection of the line $y=tx$ with the curve, and since there is always a double intersection at the origin, one gets a unique `third' intersection with the cubic curve for each $t$ by Bezout's theorem

Now you get

\begin{align*} dx&=\frac{2t(1-2t^3)+6(t^2-2)t^2}{(1-2t^3)^2}dt\\ dy&=\frac{(3t^2-2)(1-2t^3)+6t(t^2-2)t^2}{(1-2t^3)^2}dt\\ \implies\frac{dy}{dx}&=\frac{(3t^2-2)(1-2t^3)+6t(t^2-2)t^2}{2t(1-2t^3)+6(t^2-2)t^2} \end{align*}

Joshua Tilley
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