How to check if two functions only touch in one point?

Question

I have two functions $f(x,y)=x^4+y^4 -xy$ and $g(x,y) = x^2$. I know that these two functions "touch" at $(x,y,z) = (0,0,0)$

My question is, how do I know if this the only point where the two functions touch? Is it the only point or does a separation curve exist?

I thought abour setting $f(x,y) - g(x,y)=0$ and then checking thorugh the theorem of implicit function if I can express $y$ as $y(x)$ or $x$ as $x(y)$.

Would be great if someone could help me out with this one.

Answer 1

Suppose $(x,y)\in\mathbb{R}^2$ is a point other than $(0,0)$ satisfying the equation $$ x^4+y^4-xy=x^2 $$ Necessarily $x\ne 0$.

Letting $m={\large{\frac{y}{x}}}$, the equation reduces to $$ x^2(1+m^4)=1+m $$ hence for any $m > -1$ we get points $(x,y)$ given by $$ \left\lbrace \begin{align*} x&=\pm\sqrt{\frac{1+m}{1+m^4}}\\[4pt] y&=mx\\[4pt] \end{align*} \right. $$ Note that allowing $m=-1$ in the above would yield $(x,y)=(0,0)$, hence the set of points $(x,y)\in\mathbb{R}^2$ satisfying the given equation is the set of points $(x,y)$ parameterized by $m$ as specified above, where $m\ge -1$.

In particular, there are infinitely many solutions.

But note:

The phrase "the points where two surfaces touch" is not always interpreted as meaning the same as "the points where two surfaces meet". Often it's intended to mean "the points where two surfaces meet and are mutually tangent (i.e., have the same tangent plane)".

With that interpretation, suppose $(a,b)$ is a point where for the two surfaces $$ \left\lbrace \begin{align*} z&=f(x,y)=x^4+y^4-xy\\[4pt] z&=g(x,y)=x^2\\[4pt] \end{align*} \right. $$ "touch".

By standard multivariable calculus methods

• The tangent plane at $(a,b)$ to the surface $z=f(x,y)$ has normal vector $\langle{b-4a^3,a-4b^3,1}\rangle$.$\\[4pt]$
• The tangent plane at $(a,b)$ to the surface $z=g(x,y)$ has normal vector $\langle{-2a,0,1}\rangle$.

hence, in order for the surfaces to meet and the tangent planes to be the same, we must have $$ \left\lbrace \begin{align*} a^4+b^4-ab&=a^2\\[4pt] b-4a^3&=-2a\\[4pt] a-4b^3&=0\\[4pt] \end{align*} \right. $$ If $b=0$ then $a=0$, which yields the solution $(a,b)=(0,0)$.

Suppose we have a solution $(a,b)$ with $b\ne 0$.

Replacing $a$ by $4b^3$, the above system reduces to $$ \left\lbrace \begin{align*} 256b^8-3&=16b^2\\[4pt] -256b^8+1&=-8b^2\\[4pt] \end{align*} \right. $$ which sums to $$8b^2=-2$$ contradiction, since $8b^2$ can't be negative.

Thus the only solution is $(a,b)=(0,0)$.

Thus while the point $(0,0)$ is one of infinitely many points where the two surfaces meet, it's the only point where the two surfaces "touch" (with the mutually tangent interpretation of "touch").

Answer 2

Well, the statement is not true... The picture below represents the set of points where both functions have the same value. It is a curve defined implicitly by $x^4+y^4-xy-x^2=0$.

Answer 3

Let's take "polar" coordinates for intersection curve projection: $$\begin{array}{}x=r\sqrt{\cos \phi} \\ y = r \sqrt{\sin \phi} \end{array}$$ we will have $$x^4+y^4-xy-x^2=0 \to r^2 = \sqrt{\sin \phi \cos \phi} - \cos \phi$$ From where you see whole curve for intersection projection.