The curve defined implicitly by the equation $xy^3+x^3y=4$ has no horizontal tangent.

Question

The curve defined implicitly by the equation $$xy^3+x^3y=4,$$ has no horizontal tangent.

In the solution they did as follow :

$$xy^3+x^3y=4\implies y^3+x3y^2y'+3x^2y+x^3y'=0\implies y'=\frac{3x^2y-y^3}{3xy^2+x^3}.$$

Question

why they can say that $y=y(x)$ ? It looks a bit strange for me. In other word, why can I say that $$xy^3+x^3y=4\implies \exists y\in \mathcal C^1(\mathbb R): xy(x)^2+x^3y(x)=4\ \ ?$$

Answer 1

Let $f(x,y)=x^3y+xy^3$. Then $\nabla f(x,y)=\bigl(3(x^2+y^2)y,3(x^2+y^2)x\bigr)$. So, $\nabla f(x,y)=(0,0)$ at $(0,0)$ and only there. But $(0,0)$ does not belong to your curve. So, at a point $(x_0,y_0)$ of the curve with a horizontal tangent, since $\nabla f(x_0,y_0)\ne(0,0)$ but the first component of $\nabla f(x_0,y_0)$ is $0$ (since $\nabla f(x_0,y_0)$ is orthogonal to the tangent vector), then the second component is not $0$. Therefore, the implicit function theorem allows you to deduce that, near $(x_0,y_0)$, $y$ can be expressed as a function of $x$.

Answer 2

If $L\to y = \lambda$ is a tangent line to $g(x,y)=0$ then the system

$$ \cases{ y = \lambda\\ g(x,y) = 0 } $$

should have a double root regarding $x$ or

$$ \frac {dg}{dx}(x,\lambda)=0 $$

should have necessarily a real root for $x$. In our case

$$ \frac {dg}{dx}(x,\lambda) = \lambda^3+3\lambda x^2 = 0 $$

now discarding $y = \lambda= 0$ for obvious reasons, we have

$$ \lambda^2+3x^2=0 $$

with no real roots.

Answer 3

What does “the curve implicitly defined by $f(x,y)=0$” mean?

I find it misleading terminology: we're dealing with the set of points in $2$-space $\Gamma=\{(x,y):f(x,y)=0\}$. Under suitable assumptions, if $(x_0,y_0)\in\Gamma$ we can find a function $\varphi$, defined over a neighborhood $U$ of $x_0$, such that, for every $x\in U$, $f(x,\varphi(x))=0$. Such a function $\varphi$ is a function (or curve, if you prefer), implicitly defined by $f(x,y)=0$.

Under suitable assumptions, the function will exist and be differentiable, see the Implicit function theorem. But take into consideration that there can be several such functions: for instance, with $f(x,y)=x^2+y^2-1$, when considering the point $(0,1)$ we have $\varphi(x)=\sqrt{1-x^2}$; around $(0,-1)$, we have $\varphi(x)=-\sqrt{1-x^2}$; and the function does not exist around $(1,0)$ or $(-1,0)$, but we can invert using $x$ and $y$ the other way around.

In your case, we have $f(x,y)=xy^3+x^3y-4$. Therefore $$ x(\varphi(x))^3+x^3\varphi(x)-4=0 $$ and, assuming $\varphi$ is differentiable, we can differentiate both sides; since the derivative of a constant is zero, we get $$ (\varphi(x))^3+x(\varphi(x))^2\varphi'(x)+3x^2\varphi(x)+x^3\varphi'(x)=0 $$ In order to save space, it is customary to write $y$ instead of $\varphi(x)$, so $$ y^3+xy^2y'+3x^2y+x^3y'=0 $$ and thus $$ y'=-\frac{3x^2y+y^3}{x^3+xy^2} $$ Just think to have $\varphi(x)$ in place of $y$.

If you look at the implicit function theorem, you'll see that in the case of a horizontal tangent, the function exists around the point of tangency. So the derivative should be zero and the condition becomes $$ y(3x^2+y^2)=0 $$ The factor $3x^2+y^2$ doesn't vanish on the curve $\Gamma$, so the condition is $y=0$. However, no point with $y=0$ belongs to $\Gamma$.