The function $f:\mathbb{R}^2 \to \mathbb{R}$ define by $$f(x,y):= x^4 - 6x^2y^2+y^4-2x^2+2y^2.$$
Suppose a particule moves on the surface $z=f(x,y)$ as it progresses always in the same direction where $f$ decrease the faster.
If at the time $t=0$ the position of the particule is $(x,y,z)=(2,1,f(2,1))$, in which direction will it be initialy?
Show that in general the particule will follow a path along which $$xy(x^2-y^2-1)= \alpha.$$
Where $\alpha$ is a constant.
In (1), I suppose I have to calculate the gradient and evaluate it a the initial position of the particule,
$$ \nabla{f}:= (4x^3-12xy^2-4x,-12x^2y+4y^3+4y)$$
Then,
$$\nabla{f}(2,1)=(4,-40) = (1,-10)$$
How can I do the (2)?
Let's express this curve in this way
$$y=y(x)$$
If we write this curve in a parametric way, the problem will be so complicated. Instead, I avoid using parameter $t$.
$$(x'(t),y'(t))=-\vec\nabla f(x,y)$$
so
$$\frac{dy}{dt}=-\frac{\partial f}{\partial y}$$ $$\frac{dx}{dt}=-\frac{\partial f}{\partial x}$$
To remove $t$:
$$\frac{dy}{dx}={~~~\frac{\partial f}{\partial y} ~~~\over ~~~\frac{\partial f}{\partial x}~~~}$$
$$\frac{dy}{dx}=\frac{-12x^2y+4y^3+4y}{4x^3-12xy^2-4x}=\frac{-4x^2y+y^3+y}{x^3-3xy^2-x}$$
expansion and re-arranging:
$$(-3x^2y+y^3+y)=y'(x^3-3xy^2-x)$$
$$(-3x^2y-y' x^3)+(y^3+3xy^2y')+(y+y'x)=$$ $$(-y x^3)'+(y^3x)'+(yx)'=0$$ $$(-y x^3+y^3x+yx)'=0$$
$$-y x^3+y^3x+yx =C$$ $$xy(- x^2+y^2+1) =C$$
$$xy(x^2-y^2-1) =-C$$