I am told that a particle starting off at $(-1,1)$ moves parallel to $-\nabla f$ where $f(x,y)=x^2-3y^2$. So $-\nabla f=(-2x,6y)$. The question wants the path of the particle and I considered integrating the tangent vector to find the position but I dont know how to do this when we are given its position in $x,y$ coordinates.


1 Answer 1


You get two differential equations:

$$ \frac { dx } {dt} = -2x$$

$$ \frac { dy } {dt} = 6y$$

You solve these with initial conditions from the initial point

$$(x(t),y(t) )= ( -e^{-2t}, e^{6t} ) $$


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