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Imagine you have a terrain which height is determined by the function $$H(x,y)=1500e^{-x^2-2y^2}$$ If we let a ball down the top of the highest mountain, and we know it will always follow the maximum slope, prove the function it will follow will be $y=x^2$.

The highest mountain will be the point $(0,0)$, with a height of $1500$.
The parcial derivatives are : $$\frac{\partial H}{\partial x}=-1500·2x·e^{-x^2-2y^2}$$ $$\frac{\partial H}{\partial y}=-1500·4y·e^{-x^2-2y^2}$$ And so, the gradient is $$∇(x,y)=(-3000xe^{-x^2-2y^2},-6000ye^{-x^2-2y^2})$$ This is where I get stuck. Am I on the good way? What would be the next step?

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    if $y = x^2$, then the direction of the movement becomes 2x expressed as the y component of grad divided by the x component of grad. But i'm not sure how the ball ever gets moving if grad is the zero vector at the origin - what direction is it going to set off in? any way, 2x is of course the derivative of $y=x^2$ so you seem to be on good lines – 2017-01-26
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    I guess a tiny push it's given to start it moving because the excercice wouldn't have muche sense otherwise. :') – 2017-01-26
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    Anyway, I could divide both terms of the gradient by $−3000e^{−x2−2y2}$ and I would have then $∇(x,y)=(x,2y)$, but how do I reach $y=x^2$? – 2017-01-26
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    grad is a vector, if $y = x^2$ then the vector is (x, 2x^2) - then that gives a derivative value of $2x^2 / x = 2x$ if you consider the definition of the tangent to a graph – 2017-01-26
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    What if it didn't say that the function is $y=x^2$ and I had to find it from the function? – 2017-01-26
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    from your dividing by $−3000e^{−x2−2y2}$ then you can get dy / dx = 2y / x leads to (1/2)ln(y) = ln(x) which leads to y = Ax^2 as general solution - sorry I can't see what boundary condition you have to get A - as I say, I can't understand what direction it would roll off in – 2017-01-26

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