# finding position vector $r$ when $r'$ is not in terms of a single parameter

I have to find the path of quickest decent for the surface $$f(x,y)=50-x^2-2y^2$$ so I find $\nabla f$ the gradient vector and at any point ill head in the opposite direction of $\nabla f$. Now to find the curve. From the above I find $r' =2x \mathbf{i} + 4y\mathbf{j}$.

Now to find $r$ I have said assume x and y to be functions of $t$ then ye can get $$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dx}=\frac{2y}{x}$$ which I can solve given some initial point.

Two questions. Is this correct? I seen something similar and its the only way I can see how to do this. If it is right can someone explain how this works or what is going on at and after the point I assume both to be a function of t

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Can you clarify what $r$ is? –  Shiyu Oct 6 '12 at 12:34
Sorry the position vector. –  Ben Davidson Oct 6 '12 at 12:40
I think you found right direction $-\nabla f$, but now you need to find $f(x,y)$ or to say it other way, $z$, for this direction to get the curve equation on the surface. –  Mykolas Oct 6 '12 at 19:12