1. Homework Statement
Find the steepest path up a hill as an expression in terms of x and y whose height is given by f(x,y)=x^2y-2xy+5 starting at (2,1)

2. The attempt at a solution

I know that I need to get a set of parametric equations for x and y because I have done problems sort of like this before, but those always had only one variable in the partials. I'm unsure of how to get these equations when dz/dx=2xy-2y and dz/dy=x^2-2x.

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You know that the gradient will give you the path of steepest increase/decent correct? That is half of the battle already. Now just take the gradient of the equation and evaluate it at that point, which will give you your direction. If you know how to parametrize the path, aka f(x+tv), then you should be good to go. If you don't know how to parametrize the path, let us know, and we'll walk you through it.

Well, I know that the gradient will give me the gradient at that point, but I need to find the curve that will be followed continually moving along the gradient at any point. I'm not sure what you mean by f(x+vt) (it may just be that this is something we haven't covered yet in class. It's an extra credit problem and the professor likes to do that sometimes.)

So, if you could walk me through that that would be great.

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By path do you mean another real-valued function just like f(x,y) in this example, but just like a 3-D version of a derivative curve in 2-D? I'm just trying to figure out exactly what your professor wants you to do.

I think that that's pretty much what he wants. I asked him in class the other day and I think he said it could be done by solving dy/dx=(dy/dt)/(dx/dt) but I'm not sure how to get to the parametric equations with more than one variable in the partials

I believe that you would need to parametrize the curve, then compose it with the gradient vector field.

HallsofIvy
If $\nabla F= f(x,y)\vec{i}+ g(x,y)\vec{j}$, then the path having that vector as tangent vector, with parameter t, must satisfy
$$\frac{dx}{dt}= f(x,y)$$