Max rate of descent , gradient?

[wxyj]

Can anyone help me with the following question?
Find the path of the steepest descent along the surface z=x^3 + 3y^2 from the point (1, -2, 13) to (0,0,0)
Note: the general solution of the differential equation f ' (t)-kf(t) =0 is
f(t) = ce^kt, where c is an arbitary number
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I understand that max rate of change occurs in the direction of the gradient, and max descent occurs in the -gradient(f) direction

But I don't know what to do with the 2 points..and how it has anything to do with the formula given as a hint..

Can anyone help
Thanks so much

 

[viscousflow]

Research something called directional derivative.

 

[wxyj]

Research something called directional derivative.

I do know what that directional derivative is, but can u give a more concrete suggestion?

 

[zcd]

\nabla{f}= \frac{\partial f}{\partial x} \hat{i}+\frac{\partial f}{\partial y} \hat{j}, and \hat{u}=-\frac{1}{\sqrt{5}}\hat{i}+\frac{2}{\sqrt{5}}\hat{j} for all values along the path.

 

[wxyj]

\nabla{f}= \frac{\partial f}{\partial x} \hat{i}+\frac{\partial f}{\partial y} \hat{j}, and \hat{u}=-\frac{1}{\sqrt{5}}\hat{i}+\frac{2}{\sqrt{5}}\hat{j} for all values along the path.

thanks , but how did you get the answer??

I don't undestand it...

 

[zcd]

That's not the answer, but you can get direction by subtracting P-P₀ and dividing by its magnitude (to make it a unit vector).