by Juntao
 P: 45 Consider the function z=f(x,y). If you start at the point (4,5) and move toward the point (5,6), the direction derivative is sqrt(2). Starting at (4,5) and moving toward (6,6), the directional derivative is sqrt(5). Find gradient f at (4,5). Okay, this is probably a simple problem, but I don't know how to start it. Help appreciated.
 Emeritus Sci Advisor PF Gold P: 16,101 Do you know a formula that relates gradients and directional derivatives? What do you get when you plug what you know into that formula? (represent unknowns with variables)
 P: 45 Forumula...*scratches head* Um, I think this is the one, hopefully: The derivative of f at Po in the direction of u= gradient f dot with u = magnitude grad f * magnitude u * cos theta = magnitude grad f * cos theta. Or like D_u f= grad f (dot) u = |grad f| |u| cos theta = |grad f| cos theta Maybe that'll help me out?
Math
Emeritus
Thanks
PF Gold
P: 38,706

 D_u f= grad f (dot) u
Yes, that's true as long as u is a unit vector (has length 1).

Another way of saying this is that the directional derivative is
f_x u_x+ f_y u_y where f_x is the partial derivative and u_x is the x component of unit vector u.

Find unit vectors in both the given directions and write out
f_x u_x+ f_y u_y= &radic;(2) and

That gives you two equations for f_x and f_y.
Emeritus