
#1
Nov303, 08:31 PM

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. 



#2
Nov303, 09:57 PM

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) 



#3
Nov303, 11:22 PM

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? 



#4
Nov403, 06:08 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

gradient helpAnother 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= √(2) and f_x v_x+ f_y v_y= √(5). That gives you two equations for f_x and f_y. 


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