No, the problem asked for "the directional derivative of f at (3,1) in the direction towards the point (-2,3)". The vector u is <-2- 3, 1- 3>= <-5, -2>.b) u = (-2,3) ; u = 1/sqrt(2^2+3^2) = -2/sqrt(13), 3/sqrt(13) or -1/sqrt(13)
You did not answer either part of (c)! Yes, the greatest increase is in the direction of the gradient. The gradient at (3, 1) is, as you say, <13, 15>. What is a unit vector in that direction?Directional derivative: (-2/sqrt(13))(13) + (3/sqrt(13))(15) = -3.05
So, the function in the direction of (3,1) is decreasing, since the rate of change is negative.
c) gradf(3,1) = 13i+15j = (13, 15) [From what I understand, the greatest increase of instantaneous change is in the direction of the gradient, so I'm assuming this is what's being looked for in terms of an answer.]
For the rate to be zero: gradf(3,1) (dot product) u = 0, so 13(x-3)+15(y-1)=0.
13x+15y=54, so anywhere on this plane is where the instantaneous rate is 0.
Did I make any mistakes?
The rate will be 0 in a direction perpendicular to the gradient but your answer should not be a plane, but a unit vector <a, b> such that 13a+ 15b= 0. (There are actually two correct answers to this, one being the negative of the other.)