webwork problem and the rules

Hello Prof. Taylor, I have worked out this webwork problem and am doing everything right, but there only two parts of the problem which are refusing.

I thought the two part just required finding the Magnitude of the vector above them, but I am getting it wrong. Some people got theirs right, so I don't understand why my answer is rejected. I therefore need your help!

Below is the problem;

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OK, 

1) first of all, lets talk about the RULES for getting my help on a webwork problem.  The rather than take a screenshot of the problem, click the "Email Instructor" button at the bottom of the problem--this will show me what you've done.  When you do this, you'll have a field to give me a message. In that field enter a brief description of what you have already done.  "Help, I don't know what to do" is NOT useful to you or to me. You have many resources, specifically the textbook, that covers the material, and each section of the webwork corresponds to a specific section of the book, which will tell you SPECIFICALLY what to do. 

2) The gradient 

∇f = <-160*2x/(x^2+y^2+2)^2, -160*2y/(x^2+y^2+2)^2>

   = -320/(x^2+y^2+2)^2 <x, y>

so ∇f(2,2) = -3.2 <2, 2> = <-6.4, -6.4>.  This means that the direction of maximal change, AS A UNIT VECTOR is u = <-1/√2, -1/√2> .  The answer to part c) that you have is NOT A UNIT VECTOR AND NOT THE GRADIENT, but it does point in the right direction. I don't know at all why you want that vector--but I guess you are trying to normalize incorrectly.  The norm of that vector is the answer you give. What you needed, though, is the directional derivative in that direction is
D_u = u . ∇f  = <-1/√2, -1/√2> . <-6.4, -6.4> = 6.4*2/√2 = √2*6.4
which is also equal to |∇f(2,2) |.  All in all I *guess* that the mistake you made was multiplying by
1/√2 while *not* taking the dot product