Maximum directional derivative-check please?

[pockystix]

hihi,
I have another problem that I would like someone to check for me, if it isn't too much trouble...I did the work and got stuck.

The original problem is:
*Find the maximum directional derivative of f(x,y,z) =(xy^2z^3)^1/2(the square root) at (2,2,2) and the direction in which it occurs.

This is what I did-->http://img200.imageshack.us/img200/485/hwproblem3.jpg

I think I did the problem correctly, although I might be wrong...I'm just stuck because we're usually given the values of v, so that we can use the formula
u=v/llvll to get u and plug in duf(p)=grad f(p) . u to get the directional derivative. But we don't have v....so I don't know what to do...

Thanks in advance!

 

[BigGlenntheHeavy]

\(\displaystyle \text{Find the maximum directional derivative of f(x,y,z)} = (xy^{2}z^{3})^{1/2} \ at \ (2,2,2),\)
\(\displaystyle \text{and the direction in which it occurs.}\)

\(\displaystyle \text{gradient of} \ f(x,y,z) \ = \ f_x(x,y,z)i+f_y(x,y,z)j+f_z(x,y,z)k \ = \\)

\(\displaystyle \frac{y^{2}z^{3}i}{2(xy^{2}z^{3})^{1/2}} \ + \ \frac{2xyz^{3}j}{2(xy^{2}z^{3})^{1/2}} \ + \ \frac{3xy^{2}z^{2}k}{2(xy^{2}z^{3})^{1/2}}.\)

\(\displaystyle \text{Hence, it follows that the direction of maximum increase at (2,2,2) is gradient of}\)

\(\displaystyle f(2,2,2) \ = \ 2i+4j+6k\)

\(\displaystyle \text{The maximum value of D_u f(x,y,z) \ is \ ||gradient \ of \ f(x,y,z)|| \ is \ (4+16+36)^{1/2} = \sqrt56 =7.48,,,}\)