Find the directional derivative at P(0, 0, 4)

[epicjr77]

I cannot figure out even where to start with this problem, my teacher did a horrible job explianing it, if I could see this example I am sure I can figure out the rest!

Find the directional derivative of F at the point P in the indicated direction.

. . .f(x, y, z) = Z^2 tan^(-1) (x + y)

. . .P(0, 0, 4), a =<6, 0, 1)

thanks for any help

 

[pka]

First we must find: \(\displaystyle \L \nabla f(x,y,z) = \left( {\frac{{z^2 }}{{1 + \left( {x + y} \right)^2 }}} \right)i + \left( {\frac{{z^2 }}{{1 + \left( {x + y} \right)^2 }}} \right)j + 2z\arctan (x + y)k.\)

Then make a a unit vector: \(\displaystyle \L a = \left\langle {6,0,1} \right\rangle \quad \Rightarrow \quad u = \frac{a}{{\left\| a \right\|}} = \left\langle {\frac{6}{{\sqrt {37} }},0,\frac{1}{{\sqrt {37} }}} \right\rangle .\)

Thus find: \(\displaystyle \L D_u f(P) = u \cdot \nabla f(P).\)

 

[epicjr77]

Ok I see thank you