$f(x,y)=3x^2+2xy+4y^2$
This is what I have to find:
direction of fastest increase
I found the gradient vector:
$f(x,y)=(6x+2y)i+(2x+8y)j$
$f(9,2) = \langle 58,34\rangle$
Is this the direction of fastest increase? Or is it $\frac{58}{2\sqrt{1130}}$i,$\frac{34}{2\sqrt{1130}}$j
direction of fastest decrease
Is this the direction of fastest increase?$\frac{-58}{2\sqrt{1130}}$i,$\frac{-34}{2\sqrt{1130}}$j
derivative in direction of fastest increase
I know this the dot product but I dont know what vectors to use.
derivative in direction of fastest decrease
The direction of fastest increase is in the same direction of the gradient vector at that point. If you think about it geometrically, you'll know that the $\nabla F$ at a point is perpendicular to the level surface/contour path. So it is correct that $\nabla$ also indicates that direction. The direction of the fastest decrease is just $-\nabla$ at $P$.
For the second part of you question use the formula of Directional Derivatives where $$D_u f(x,y)= f_x(x,y)a+f_y(x,a)b$$ where $<a,b>$ is the unit vector of your direction.
So that means $$D_u f(x,y)= (6x+2y)\frac{58}{2 \sqrt{1130}}+(2x+8y)\frac{34}{2 \sqrt{1130}}$$
You can substitute your Point in for $x$ and $y$ and solve. You should get $$58(\frac {58} {2\sqrt{1130}}) + 34(\frac {34} {2\sqrt{1130}})$$
Similarly you use the same method to determine the derivative in the direction of the fastest decrease.
NB: you can only substitute a point in for $x$, and $y$ if the question specifically states that it only needs to be calculated at that point. If not, the general solution (leaving it in $x$ and $y$) has to be used as that would give the derivative of greatest increase/decrease for all points.