Question

Directional Derivatives:
Let f(x,y) = 2xy - y^2 - 3x + 5 ...

Answer 1

Let \[f(x,y) = 2xy - y^2 - 3x + 5\]Find the unit vector in the direction of the maximum increase of f at the point (1,3).
Find a unit vector giving \[D_uf(1,3)=0\]

Answer 2

Okay, so the directional derivative, if I am remembering this correctly and not completely misreading my text book is the dot product of the gradient vector and the unit vector in the direction of the directional derivative.
As such, the unit vector for maximum increase should be parallel with the gradient vector, since the gradient is perpendicular to the level curve at a given point. Thus if U = unit vector and G = the gradient vector, U x G = 0 (sorry about the poor notation, couldn't figure out how to do vectors or gradient symbol)
Similarly, when the directional derivative equals 0, U and G must be perpendicular, and so the dot product, U * G = 0

Now, to find the cross product, I did \[\det \left[\begin{matrix}f_x & f_y \\ a & b\end{matrix}\right]\]where the top row are the components of the gradient vector and the bottom the components of the unit vector, and found the unit vector to equal \[\<frac{4}{5}, \frac{3}{5}>\]Except then I also got the same answer for finding when the directional derivative equals 0...

Answer 3

last eq there should be \[<\frac{4}{5}, \frac{3}{5}>\] sorry