Question

find the directional derivative of f at the point p in the indicated direction.
f(x,y) = x^2-5xy+3y^2
p (3,-1)
u= sqrt(2)/2 (i+j)

Answer 1

so fx=2x-5y, fy=-5x+6y
gradient of f = 11i + -21k
... i think :)

Answer 2

Now, (11, -21) dot product (sqrt(2)/2, (sqrt(2)/2)

Answer 3

11sqrt(2)/2 +-21sqrt(2(/2? ?

Answer 4

Yes, exactly that, (a,b) dot (c,d) = ac + bd

Answer 5

i got -7.06, which is pretty much the right answer.. i didn't realize it was so simple! thank you so much :D

Answer 6

Besides the name, directional derivatives are easy! Good luck

Answer 7

how about f(x,y)= arctan (y/x) i'm not even sure how to get the partial derivatives of that :(

Answer 8

d/dx (arctan(x)) = 1/1+x^2

So...u = y/x

partial x (u) = -y/x^2

d/du (arctan(u)) = 1/(1+u^2)*(-y/x^2)

Replace correctly the things, CHECK my answer because im not completely sure if this makes sense, and finally repeat for partial y.

Answer 9

i mean this d/du (arctan(u)) = (-y/x^2)/(1+u^2)

Answer 10

forgot to replace u lol