Question

Let h(x, y) = ycos(x^2 + y)

If u = (3/5)i - (4/5)j, calculate h_u(1,1).

Answer 1

Hmm I'm a little confused on what's going on here.
Is the h_u a partial derivative we're taking?
Or is this like a directional derivative?

Answer 2

Sorry, this text editor doesn't let me format the way I need to.

u is a directional vector. h_u should read as h with the subscript u. It's a derivative of the function h(x, y) in the direction u.

Answer 3

\[\Large\bf\sf Dir_u\;h(1,1)\quad=\quad \nabla h\cdot u\]I dunno if I used the correct notation for the directional derivative :P
I kinda forgot this stuff hehe.

Isn't it something like this?
The gradient of h dotted with our u?

Answer 4

Yes, that's it. Okay, I see. So I should evaluate grad h at (1, 1), then dot it with u?

Answer 5

Mmmm ya that sounds right :o
I'm gonna run through it really quick also.

Answer 6

Okay, thank you. My result is \[h _{u} = -2.1515\]

Answer 7

Hmm Wolfram is coming up with something else.
http://www.wolframalpha.com/input/?i=directional+derivative+of+y*cos%28x%5E2%2By%29+in+the+direction+%283%2F5%2C+-4%2F5%29%2C+where+x+%3D+1%2C+y+%3D+1

Did you remember to product rule on the y component when you did your gradient?

Answer 8

Oh I forgot the negative on the 4/5. There we are. Great, thank you so much!

Answer 9

Cool! :)