Question

find the directional derivative of f at point p

f(x,y)= arctan (y/x)
p(4,-4)
u=2i-3j

Answer 1

First step, find the unit vector in the direction of u.

Answer 2

but isn't the unit vector already given to you?

Answer 3

or is it just 1/||u||, in which case its 1/4 (2i-3j) which would be 1/2i - 3/4j?

Answer 4

wait.. 4 + 9 is not 16...

Answer 5

So let \(\vec u_0\) be directional derivative in the direction of \(\vec u\). Then \[\vec u_0={2 \over \sqrt{13}}\vec i - {3 \over \sqrt{13}}\vec j\]

Answer 6

2/sqrt(13)i+3/sqrt(13)j??

Answer 7

i'm not quite sure how to find Fx or Fy
and btw, HI GEORGE haha

Answer 8

It should be -3 since your original vector has a -3.

Answer 9

ohh that's right..

Answer 10

Hi :) I had to go to bed last night so I missed the question you tagged me in.

Answer 11

no worries haha

Answer 12

Anyways, next step is to find \(f_x\) and \(f_y\).

Answer 13

this is my guess:
Fx = -(sec^2)^-2(xy^-1)y^-1
Fy= -(sec^2)^-2(xy^-1)x

Answer 14

We can use the fact that \[{ d \over dx}\;\;\arctan(x)={1 \over {1+x^2}}\]

Answer 15

wait, is that a rule?

Answer 16

Thus, \[\large f_x={\partial \over \partial x} \;\;\arctan\left({y \over x}\right)={1 \over 1+\left({y \over x}\right)^2}\cdot ({\partial \over \partial x}\;\;\; {y \over x})={-y \over {x^2+y^2}}\]

Answer 17

And yes that is a rule. Feel free to use it whenever.

Answer 18

Likewise, \[\large f_y= {\partial \over \partial y}\;\;\arctan\left({y \over x}\right)={1 \over 1+\left({y \over x}\right)^2}\cdot \left({\partial \over \partial y} \;\;\;{y \over x}\right)={x \over x^2+y^2}\](I skipped some steps in the simplification)

Answer 19

are you multiplying it by ∂/∂y yx because of chain rule?

Answer 20

Yes. In this case, we don't have to do any complicated sort of chain rule like we did last night.

Answer 21

so then
gradf = -4/32 x (2/sqrt(13) + 4/32 x (-2/sqrt(13)?!

Answer 22

In the second term, you should have a -3 instead of -2.

However, that is the directional derivative. The gradient is slightly different.

Answer 23

oops, that was a typo, but aren't i supposed to get the directional derivative by getting the gradient and dot product'ing it with the unit vector?

Answer 24

That's basically what we did here, just in a more direct way. We calculated the coefficients of the unit vector first, and then the coefficients of the gradient. By multiplying them together the way we did, it was like we found the gradient and used the dot product on the unit vector.

Answer 25

ohh i see what you're saying, so you weren't correcting my answer, just my term okay that makes sense

Answer 26

Precisely.

Answer 27

ohh okay, cool! thanks georige!!

Answer 28

You're welcome.