Question

find the directional derivative of f at the point P in the indicated direction

f(x,y) = sqrt(9x^2-4y^2-1)
p(3, -2)
a= i+5j

Answer 1

i'm pretty sure
\[Fx = 9x(9x^2-4y^2-1)^-(1/2)\]
\[Fy= -4y(9x^2-4y^2-1)^-(1/2)\]

Answer 2

You're derivatives look correct to me.

Answer 3

Now you need to find the unit vector \(\vec u_0\) in the indicated direction.

Answer 4

and my unit thingy is 1/6i+5/6j

Answer 5

I'm getting a different unit vector.

Answer 6

but when i try to find the directional derivative, and plug in numbers, i get something very different from the correct answer.. i thought maybe i was doing the derivation wrong but now.. i dunno

Answer 7

its sqrt of (1^2+5^2), right? which is sqrt of 36, which is 6, no?

Answer 8

Instead of dividing by \(1+5\), divide by \(\sqrt{1^1+5^2}=\sqrt{1+25}=\sqrt{26}\)

Answer 9

............ oh.
-_-"

Answer 10

so its 1/sqrt(26) + 5/sqrt(26) right?

Answer 11

I've done worse.

And that looks correct.

Answer 12

after plugging in all the numbers, should it be 27(1/8)(1/sqrt(26)? or am i doing something wrong there?

Answer 13

my denominator is correct, 8sqrt(26) but i'm getting 27 for my numerator, and according to the book at it's 67 :(

Answer 14

I'm also getting \[67 \over 8\sqrt{26}\]Did you remember to add\[f_y(3, -2)\cdot{5 \over \sqrt{26}}\]

Remember that \[{5 \over \sqrt{26}}={40 \over 8\sqrt{26}}\]

Answer 15

noo i still didn't get it right :( this is what i did..
9(3) x (9(3)^2-4(-2^2)-1)^-1/2 = 1/sqrt(64) which = 1/8x27, x 1/sqrt(26)
to get 27/8sqrt(26)

then for Fy
8(1/8)x5/sqrt(26) = 1x 5/sqrt(26)

so then its
27/8sqrt(26) + 5/sqrt(26)
which isn't 67 =\
where did i go wrong? :(

Answer 16

All your math looks correct, you just didn't simplify as much as the solution wants you to.\[{27 \over 8\sqrt{26}}+{5 \over \sqrt{26}}={27 \over 8\sqrt{26}}+{40 \over 8\sqrt{26}}={67 \over 8\sqrt{26}}\]

Answer 17

ohhh, cuz it needed a common denominator, right? i completely forgot...! lol, thank you!!!! you're my hero lol

Answer 18

You're very welcome.

Answer 19

@Zarkon

If I were to ask some questions about representation theory sometime in the near future, would you be able to help me with them?

Answer 20

@itzmashy sorry that I'm using your thread as a place to ask a question :(

Answer 21

no worries ^.^, wish i could help with representation theory though!

Answer 22

if i have another question, should i just ask here, or do you want me to create a new thread?

Answer 23

A new thread would be preferred. If a certain thread gets too long, it starts lagging.

Answer 24

kk ^.^