Question

Find the directional derivaitve of the hemisphere z=sqrt(a^2-x^2-y^2) at the point (a/sqrt(3),a/sqrt(3),a/sqrt(3)) in the direction making an angle pi/4 with the positive x-axis

Answer 1

i am confident that @Kreshnik knows this so i leave you in his most capable hands :P

Answer 2

lol @lgbasallote you got to be kidding.. :A

Answer 3

There is so much going on in this problem. It scares me.

Answer 4

well we have a unit vector rightL sqrt(2)/2(i)+sqrt(2)/2(j)

Answer 5

but this hemisphere is what is bothering me

Answer 6

how do i compute the gradient

Answer 7

The directional derivative at a point is
gradient(z)(point) dot product with a unit vector in the asking direction.

Answer 8

\[ \sqrt(2)/2(i)+\sqrt(2)/2(j)\]

Answer 9

okay, so then that a^2 is a constant number, and we would get a gradeient in the i and j directions only

Answer 10

They should specify that the vector is in the xy-plane.

Answer 11

so whats the answer

Answer 12

what do i plug in for a when i get the gradient

Answer 13

boy, and i thought this place was full of smart people

Answer 14

so the directional deriavitve is -sqrt(2)

Answer 15

Yes

Answer 16

That is the gradient of z
\[
\left\{-\frac{x}{\sqrt{a^2-x^2
-y^2}},-\frac{y}{\sqrt{a^2-
x^2-y^2}}\right\}
\]
at the given point
\[
\left\{-\frac{a}{\sqrt{a^2}},-
\frac{a}{\sqrt{a^2}}\right\}
\]
Which is {-1,-1}