Question

will award medal. question posted below

Answer 1

start by finding the gradient

Answer 2

just to start since the function contains all the variables and its not in any type of vector form i do the partial for x y and z using the whole equaiton

Answer 3

gradient vector is just a package of partial derivatives :

\[gradient(f) = \left\langle \dfrac{\partial f}{\partial x}, ~\dfrac{\partial f}{\partial y}, ~\dfrac{\partial f}{\partial z}\right\rangle \]

Answer 4

yea so the i have the equation \[(e^y+ze^x)i+(xe^y+e^z)j+(ye^z+e^x)k\]

Answer 5

Yep! evaluate the gradient at given point (0, 0, 0)

Answer 6

the you just get the unit vector equation, of I+j+k

Answer 7

so the gradient at the given point (0,0,0) is i+j+k

Answer 8

for directional derivaitve along <9, -7, 3>, you simply take the dot product of unit vector along this vector and the gradient

Answer 9

thats what i got at least
then multiply and get 6

Answer 10

nvm i did algebra...addition like a 5 year old
so wrong.

Answer 11

oh its not 5 either

Answer 12

<1, 1, 1> . <9, -7, 3>/sqrt(9^2+7^2+3^2)

Answer 13

5/sqrt(139)

Answer 14

see if that works

Answer 15

yooooh, thats tight. what did i do wrong?

Answer 16

remember you need to dot with "unit vector"

Answer 17

ohhh.

Answer 18

you get unit vector along <9, -7, 3> by dividing it by the magnitude : sqrt(9^2+7^2+3^2)