what's the directional derivative of f(x, y, z) = x2 + y2 + z2 at P(2, 1, 3) in the direction of the origin?

Question

what's the directional derivative of f(x, y, z) = x2 + y2 + z2 at P(2, 1, 3) in the direction of the origin?

i have tried so many times for this problem, i still couldn't get the right answer.
well i know the directional vector would be [(0,0,0)-(2,1,3)] / sqrt(14) this is normalized
and i found the gradient of the function f which is 2x+2y+2z
so i just need to dot those two together to get the directional derivative.
but i still got the wrong answer...
help please~~

Answer 1

Isn't your direction vector going the wrong way?
It should be (2,1,3) -(0,0,0) = (2,1,3).
Normalising it gives (2,1,3)/sqt(14).
Now evaluate the gradient at (2,1,3). It is (4,2,6).
Now take the dot product of (4,2,6) and (2,1,3)/sqrt(14).
You get 2*14/sqrt(14) = 2*sqrt(14).
Is this the answer you are seeking?

Answer 2

Am told this should help!

(-2 * 2 * 2 – 1 * 2 * 1 – 3 * 2 * 1)/sqrt(14) =

-4.27618

So function would slope downwards towards the origin!!

Answer 3

I x+y+z=10 and x2+y2+z2=40 find xy+yz+zx [ x+y+z]^2=100=40+2[xy+yz+zx] xy+yz+zx =[100--40]/2=30