Question

ok, now what in the heck am i doing wrong? i really cannot seem to get a problem right on this section.....

Find the directional derivative of f(x,y,z)=z3−x2y at the point (-5, 2, -1) in the direction of the vector v=⟨−5,5,−1⟩.

Answer 1

grad = <-2xy, -x^2,3z^2> making the grad at said point = <20,25,3>....unit vector is u=v/magv or <-5/sqrt(51), 5/sqrt(51), -1/sqrt(51).....making the directional be (-5/sqrt51)(20)+(5/sqrt51)(25)+(-1/sqrt51)(3), but again, webwork says no.

Answer 2

how did you get <20,25,3> from \( <-2xy, -x^2,3z^2> \) ?

Answer 3

here's my work....

Answer 4

that was meant to be rhetorical, as in "check your arithmetic". in particular
- x^2 is not +25

Answer 5

oh, the negative isn't part of the x? i thought it was (-5)^2 not -(5^2). ok thanks

Answer 6

it is - (-5)^2 = -1 * -5 * -5

Answer 7

to be more clear

\[ \frac{\partial}{\partial y} (−x^2y ) = - \frac{\partial x^2 y}{\partial y}= -1 \cdot x^2\]

Answer 8

mod, can you please go to math chat and scroll up