0
$\begingroup$

Alright, I'm trying to fine the gradient of the function and the maximum value of the directional derivative.

$$z = \frac{y}{(x^2 + y^2 )}\text{ at }(1,1)$$

So far, I've gotten the following values for partial derivates, but I can't guarantee they are correct:

$$F_x = \frac{-2xy}{(x^2 + y^2 )^2}$$ $$F_y = \frac{-2y^2}{( x^2 + y^2 )^2} + \frac1{x^2 + y^2}$$

GradientF(x,y) = \frac{-2xy}{( x^2 + y^2 )^2 + (-2y^2)/( x^2 + y^2 )^2} + 1/( x^2 + y^2 )^2}$$ GradientF(1,1) = -2/4i + (-2/4)j + (1/2)k = <-2/4, -2/4, 1/2> ||F(1,1)|| = <-2/4, -2/4, 1/2> (1/sqrt(3/4))

I get <-2/3, -2/3, 2/3> but this is wrong...

I should be getting <1/2, 0> at max value of 1/2 according to my friend who ceases to explain anything to me.

Any help? Thanks.

  • 0
    Your partial with respect to $x$ is missing a factor of $\frac2{x^2+y^2}$...2011-05-11
  • 0
    How? I just did it again and got the same answer?2011-05-11
  • 0
    Maybe you can post your solution in detail? Then we can debug...2011-05-11
  • 0
    OK, mind editing in my mistakes that I make when formatting? Sorry, I'm new here =(2011-05-11
  • 0
    Edited in. Now for formatting2011-05-11
  • 0
    Nono, I meant what steps did you take to go from, say $\frac{\partial}{\partial x} \frac{y}{(x^2 + y^2 )^2}$ to your result?2011-05-11
  • 0
    I actually mis-wrote that. I edited it now, it wasn't squared on the bottom. Sorry!2011-05-11

1 Answers 1