Question #712b3

1 Answer
Oct 2, 2017

See the explanation below

Explanation:

The function is #f(x,y)=(2xy)/(x^2+y^2)#

and #(x,y)!=(0,0)#

The gradient is

#gradf(x,y)=((delf)/(delx),(delf)/(dely))#

#(delf)/(delx)=(2y(x^2+y^2)-2xy(2x))/(x^2+y^2)^2=(2y(-x^2+y^2))/(x^2+y^2)^2#

#(delf)/(dely)=(2x(x^2+y^2)-2xy(2y))/(x^2+y^2)^2=(2x(x^2-y^2))/(x^2+y^2)^2#

At the point #(0,0)#

#gradf(0,0)# is undefined

#(1)# The direction is #vecu=(0,1)#, #theta=pi/2#

The magnitude of #||vecu||=sqrt(0^2+1^2)=1#

#vecu# is already normalized

#hatu=<0,1>#

The directional derivative is the dot product of the gradient and the unit vector in the specified direction.

# D( (2xy) / (x^2+y^2))_(hatu) (0,0) = < * *, >. < 0,1 > # is undefined

#(2)# The direction is #vecv=(1,1)#, #theta=pi/4#

The magnitude of #||vecv||=sqrt(1^2+1^2)=sqrt2#

The unit vector is #hatv=1/||vecv||vecv=1/sqrt2<1,1>#