Question

For the function f(x,y) = [sin(x-y)]^2, find the maximum and minimum rate of change and in which direction it occurs. Also find the direction for which Du f is zero.

I've got the gradient function from there, which is 2sin(x-y)cos(x-y) i - 2sin(x-y)cos(x-y) j but no sure how to progress from there...

Answer 1

\[\left| \Delta f \right|=\sqrt{2} \ \left| \sin(2x-2y) \right|\]
Delta is gradiant

Answer 2

maximum of gradient is for when
\[\sin (2x-2y)=\pm 1\]
and the minimum of it for
\[\sin (2x-2y)=0\]

Answer 3

Thanks, how do you find the direction in which it occurred? (i.e. the vector?)

Answer 4

i think depends on (x,y) for example maximum of it occurs in direction
\[\pm(i-j) \\ since \\ \sin(2x-2y)=\pm1\]

Answer 5

can u imagine the places for which the minimums take place look likes to what?

Answer 6

Like a local (more like, global) minimum point as in a hyperbolloid?