What is the maximum rate of change of f(x,y) = ln(x^2 + y^2) at the point (-1, 1) and the direction in which it occurs?

Question

I got #grad#f = #(2/x, 2/y)#

#gradf(-1,1)=(-2,2)#

I thought #abs(gradf(-1,1))# was maximum gradiant change...Which would be #sqrt(8)#?

Check your(my) math...#gradf=((2x)/(x^2+y^2), (2y)/(x^2+y^2))#

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Answer 1 · 2018-01-25T17:07:07

When given two dimensional function #f(x,y)# and a point #(x_0,y_0)#, the maximum rate of change is:

#|gradf(x_0,y_0)|#

and the direction is:

#gradf(x_0,y_0)#

Compute #gradf(x,y)#:

#gradf(x,y) = (del(f(x,y)))/(delx)hati+(del(f(x,y)))/(dely)hatj#

#gradf(x,y) = (2x)/(x^2+y^2)hati+(2y)/(x^2+y^2)hatj#

Evaluate at the point #(-1,1)#:

#gradf(-1,1) = (2(-1))/((-1)^2+(1)^2)hati+(2(1))/((1)^2+(1)^2)hatj#

#gradf(-1,1) = -hati+hatj#

Compute the magnitude:

#|gradf(-1,1)| = |-hati+hatj|#

#|gradf(-1,1)| = sqrt((-1)^2+1^2)#

#|gradf(-1,1)| = sqrt2#

The maximum rate of change is #sqrt2# and its direction is #-hati+hatj#