Answer:
Maximum rate change |[tex]\nabla[/tex]f| = √65
Direction = ( 1, 8 ) / √65
i.e Maximum rate change is √65 and it occurs in a direction of ( 1, 8 )
Step-by-step explanation:
Given that;
f(x, y) = 4y √x, (4, 1)
{ x=4 and y=1 }
Maximum rate of change occurs with the gradient vector;
[tex]\nabla[/tex]f = [ df/dx, df/dy ] = [ 2y/√x, 4√x ]
we substitute in our value of x and y
[tex]\nabla[/tex]f = 2(1)/√4, 4√4
[tex]\nabla[/tex]f = 2/√4, 4√4
[tex]\nabla[/tex]f = ( 1, 8 )
Maximum rate change |[tex]\nabla[/tex]f| = | ( 1, 8 ) | = √( 1² + 8² ) = √(1 + 64)
Maximum rate change |[tex]\nabla[/tex]f| = √65
Direction = [tex]\nabla[/tex]f / |[tex]\nabla[/tex]f|
we substitute
Direction = ( 1, 8 ) / √65
i.e Maximum rate change is √65 and it occurs in a direction of ( 1, 8 )
