let f(x,y) = 4x^3y^2.

a)Find a unit vector in the direction in which f decreases most rapidly from the point (2,1).

b) what is the rate of change of f in the direction given in (a).

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- Apr 29th 2007, 10:32 AMlittlemudirectional derivative
let f(x,y) = 4x^3y^2.

a)Find a unit vector in the direction in which f decreases most rapidly from the point (2,1).

b) what is the rate of change of f in the direction given in (a). - Apr 29th 2007, 11:57 AMJhevon
a function f(x,y) increases most rapidly at a point (x0,y0) in the direction of gradf(x0,y0) and decreases most rapidly in the direction of -gradf(x0,y0)

gradf(x,y) = <fx , fy> = <12(y^2)(x^2) , 8(x^3)y>

gradf(2,1) = <48 , 64>

so -gradf(2,1) = <-48, -64>

u = <-48, -64>/(sqrt(48^2 + 64^2)) = <-48/80 , -64/80> = <-3/5 , -4/5> ....unit vector in the direction of most rapid decrease

Quote:

b) what is the rate of change of f in the direction given in (a).

|-gradf(2,1)| = 80 - Apr 29th 2007, 12:29 PMlittlemu
Hi Jhevon,

I got -80 for my part B answer which is slightly different from your answer 80. I am checking why?

Thank you again. - Apr 29th 2007, 12:38 PMJhevon
i guess you used -|gradf(2,1)|

you are probably correct. i know that the rate of change is concerned with the magnitude of the gradient vector, so i said to myself, it couldn't be negative and put the minus sign inside the absolute values. but now that i think about it, it is valid to have a negative rate of change, as it denotes the rate of change in the opposite direction - Apr 29th 2007, 12:47 PMlittlemu
I agree with you. :)