Thread: directional 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).

Originally Posted by littlemu

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).

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

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

the rate of change in the direction of u is given by
|-gradf(2,1)| = 80

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.

Originally Posted by littlemu

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.

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

I agree with you.