Question

What is the directional derivative of the function Q = xy+yz at the point (2,-1, 1) in the direction of the

vector i +23+3R?​

Answer 1

Answer:

The maximum magnitude of the directional derivative of Ø = x2 - y2 + 2z2

the point (1,2,3) isThe gradient is just the vector of partial derivatives. The partial derivatives of f at the point (x,y)=(3,2) are:

∂f∂x(x,y)∂f∂x(3,2)=2xy=12∂f∂y(x,y)∂f∂y(3,2)=x2=9

Therefore, the gradient is

∇f(3,2)=12i+9j=(12,9).

(b) Let u=u1i+u2j be a unit vector. The directional derivative at (3,2) in the direction of u is

Duf(3,2)=∇f(3,2)⋅u=(12i+9j)⋅(u1i+u2j)=12u1+9u2.(1)We simply divide by the magnitude of (1,2).

u=(1,2)∥(1,2)∥=(1,2)12+22−−−−−−√=(1,2)5√=(1/5√,2/5√).Plugging this expression for u=(u1,u2) into equation (1) for the directional derivative, and we find that the directional derivative at the point (3,2) in the direction of (1,2) is

Duf(3,2)=12u1+9u2=125√+185√=305√.