Question

f(x,y) x^2y find
the derivative off
in the direction of (1,2)
at the point (3,2) is

Answer 1

Answer:

Solution: (a) The 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)

To find the directional derivative in the direction of the vector (1,2), we need to find a unit vector in the direction of the vector (1,2). 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√.

Step-by-step explanation:

I hope this helps u