Final answer:
The directional derivative of f at the given point in the direction indicated by the angle θ is -1.
Step-by-step explanation:
To find the directional derivative of f at the given point in the direction indicated by the angle θ, we can use the formula:
∇f(a,b) • u = ||∇f(a,b)||cos(θ)
In this formula, ∇f(a,b) is the gradient of f, (a,b) is the given point, u is the unit vector in the direction of θ, and ||∇f(a,b)|| is the magnitude of the gradient. First, we need to find the gradient of f and the unit vector in the direction of θ.
The gradient of f is given by:
∇f(x,y) = (∂f/∂x, ∂f/∂y)
In this case, ∂f/∂x = -2ye^(-x) and ∂f/∂y = 2e^(-x).
At the point (0,9), ∂f/∂x(0,9) = 0 and ∂f/∂y(0,9) = 2e^(-0) = 2.
Therefore, the gradient at the given point is ∇f(0,9) = (0,2).
The unit vector in the direction of θ can be found using:
u = (cos(θ),sin(θ))
In this case, θ = 2π/3, so u = (cos(2π/3),sin(2π/3)) = (-0.5,√3/2).
Now, we can substitute the values into the formula:
∇f(0,9) • u = ||∇f(0,9)||cos(2π/3)
||∇f(0,9)|| = √((0)^2+(2)^2) = √4 = 2
cos(2π/3) = (-0.5)
Substituting these values, we get:
∇f(0,9) • u = 2 * (-0.5) = -1