Final answer:
The directional derivative of f(x, y) = x^2 - xy^3 at the point (1, 2) in the direction of <3, 5> is -59/√34.
Step-by-step explanation:
The directional derivative of a function at a point in the direction of a given vector can be found using the gradient of the function and the dot product with the unit vector in the given direction.
First, we need to find the gradient of the function f(x, y) = x^2 - xy^3. The gradient can be found by taking the partial derivatives of the function with respect to x and y. We have:
∇f(x, y) = (∂f/∂x)i + (∂f/∂y)ĵ = (2x - y^3)i + (-3xy^2)ĵ
Next, we need to find the unit vector in the direction of <3,5>. The magnitude of the vector is √(3^2 + 5^2) = √34. So, the unit vector is u = (3/√34)i + (5/√34)ĵ.
Finally, the directional derivative of f(x, y) at the point (1, 2) in the direction of <3, 5> is given by the dot product of the gradient and the unit vector:
Duf(1, 2) = ∇f(1, 2) · u = (2(1) - 2^3) * (3/√34) + (-3(1)(2)^2) * (5/√34) = 1/√34 - 60/√34 = -59/√34