Question

Find the gradient vector field of f. f(x, y) = tan(2x − 3y)

Answer 1

To find the gradient vector field of f(x, y) = tan(2x - 3y), we need to calculate the partial derivatives with respect to x and y. The gradient vector field is given by ∇f = (∂f/∂x, ∂f/∂y) = (2sec²(2x - 3y), -3sec²(2x - 3y)).

Step-by-step explanation:

To find the gradient vector field of f(x, y) = tan(2x - 3y), we need to calculate the partial derivatives with respect to x and y.

∂f/∂x = 2sec²(2x - 3y)

∂f/∂y = -3sec²(2x - 3y)

Therefore, the gradient vector field is given by ∇f = (∂f/∂x, ∂f/∂y) = (2sec²(2x - 3y), -3sec²(2x - 3y))

Answer 2

The gradient vector field of the function f(x, y) = tan(2x - 3y) is found by computing the partial derivatives with respect to x and y, resulting in grad f = <2sec^2(2x - 3y), -3sec^2(2x - 3y)>.

Step-by-step explanation:

To find the gradient vector field of the function f(x, y) = tan(2x − 3y), we need to take the partial derivatives of f with respect to each variable, x and y.

The partial derivative of f with respect to x is given by:

fx = sec2(2x − 3y) × (2).

This represents the rate of change of the function in the direction of the x-axis.

The partial derivative of f with respect to y is given by:

fy = sec2(2x − 3y) × (−3).

This represents the rate of change of the function in the direction of the y-axis.

Therefore, the gradient vector field of the function f is the vector field composed of these two partial derivatives:

grad f = <fx, fy> = <2sec2(2x − 3y), −3sec2(2x − 3y)>.