Question

Если z=x^2−2xy+3y−1 , то значение 3√5 |градиент| в точке M0 (1;2) равна?

Answer 1

Ответ:

Пошаговое объяснение:

Answer 2

Calculating the Gradient at Point M0(1, 2)

To calculate the gradient at the point M0(1, 2) for the function z = x^2 - 2xy + 3y - 1, we can use the following steps:

1. Calculate the Partial Derivatives: - The gradient of a function in two variables (x, y) is given by the vector (∂z/∂x, ∂z/∂y). - Calculate the partial derivatives of z with respect to x and y.

2. Evaluate the Gradient at Point M0(1, 2): - Substitute the values x = 1 and y = 2 into the partial derivatives to find the gradient at the point M0(1, 2).

Calculation Steps

1. Calculate the Partial Derivatives: - The partial derivative of z with respect to x, denoted as ∂z/∂x, is given by the derivative of z with respect to x while treating y as a constant. - The partial derivative of z with respect to y, denoted as ∂z/∂y, is given by the derivative of z with respect to y while treating x as a constant.

The partial derivatives are: - ∂z/∂x = 2x - 2y - ∂z/∂y = -2x + 3

2. Evaluate the Gradient at Point M0(1, 2): - Substitute x = 1 and y = 2 into the partial derivatives to find the gradient at the point M0(1, 2).

Substituting x = 1 and y = 2: - ∂z/∂x = 2(1) - 2(2) = 2 - 4 = -2 - ∂z/∂y = -2(1) + 3 = -2 + 3 = 1

Therefore, the gradient at the point M0(1, 2) is given by the vector (-2, 1).

Conclusion

The value of the gradient at the point M0(1, 2) for the function z = x^2 - 2xy + 3y - 1 is (-2, 1).