Question

Find the gradient of the function at the given point. Then sketch the gradient together with the level curve that passes through the point f(x, y)= ln (x²+y²), (1,1).

Answer 1

The gradient of the function f(x, y) = ln(x²+y²) at the point (1, 1) is (∂f/∂x, ∂f/∂y) = (2/2, 2/2) = (1, 1).

Step-by-step explanation:

To find the gradient of the function f(x, y) = ln(x²+y²), we need to calculate its partial derivatives with respect to x and y separately. Let's denote the function as u = x² + y² for ease of computation. Taking the partial derivative of f with respect to x (∂f/∂x) involves differentiating ln(u) with respect to x. Applying the chain rule, ∂f/∂x = (∂f/∂u) * (∂u/∂x). The derivative of ln(u) with respect to u is 1/u, and the derivative of u = x² + y² with respect to x is 2x. Thus, (∂f/∂x) = (1/u) * (2x). At the point (1, 1), where x = 1 and y = 1, u = 1² + 1² = 2. Therefore, (∂f/∂x) = (1/2) * (2) = 1.

Next, to find (∂f/∂y), we follow a similar procedure. (∂f/∂y) = (∂f/∂u) * (∂u/∂y). The derivative of u = x² + y² with respect to y is 2y. So, at the point (1, 1), (∂f/∂y) = (1/u) * (2y) = (1/2) * (2) = 1.

Therefore, the gradient of f(x, y) = ln(x²+y²) at the point (1, 1) is (∂f/∂x, ∂f/∂y) = (1, 1).

To sketch the gradient together with the level curve passing through (1, 1) for ln(x²+y²), we know that the gradient vector (1, 1) indicates the direction of the steepest ascent. The level curve for ln(x²+y²) at (1, 1) represents points where the function is constant. The gradient vector (1, 1) at this point signifies that the function increases equally in the x and y directions. This information helps sketch the gradient direction and the level curve intersecting at (1, 1).

This combination of the gradient vector and the level curve offers insights into the function's behavior around the point (1, 1).

Answer 2

To find the gradient of the function f(x, y)= ln (x²+y²) at the point (1,1), calculate the partial derivatives, which are (2x)/(x²+y²) and (2y)/(x²+y²), and evaluate them at (1,1). The gradient is (1,1) and the level curve is a circle with radius √2. The gradient can be sketched as a tangent vector pointing in the direction of greatest increase of f at the point (1,1).

Step-by-step explanation:

To find the gradient of the function at the given point (1,1) for the function f(x, y)= ln (x²+y²), we must first calculate the partial derivatives with respect to x and y and then evaluate them at the point (1,1).

The partial derivative with respect to x is given by the derivative of ln (x²+y²) with respect to x, which is (2x)/(x²+y²). Similarly, the partial derivative with respect to y is given by (2y)/(x²+y²).

At the point (1,1), these evaluations give us 2/(1+1) for both partial derivatives, which is 1.

Therefore, the gradient at (1,1) is the vector (1,1).

To sketch the gradient together with the level curve that passes through the point, remember that a level curve of a function f(x, y) is a curve along which the function has constant value.

At the point (1,1), the function value is ln(2). So the level curve through (1,1) is the set of all points (x, y) where ln(x²+y²) = ln(2), which simplifies to x²+y² = 2.

This represents a circle with radius √2 centered at the origin.

The gradient vector at (1,1) can be sketched as a tangent vector to this circle at that point, pointing in the direction of the greatest increase of f.