Final answer:
To find a normal vector to the curve at (1, 2), calculate the gradient at that point by finding the partial derivatives, evaluate them at (1,2) to get the gradient (5, 13), and then find a perpendicular vector such as (-13, 5).
Step-by-step explanation:
To find a vector normal to the curve x³ + xy + y³ = 11 at the point (1,2), we must first calculate the gradient of the curve at that point. The gradient is given by the vector of partial derivatives, which represents the slope of the tangent plane to the surface at a particular point, and the normal vector is perpendicular to this gradient.
The partial derivatives of the given function with respect to x and y are:
- ∂f/∂x = 3x² + y
- ∂f/∂y = x + 3y²
Evaluating these at the point (1,2) gives:
- ∂f/∂x (1,2) = 3(1)² + 2 = 5
- ∂f/∂y (1,2) = 1 + 3(2)² = 13
Thus, the gradient vector at the point (1,2) is (5, 13), and a vector normal to the curve at this point would be perpendicular to this vector.
Therefore, a normal vector to the curve at (1,2) could be taken to be (-13, 5) or any scalar multiple of this vector. We can simply negate one of the components and swap them to find a vector that is orthogonal (perpendicular) to the given gradient vector.