Final answer:
To find the gradient of the function g(x,y) at the point (3,1), the partial derivatives are computed and evaluated at that point, resulting in the gradient vector ∇g(3,1) = (3/5, 1/5).
Step-by-step explanation:
To find the gradient of g(x,y) = √(x²+2y+14) at the point (3,1), denoted as ∇g(3,1), we first need to calculate the partial derivatives of g with respect to x and y at the given point.
The partial derivative of g with respect to x is ∂g/∂x = x/√(x²+2y+14), and similarly, the partial derivative with respect to y is ∂g/∂y = 1/√(x²+2y+14).
Substituting the point (3,1) into these derivatives, we get:
- ∂g/∂x (3,1) = 3/√(3²+2*1+14)
- ∂g/∂y (3,1) = 1/√(3²+2*1+14)
Calculating these values, we find:
- ∂g/∂x (3,1) = 3/√(9+2+14) = 3/√25 = 3/5
- ∂g/∂y (3,1) = 1/√(9+2+14) = 1/√25 = 1/5
Therefore, the gradient vector at the point (3,1) is
∇g(3,1) = (3/5, 1/5)