Final answer:
To find the smallest angle between the xy-plane and the tangent plane at a given point on a surface, we calculate the gradient vector of the function at that point and determine the angle between this vector and the z-axis.
Step-by-step explanation:
To find the smallest angle between the xy-plane and the tangent plane to the graph of f(x,y)=ln(x²+y²+1) at the point (2,2,2ln(3)), we need to calculate the gradient of f at the point (2,2). The gradient vector at a point gives the direction of the steepest ascent, while its negative gives the direction of the steepest descent. Therefore, the normal vector to the tangent plane at this point is aligned with the gradient vector. The angle between two planes can be found by calculating the angle between their normal vectors, which corresponds to the angle between the gradient vector of f and the z-axis (since the z-axis is perpendicular to the xy-plane). Remember, to get the smallest angle, we consider the acute angle formed.
First, we compute the partial derivatives of f with respect to x and y to get the gradient vector at the point (2,2): ∂f/∂x and ∂f/∂y, and evaluate them at (2,2). We find that the normal vector to the tangent plane is (4/3, 4/3, -1). The z-component of the gradient vector at our point of interest will be directly compared to a unit vector on the z-axis, (0,0,1), to find the smallest angle using the dot product because the xy-plane is parallel to the vector (0,0,1).
The cosine of the angle θ can be found using the dot product of the two vectors: cos(θ) = N ⋅ k / (|N||k|), where N is the gradient vector and k is the unit vector along the z-axis. After computing this, we take the arccos to find the angle θ. The smallest angle is the complement of θ, which is 90° - θ.