Final answer:
The equation of the tangent plane to the surface z = cos(5x)cos(6y) at the point (-2π/2, 3π/2, 1) is z = 1, which is a horizontal plane at the height of 1 unit.
Step-by-step explanation:
To find the equation of the tangent plane to the surface z = cos(5x)cos(6y) at the point (-2π/2, 3π/2, 1), we need to calculate the partial derivatives of z with respect to x and y at the given point and use them in the tangent plane equation formula.
The partial derivative of z with respect to x is -5sin(5x)cos(6y), and the partial derivative of z with respect to y is -6cos(5x)sin(6y). Evaluating these at the point (-2π/2, 3π/2), we get:
- dz/dx at the point is 0,
- dz/dy at the point is 0.
Since the gradients are both zero, the tangent plane equation at the point (-2π/2, 3π/2, 1) is z = 1, which is a horizontal plane at the height of 1 unit.