using tangent planes to approximate

Question

Use a tangent plane to approximate the value of the following function at the point (-4.9,-16.1)

f(x,y) = arctan(xy^2+1281).

Present your answer accurate to four decimal places.

Answer 1

z = f(x,y) = arctan(xy²+1281)

f(-5, -16) = arctan(1) = π/4

The tangent plane at point (x₀, y₀, z₀) = (-5, -16, π/4) is given by:

z = z₀ + f_x(x₀, y₀)(x - x₀) + f_y(x₀,y₀)(y - y₀)

Plug in the values of (x₀, y₀, z₀), evaluate the partial derivatives at (x₀, y₀), and calculate z for x = -4.9 and y = -16.1. That z of the tangent plane approximates the value of f(-4.9, -16.1).