Question

Find the equation of the tangent plane to the surface z = 55 - x²y² at the point (4, 2).

Answer 1

The equation of the tangent plane to the surface z = 55 - x²y² at the point (4, 2) is z - 39 = -32(x - 4) - 64(y - 2).

Step-by-step explanation:

To find the equation of the tangent plane to the surface given by z = 55 - x²y² at the point (4, 2), we can use the formula for the tangent plane at a point on a surface defined by f(x, y, z) = 0. For the function z = f(x, y), the formula for the tangent plane is z - z0 = fx(x0, y0)(x - x0) + fy(x0, y0)(y - y0), where fx and fy are the partial derivatives of f with respect to x and y, respectively, and (x0, y0, z0) is the point of tangency.

First, we calculate the partial derivatives of f(x, y) = 55 - x²y²:

• fx = -2xy²
• fy = -2yx²

Evaluating these at the point (4, 2), we get:

• fx(4, 2) = -2 × 4 × 2² = -32
• fy(4, 2) = -2 × 2 × 4² = -64

And since z = 55 - x²y², z0 = 55 - 4² × 2² = 39. So, the equation of the tangent plane is:

z - 39 = -32(x - 4) - 64(y - 2)