Final answer:
The equation of the tangent plane to the surface 2x² + 3y² + 3z² = 62 at the point (4, -3, 1) is 16x - 18y + 6z = 50.
Step-by-step explanation:
To find the equation of the tangent plane to the given surface at a point, you need to establish the surface's equation and then use the point to find the plane's specific equation. The surface given is 2x² + 3y² + 3z² = 62. To find the gradient, which provides the perpendicular vector (normal) to the surface at the point, we calculate the partial derivatives with respect to each variable. The partial derivatives are 4x, 6y, and 6z, respectively. At the point (4, -3, 1), the gradient vector is (16, -18, 6).
Now, the general equation for a tangent plane at a point (x0, y0, z0) with a normal vector (a, b, c) is a(x - x0) + b(y - y0) + c(z - z0) = 0. Plugging in our point and normal vector, we get 16(x - 4) - 18(y + 3) + 6(z - 1) = 0.
Simplifying, the final equation of the tangent plane is 16x - 18y + 6z = 50.