Final answer:
To find the equation of the tangent plane, we calculated the partial derivatives, found the gradient vector, and applied the point-normal form of the plane equation using the specified point and the gradient vector as the normal.
Step-by-step explanation:
To find the equation of the tangent plane to the given surface at the specified point, we need to find the gradient (also known as the normal vector) at that point.
The surface is given by z = 4(x - 1)² + 5(y - 3)² - 1. The point of tangency is (2, -2, 10). To find the gradient, we take the partial derivatives of z with respect to x and y:
- ∂z/∂x = 8(x - 1)
- ∂z/∂y = 10(y - 3)
At the point (2, -2), these partial derivatives are:
- ∂z/∂x evaluated at x=2 is 8(2 - 1) = 8
- ∂z/∂y evaluated at y=-2 is 10(-2 - 3) = -50
Now, we have the gradient vector ∇z = <8, -50, -1>.
To write the equation of the tangent plane, we use the point-normal form:
A(x - x₀) + B(y - y₀) + C(z - z₀) = 0, where <A, B, C> is the normal vector and (x₀, y₀, z₀) is the given point.
Substituting the normal vector and the point, we get:
8(x - 2) - 50(y + 2) - (z - 10) = 0
Rearranged, the equation of the tangent plane is:
8x - 50y - z = 86