Final answer:
To find the equation of the tangent plane at the point (1, 2) for the function F(x, y) = sqrt(3y + 2x^2), calculate the partial derivatives, evaluate them at the point to get the normal vector, and plug into the plane equation with the given point.
Step-by-step explanation:
To find the equation of the tangent plane to the surface represented by the function F(x, y) = sqrt(3y + 2x^2) at the point (1, 2), we first need to calculate the partial derivatives of F with respect to x and y at that point. The partial derivative with respect to x is Fx(x, y) = 2x/(2sqrt(3y + 2x^2)) and with respect to y is Fy(x, y) = 3/(2sqrt(3y + 2x^2)). Evaluating these derivatives at point (1, 2) gives us the slopes (gradients) in the x and y directions, which are components of the normal to the tangent plane.
At (1, 2):
Fx(1, 2) = 2(1)/(2sqrt(3(2) + 2(1)^2)) = 1/sqrt(8)
Fy(1, 2) = 3/(2sqrt(3(2) + 2(1)^2)) = 3/(2sqrt(8))
The normal vector to the plane at (1, 2) is then N = <(1/sqrt(8)), (3/(2sqrt(8))), -1>, where the -1 comes from the implicit function theorem, including the negative sign to account for the z-component (in this case F(x, y)) in the normal vector. The equation of the tangent plane can be found by plugging the normal vector components and the point (1, 2, F(1, 2)) into the plane equation:
a(x - x0) + b(y - y0) + c(z - z0) = 0, where (a, b, c) are components of the normal vector, and (x0, y0, z0) is the point of tangency.
By plugging in the values, we get the equation of the tangent plane:
(1/sqrt(8))(x - 1) + (3/(2sqrt(8)))(y - 2) - (z - sqrt(8)) = 0.