91Ó°ÊÓ

First, we need to find the gradient of the given surface function: \(f(x,y,z) = xy^2 + 2yz -16\). The gradient of a scalar function is a vector that points in the direction of the greatest rate of increase of the function, and its components are the partial derivatives with respect to each coordinate. The gradient is given by: \( \nabla f = \begin{bmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \\ \frac{\partial f}{\partial z} \end{bmatrix} \) Now, let's find the partial derivatives: \( \frac{\partial f}{\partial x} = y^2 \) \( \frac{\partial f}{\partial y} = 2xy + 2z \) \( \frac{\partial f}{\partial z} = 2y \) The gradient is then: \( \nabla f = \begin{bmatrix} y^2 \\ 2xy + 2z \\ 2y \end{bmatrix} \)

Now that we have the gradient, we can find the normal vector to the surface at the point P(1, 2, 3) by substituting the given point into the gradient: \( \mathbf{N}(1, 2, 3) = \begin{bmatrix} (2)^2 \\ 2(1)(2) + 2(3) \\ 2(2) \end{bmatrix} = \begin{bmatrix} 4 \\ 10 \\ 4 \end{bmatrix} \) So, the normal vector to the surface at point P(1, 2, 3) is \(\mathbf{N} = \langle 4, 10, 4 \rangle\).

Now that we have the normal vector to the surface node at point P(1, 2, 3), we can construct the equation for the tangent plane. The general equation for the tangent plane to a surface at a point is: \( A(x-x_0) + B(y-y_0) + C(z-z_0) = 0\), where (xâ‚€, yâ‚€, zâ‚€) is the given point and A, B, C are the components of the normal vector. In our case, the normal vector is \( \langle 4,10,4 \rangle\), and the given point is (1, 2, 3). So, we have: \( 4(x-1) + 10(y-2) + 4(z-3) = 0 \) which simplifies to: \( 4x + 10y + 4z = 38 \) So, the equation of the tangent plane H to the surface S at the point P(1, 2, 3) is \( 4x + 10y + 4z = 38\).