Question

find the equation of the tangent plane to the surface ( z = 8x^{2}+y^{2}+2xy ) at the point ( (0,3,9) ).

Explanation:

Step1: Find the partial derivatives

Let \(F(x,y,z)=8x^{2}+y^{2}+2xy - z\).
The partial derivative with respect to \(x\):
\(F_{x}=\frac{\partial F}{\partial x}=16x + 2y\).
Evaluate at \((0,3)\): \(F_{x}(0,3)=16\times0+2\times3 = 6\).
The partial derivative with respect to \(y\):
\(F_{y}=\frac{\partial F}{\partial y}=2y+2x\).
Evaluate at \((0,3)\): \(F_{y}(0,3)=2\times3+2\times0 = 6\).
The partial derivative with respect to \(z\): \(F_{z}=\frac{\partial F}{\partial z}=- 1\).

Step2: Use the tangent - plane formula

The equation of the tangent plane is \(F_{x}(x_{0},y_{0})(x - x_{0})+F_{y}(x_{0},y_{0})(y - y_{0})+F_{z}(x_{0},y_{0})(z - z_{0})=0\).
Here \(x_{0}=0,y_{0}=3,z_{0}=9\).
Substitute the values: \(6(x - 0)+6(y - 3)-1(z - 9)=0\).
Expand: \(6x+6y-18 - z + 9=0\).
Rearrange to get \(z = 6x+6y - 9\).

Answer:

\(z = 6x+6y - 9\)