Question

Find an equation of the tangent plane to the given surface at the specified point..

\(z = 3{(x - 1)^2} + 2{(y + 3)^2} + 7,(2, - 2,12)\)..

Short answer

The tangent plane Equation is \(z = 6x + 4y + 8\)..

Step-by-step solution

Step-1(Given information and Formula Used)

The equation of the surface is\(z = 3{(x - 1)^2} + 2{(y + 3)^2} + 7\)or\(f(x,y) = 3{(x - 1)^2} + 2{(y + 3)^2} + 7\)

Let\(({x_0},{y_0},{z_0}) = (2, - 2,12)\)..

We have to find the equation of tangent plane to the surface at the point\((2, - 1, - 3)\).

Using the formula:

The equation of Tangent plane to the Surface\(z = f(x,y)\)at the point\(({x_0},{y_0},{z_0})\)is

\(z - {z_0} = {f_x}({x_0},{y_0})(x - {x_0}) + {f_y}({x_0},{y_0})(y - {y_0})\)..

Step-2(Finding Partial Derivative of \(f(x,y) = 3{(x - 1)^2} + 2{(y + 3)^2} + 7\)..)

Partial Derivative of\(f(x,y) = 3{(x - 1)^2} + 2{(y + 3)^2} + 7\)with respect to\(x\)and\(y\).

\({f_x}(x,y) = \frac{\partial }{{\partial x}}(3{(x - 1)^2} + 2{(y + 3)^2} + 7)\)

\(2(3(x - 1))\)

\( = 6(x - 1)\)..

\({f_y}(x,y) = \frac{\partial }{{\partial x}}(3{(x - 1)^2} + 2{(y + 3)^2} + 7)\)

\( = 2(2(y + 3))\)

\({f_y}({x_0},{y_0}) = 4(y + 3)\)

Find,\({f_x}\)at\((2, - 2,12)\)

\({f_x}(2, - 2) = 6(2 - 1)\)

\( = 6(1)\)

=6

\({f_y}(2, - 2) = 4(( - 2) + 3)\)

\( = 4(1) = 4\).

Step-3(Calculating The Equation of Tangent Plane)

\(z - {z_0} = {f_x}({x_0},{y_0})(x - {x_0}) + {f_y}({x_0},{y_0})(y - {y_0})\)

\(z = {f_x}({x_0},{y_0})(x - {x_0}) + {f_y}({x_0},{y_0})(y - {y_0}) + {z_0}\)

\( = 6(x - 2) + 4(y + 2) + 12\)

\( = 6x + 4y - 12 + 8 + 12\)

\(z = 6x + 4y + 8\)

Hence,The Tangent Plane Equation is \(z = 6x + 4y + 8\)…