Question

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

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

Short answer

The tangent plane Equation is \(z = - 7x - 6y + 5\)..

Step-by-step solution

Step-1(Given information and Formula Used)

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

Let\(({x_0},{y_0},{z_0}) = (2, - 1, - 3)\)..

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 \(f(x,y) = 3{y^2} - 2{x^2} + x\)..)

For Partial Derivative of\(f(x,y) = 3{y^2} - 2{x^2} + x\)

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

\( = - 4x + 1\)

\({f_x}({x_0},{y_0}) = {f_x}(2, - 1)\)

\( - 8 + 1 = - 7\)

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

\( = 6y\)

\({f_y}({x_0},{y_0}) = {f_y}(2, - 1)\)

\(6( - 1) = - 6\)

Step-3(Calculating The Equation of Tangent Plane)

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

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

\( = - 7(x - 2) - 6(y + 1) - 3\)

\( = - 7x - 6y + 14 - 6 - 3\)

\( = - 7x - 6y + 5\)

Hence,The Tangent Plane Equation is \(z = - 7x - 6y + 5\)…