Question

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

\(z = x{e^{xy}},(2,0,2)\)..

Short answer

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

Step-by-step solution

Step-1(Given information and Formula Used)

The equation of the surface is\(z = x{e^{xy}}\)or\(f(x,y) = x{e^{xy}}\)

Let\((x,{y_0},{z_0}) = (2,0,2)\)..

We have to find the equation of tangent plane to the surface at the point\((2,0,2)\).

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 \(f(x,y) = x{e^{xy}}\)..)

Partial Derivative of\(f(x,y) = x{e^{xy}}\)with respect to\(x\)and\(y\).

\({f_x}(x,y) = \frac{\partial }{{\partial x}}(x{e^{xy}})\)

\(x{e^{xy}}(y) + {e^{xy}}\)

\(xy{e^{xy}} + {e^{xy}}\)

\({f_x}(2,0) = 2(0){e^{2(0)}} + {e^{2(0)}} = 1\)..

\({f_y}(x,y) = \frac{\partial }{{\partial x}}(x{e^{xy}})\)..

\( = {x^2}{e^{xy}}\)

\({f_y}(2,0) = ({2^2}{e^{2(0)}}) = 4\)..

Step-3(Calculating The Equation of Tangent Plane)

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

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

\(z = 1(x - 2) + 4(y - 0) + 2\)

\(z - 2 = x - 2 + 4y\)

\(x + 4y = z\)

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