Question

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

\(z = \sqrt {xy} ,(1,1,1)\)..

Short answer

The tangent plane Equation is \(x + y - 2z = 0\)..

Step-by-step solution

Step-1(Given information and Formula Used)

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

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

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

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 \(f(x,y) = \sqrt {xy} \)..)

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

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

\(\frac{y}{{2\sqrt {xy} }}\)

\({f_x}(1,1) = \frac{1}{{2\sqrt {1(1)} }} = \frac{1}{2}\)..

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

\( = \frac{x}{{2\sqrt {xy} }}\)

\({f_y}(1,1) = (\frac{1}{{2\sqrt {1(1)} }}) = \frac{1}{2}\)..

Step-3(Calculating The Equation of Tangent Plane)

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

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

\(z = \frac{{(x - 1)}}{2} + (\frac{{y - 1)}}{2} + 1\)

\(2z = x + y\)

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

Hence,The Tangent Plane Equation is \(x + y - 2z = 0\)