Question

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

\(z = x\sin (x + y),( - 1,1,0)\)..

Short answer

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

Step-by-step solution

Step-1(Given information and Formula Used)

The equation of the surface is\(z = x\sin (x + y)\)or\(f(x,y) = x\sin (x + 2y)\)

Let\((x,{y_0},{z_0}) = ( - 1,1,0)\)..

We have to find the equation of tangent plane to the surface at the point\(( - 1,1,0)\).

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 \(f(x,y) = x\sin (x + y)\)..)

Partial Derivative of\(f(x,y) = x\sin (x + y)\)with respect to\(x\)and\(y\).

\({f_x}(x,y) = \frac{\partial }{{\partial x}}(x\sin (x + y))\)

\(x\cos (x + y) + \sin (x + y)\)

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

\({f_x}( - 1,1) = - 1\cos ( - 1 + 1) + \sin ( - 1 + 1)\)..

\( = - \cos 0 + \sin 0\)

\( = - 1\)

\({f_y}(x,y) = \frac{\partial }{{\partial x}}(x\sin (x + y))\)..

\( = x\cos (x + y)\)

\({f_y}( - 1,1) = ( - 1\cos ( - 1 + 1)) = - \cos 0\)

\( = - 1\)

Step-3(Calculating The Equation of Tangent Plane)

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

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

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

\(z = - x - y\)

\(x + y + z = 0\)

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