Question

Explain why the function is differentiable at the given point.Then find the linearization\(L(x,y)\) of the function at that point..

\(f(x,y) = {e^{ - xy}}\cos y,,(\pi ,0)\)

Short answer

The linearization of the function at that point is \(1 - \pi y\)

Step-by-step solution

Step-1(Equation used in finding the Linearization)

The equation use to find linearization is:-

\(L(x,y) = f(a,b) + {f_x}(a,b)(x - a) + {f_y}(a,b)(y - b)\),where \({f_x}\)and\({f_y}\) are partial derivatives..

Step-2(Calculation of Partial Derivative)

\(f(x,y) = {e^{ - xy}}*\cos y\)

\({f_x}(x,y) = - y{e^{ - xy}}\cos y\)

\({f_x}(\pi ,0) = - (0){e^{ - \pi (0)}}*\cos (0) = 0\)

Now, \(f(x,y) = {e^{ - xy}}*\cos y\)

\({f_y}(x,y) = - {e^{ - xy}}(\sin y + x\cos y)\)

\(\)\({f_y}(\pi ,0) = - {e^{ - \pi .0}}*(\sin y + x\cos y) = - \pi \)

Partial Derivative exist at the point\((\pi ,0)\) and are continuous at \((\pi ,0)\)..

Therefore, Linearization is possible..

Step-3(Calculation of Linearization)

The function\(f(x,y) = {e^{ - xy}}*\cos y\) at the point \((\pi ,0)\)is

\(f(\pi ,0) = {e^{ - \pi .0}}\cos (0) = 1\)

\(L(x,y) = f(a,b) + {f_x}(a,b)(x - a) + {f_y}(a,b)(y - b)\)

\(L(x,y) = 1 + 0(x - \pi ) - \pi (y - 0)\)

\( = 1 - \pi y\)

Hence, The linearization of the function at point \((\pi ,0)\) is \(1 - \pi y\)..