# find the tangent plane to the surface z = tan-1 y/x at the point (1,1)

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You need to remember the equation of tangent plane, at a point `(x_0,y_0), ` such that:

`z = f(x_0,y_0) + f_x(x_0,y_0)(x - x_0) + f_y(x_0,y_0)(y - y_0)`

You need to evaluate the partial derivatives of the function `z = tan^(-1) y/x`  such that:

`f_x = 1/(1 + (y/x)^2)*(-y/x^2)`

`f_y = 1/(1 + (y/x)^2)*(1/x)`

You need to evaluate `f_x(x_0,y_0)`  and `f_y(x_0,y_0)`  at `x_0 = 1`  and `y_0 = 1` , such that:

`f_x(1,1) = 1/(1 + (1/1)^2)*(-1/1^2) = -1/2`

`f_y(1,1) = 1/(1 + (1/1)^2)*(1/1) = 1/2`

`f(1,1) = arctan(1/1) = pi/4`

`z = pi/4 -(x - 1)/2 + (y - 1)/2`

Hence, evaluating the equation of the tangent plane, at the point (1,1), yields `z = pi/4 -(x - 1)/2 + (y - 1)/2.`

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