Question

Find an equation of the tangent plane and find symmetric equations of the normal line to the surface at the given point. $$ z=\arctan \frac{y}{x}, \quad\left(1,1, \frac{\pi}{4}\right) $$

Short answer

The equation of the tangent plane at the point (1,1, \( \pi/4 \)) is \( -\frac{1}{2}(x-1) + \frac{1}{2}(y-1) + (z - \pi/4) = 0 \). The symmetric equations of the normal line to the surface at this point are \( \frac{x - 1}{-1/2} = \frac{y - 1}{1/2} = \frac{z - \pi/4}{1} \).

Step-by-step solution

Calculation of Gradient

First, compute the gradient of the given scalar field. This will give us normal vectors to the tangent plane. The given scalar field is \( z = \arctan(y/x) \) so it needs to be converted into \( f(x,y,z) = 0 \) form. Hence, equation becomes \( f(x,y,z) = z - \arctan(y/x) = 0 \). Compute the partial derivatives \( f_x = -\frac{y}{x^2+y^2} \), \( f_y = \frac{x}{x^2+y^2} \), and \( f_z = 1 \). At the point (1,1,\( \pi/4 \)) these become \( f_x(1,1,\pi/4) = -1/2 \), \( f_y(1,1,\pi/4) = 1/2 \), and \( f_z(1,1,\pi/4) = 1 \). Therefore, gradient of \( f \) at the given point is ( -1/2, 1/2, 1 ).

Equation of Tangent Plane

The equation of the tangent plane is given by \( (f_x, f_y, f_z) \cdot (x - x_0,y - y_0, z - z_0) = 0 \). Substituting the values of \( f_x \), \( f_y \), \( f_z \), and the coordinates of the given point (1,1,\( \pi/4 \)) into the equation gives \( -\frac{1}{2}(x-1) + \frac{1}{2}(y-1) + (z - \pi/4) = 0 \).

Symmetric Equations of Normal Line

The symmetric equations of the normal line to the surface at the given point are given by \( \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} \), where \( a, b, c \) are the components of the normal vector. Substituting the values into the equation yields \( \frac{x - 1}{-1/2} = \frac{y - 1}{1/2} = \frac{z - \pi/4}{1} \).