Find the equation of the tangent plane to \( z=\sqrt{x^{2}+y^{2}} \) at the point \( (5,-12,13) \). DO NOT MULTIPLY THROUGH BY A LEADING NEGATIVE SIGN. DDO MULTIPLY THROUGH BY DENOMIATORS.

To find the equation of the tangent plane to the surface \( z = \sqrt{x^2 + y^2} \) at the point \( (5, -12, 13) \), we first need to compute the partial derivatives at that point. The partial derivatives of \( z \) are: \[ \frac{\partial z}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} \quad \text{and} \quad \frac{\partial z}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}}. \] Now, substituting the coordinates \( (5, -12) \) into these formulas: \[ \frac{\partial z}{\partial x} \bigg|_{(5, -12)} = \frac{5}{\sqrt{5^2 + (-12)^2}} = \frac{5}{\sqrt{25 + 144}} = \frac{5}{13}, \] \[ \frac{\partial z}{\partial y} \bigg|_{(5, -12)} = \frac{-12}{\sqrt{5^2 + (-12)^2}} = \frac{-12}{\sqrt{25 + 144}} = \frac{-12}{13}. \] The general equation for the tangent plane at point \( (x_0, y_0, z_0) \) is given by: \[ z = z_0 + \frac{\partial z}{\partial x}\bigg|_{(x_0, y_0)} (x - x_0) + \frac{\partial z}{\partial y}\bigg|_{(x_0, y_0)} (y - y_0). \] Substituting the values we have: \[ z = 13 + \frac{5}{13}(x - 5) + \frac{-12}{13}(y + 12). \] To put it in the standard form, we can eliminate the fractions by multiplying through by 13: \[ 13z = 169 + 5(x - 5) - 12(y + 12). \] Expanding this: \[ 13z = 169 + 5x - 25 - 12y - 144. \] Simplifying: \[ 13z = 5x - 12y + 0. \] Thus, the equation of the tangent plane can be written as: \[ 13z = 5x - 12y. \]