Find an equation for the tangent plane to the surface \[ z+5=x y^{3} \cos (z) \] at the point $(5,1,0)$.
Solution
Step 1 ：Given the surface equation \(z+5=xy^3\cos(z)\), we want to find the equation of the tangent plane at the point \((5,1,0)\).
Step 2 ：The equation of the tangent plane to a surface given by \(f(x, y, z) = 0\) at a point \((x_0, y_0, z_0)\) is given by \(f_x(x_0, y_0, z_0)(x - x_0) + f_y(x_0, y_0, z_0)(y - y_0) + f_z(x_0, y_0, z_0)(z - z_0) = 0\), where \(f_x\), \(f_y\), and \(f_z\) are the partial derivatives of \(f\) with respect to \(x\), \(y\), and \(z\), respectively.
Step 3 ：First, we need to find the partial derivatives of the function \(f(x, y, z) = xy^3\cos(z) + 5 - z\).
Step 4 ：The partial derivative of \(f\) with respect to \(x\) is \(y^3\cos(z)\), with respect to \(y\) is \(3xy^2\cos(z)\), and with respect to \(z\) is \(-xy^3\sin(z) - 1\).
Step 5 ：Substituting the point \((5,1,0)\) into the partial derivatives, we get \(f_x = 1\), \(f_y = 15\), and \(f_z = -1\).
Step 6 ：Substituting these values into the equation of the tangent plane, we get the final equation \(x + 15y - z - 20 = 0\).
Step 7 ：So, the equation of the tangent plane to the surface \(z+5=xy^3\cos(z)\) at the point \((5,1,0)\) is \(\boxed{x + 15y - z - 20 = 0}\).
