Question

Find an equation of the tangent plane to the surface at the given point. $$ x y^{2}+3 x-z^{2}=4, \quad(2,1,-2) $$

Short answer

The equation of the tangent plane to the given surface \(x y^{2}+3 x-z^{2}=4\) at the point (2,1,-2) is \(x + y + z = 0\).

Step-by-step solution

Calculate the Gradient Function

First we must compute the gradient of the function, which involves taking the partial derivatives of the function with respect to x, y, and z. The function is \(f(x, y, z)=x y^{2}+3 x-z^{2}-4\). So its gradient ∇f is given as: \(\nabla f = [f_x, f_y, f_z]\), where \(f_x = y^{2} + 3\), \(f_y = 2xy\), and \(f_z = -2z\)

Evaluate the Gradient

Next we substitute the given point (2,1,-2) into the above partial derivatives. Thus we get: \(f_x(2, 1, -2) = 1^{2} + 3 = 4\), \(f_y(2, 1, -2) = 2*2*1 = 4\), and \(f_z(2, 1, -2) = -2*-2 = 4\)

Form The Tangent Plane Equation

The equation of the tangent plane at a given point is \(f_x(x_0, y_0, z_0)(x - x_0)+f_y(x_0, y_0, z_0)(y - y_0)+f_z(x_0, y_0, z_0)(z - z_0)=0\), substituting \(f_x, f_y, f_z \mathrm{and} (x_0, y_0, z_0) = (2, 1, -2)\) we obtain \(4(x - 2) + 4(y - 1) + 4(z + 2)=0 \) which simplifies to \(x + y + z = 0\)