Question

Find the point on the surface where the tangent plane is horizontal. Use a computer algebra system to graph the surface and the horizontal tangent plane. Describe the surface where the tangent plane is horizontal. $$ z=3 x^{2}+2 y^{2}-3 x+4 y-5 $$

Short answer

The point on the surface \(z = 3x^2 + 2y^2 - 3x + 4y - 5\) at which the tangent plane is horizontal is (0.5, -1, -4).

Step-by-step solution

Calculation of Partial Derivatives

Firstly, the goal is to calculate the partial derivative of the function \(z = 3x^2 + 2y^2 - 3x + 4y - 5\) with respect to x and y. The partial derivative with respect to x is \(\frac{\partial z}{\partial x} = 6x -3 \) and the partial derivative with respect to y is \(\frac{\partial_z}{\partial y} = 4y + 4 \)

Determination of the Points for the Horizontal Tangent Plane

The next step is to set both partial derivatives equal to zero since for the tangent plane to be horizontal, both rate of changes in the x and y-directions should be zero. This results in the system of equations: \\[6x -3 = 0 \\4y + 4 = 0\\] Solving this system of equations for x and y gives the point as \\[x = 0.5\\y = -1\\]

Calculation of z-coordinate for the point

To locate the exact point on the surface, substitute x and y into the given equation to find the corresponding z-coordinate. This results in \(z = 3(0.5)^2+ 2(-1)^2 - 3(0.5) + 4(-1) - 5\), which simplifies to \(z = -4\). So the point at which the tangent plane is horizontal is (0.5, -1, -4)

Graphing the surface and the tangent plane

The final step involves graphing the surface and the horizontal plane using a computer algebra system. The surface graph would be produced using the function \(z = 3x^2 + 2y^2 - 3x + 4y - 5\). The graph of the horizontal plane is a constant on the z-axis, at \(z = -4\)