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. $$ z=3-x^{2}-y^{2}+6 y $$

Short answer

The point on the surface where the tangent plane is horizontal is the solution from Step 3, with the z-coordinate calculated in Step 4. This point and the plane are described in Step 5.

Step-by-step solution

Calculate the first derivatives

To find the horizontal tangent on the surface of z, the first step is to calculate the first derivatives of the function. To do this, differentiate the function with respect to \(x\) and \(y\): Find \(\partial z/ \partial x\) and \(\partial z/ \partial y\) by applying the standard rules of differentiation.

Set the first derivatives equal to zero

The tangent plane is horizontal at a point where the slope in both the \(x\) and \(y\) directions are zero. That corresponds to the values of \(x\) and \(y\) that make both the \(\partial z/ \partial x\) and \(\partial z/ \partial y\) equal to zero. Therefore, setting both first derivative equal to zero and solve the resulting system of equations.

Solve the first derivative equal to zero

This step involves solving equations for each of the variables, \(x\) and \(y\). This might require skills in algebra, particularly if the equations are not simple linear equations.

Substitute \(x\) and \(y\) into original function

Now that the solution has been calculated, it needs to be verified. Substitute the \(x\) and \(y\) values into the original function \(z=3-x^{2}-y^{2}+6 y\) to find the value of \(z\) at this point.

Visualize the surface graph and tangent plane

The final step involves using a computer algebra system to visualise the the 3D surface described by the function, the point where the tangent plane is horizontal, and the plane itself. You may see how the tangent is horizontal by looking at the plane really hitting the surface just at one point.