Paraboloid Family

What it draws

The height is x²/9 + s·y²/4 over a square 6 on a side, with s on its own slider. The first term is a parabola along x that reaches 1 at the edges x = ±3. The second is a parabola along y, scaled by s, which at s = 1 reaches 2.25 at y = ±3.

Nothing moves on its own. The surface changes only when s does, and the one thing that decides its kind is the sign of s.

Bowl, trough, saddle

For s above zero both parabolas open upward and the surface is an elliptic paraboloid, a bowl. Its level curves x²/9 + s·y²/4 = c are ellipses, and at s = 1 the corners stand at 3.25.

At s = 0 the second term vanishes. The height no longer depends on y, and every slice across the square is the same parabola: a parabolic cylinder, a trough.

For s below zero the parabola along y turns over and the surface is a hyperbolic paraboloid, a saddle. It rises along x and falls along y, and at s = −1 the height is zero on the two lines y = ±2x/3.

This is the second derivative test drawn out. At the origin the determinant of the second derivatives is (2/9)·(s/2) = s/9. Positive means a minimum, zero means the test is silent, and negative means a saddle point.

Try

• Drag s from 1 down to −1 and watch the bowl flatten into a trough at 0, then turn into a saddle.
• Set s to 0.4. Then 0.4/4 = 0.1 is close to 1/9 ≈ 0.11, so the bowl is nearly round, a paraboloid of revolution.
• Replace s with sin(t), as in x²/9 + sin(t)·y²/4. The surface passes through all three kinds every 2π ≈ 6.28 seconds at Speed 1.
• Press Top in the deck at s = −1. The colours split the square into two rising and two falling wedges, parted by the lines where the height is zero.

Read more

• Paraboloid
• Parabolic cylinder
• Second partial derivative test
• Quadric