Quadric Functions

Question

The directrix of a cylinder is a circle in the xz-plane with generators perpendicular to that plane.

The directrix is tangent to the x-axis at x=3

The directrix is tangent to the z-axis at z=3

Find an equation of the cylinder.

Sketch the portion of the cylinder between the xz-plane and the plane y=5.

Answer 1

While this might not be a complete answer, hopefully it will be enough to get you going in the right direction...

Because the directrix of this right cylinder (as stated by the words "with generators perpendicular to [the xy] plane"), first consider the general equation for a circle: (x-h)²+(y-h)²=r², (h,k)=center & r=radius. In this case, the circle is in the xz plane, not xy, so simply replace the y with z in the equation. Then if you draw a sketch of said circle (tangent to x-axis @ x=3 & z-axis @ z=3), you see that the center of the circle lies at (3,0,3), with radius=3. So the equation of the circle alone is (x-3)²+(z-3)²=9, or, f(x,z)=(x-3)²+(z-3)²-9.

The formula, or equation for the associated cylinder - which is created by layering one circle directly on top of another - can then be found by finding the integral from one y-value to the next, as the cylinder rises parallel to the y-axis.

I hope this helps!

Answer 2

The center of the circle in the zx-plane is (x, y, z) = (3, y, 3)

The radius of the above is thus 3.

So, the equation is:

(x-3)^2+(z-3)^2 = 3^2