Parameterization of Circles in Space

The video tutorial explains how to parameterize a circle of radius six with a center at (4, 5, 3) in a plane parallel to the XY plane. It details the constant Z-component and derives the parametric equations for the circle's XY trace. The final parameterization in space is given as x = 4 + 6 cos(t), y = 5 + 6 sin(t), and z = 3.

What is the radius of the circle for which we need to find a parameterization?

• 3

• 4

• 5

• 6

Why is the z component of the parameterization constant?

• Because the circle is on the yz-plane

• Because the circle is on the xz-plane

• Because the circle is a sphere

• Because the circle is parallel to the xy-plane

What is the z-coordinate of the circle's center?

• 6

• 3

• 5

• 4

How many units above the xy-plane is the circle located?

• 5

• 4

• 3

• 2

What is the center of the circle in the xy-plane?

• (4, 5)

• (5, 6)

• (3, 4)

• (6, 7)

What is the radius of the circle in the xy-plane?

• 4

• 5

• 6

• 7

What is the parametric equation for x in the circle's parameterization?

• x = 4 + 6 sin(t)

• x = 5 + 6 cos(t)

• x = 5 + 6 sin(t)

• x = 4 + 6 cos(t)

What is the parametric equation for y in the circle's parameterization?

• y = 4 + 6 sin(t)

• y = 5 + 6 cos(t)

• y = 4 + 6 cos(t)

• y = 5 + 6 sin(t)

What is the constant value of z in the circle's parameterization?

• 4

• 5

• 3

• 6

Which of the following correctly describes the parameterization of the circle in space?

• x = 5 + 6 sin(t), y = 4 + 6 cos(t), z = 3

• x = 5 + 6 cos(t), y = 4 + 6 sin(t), z = 3

• x = 4 + 6 cos(t), y = 5 + 6 sin(t), z = 3

• x = 4 + 6 sin(t), y = 5 + 6 cos(t), z = 3