Parametric Equations for Spiral on Radius 3 Sphere

Answer

The curve $\mathcal{C}$ can be parametrized by using $z$ as the parameter over the interval $[-3, 3]$. By combining the periodic nature of the spiral with the spherical geometry, the equations are $x = \sqrt{9-z^2}\cos(\pi z)$ and $y = \sqrt{9-z^2}\sin(\pi z)$.

Explanation

The first image shows a sphere of radius $R=3$ with a spiral curve $\mathcal{C}$ starting at $(0,0,-3)$ and ending at $(0,0,3)$, circling the $z$-axis exactly 3 times. The second image provides two functions: $g(z) = \sqrt{9-z^2}$, which represents the radius of the horizontal cross-section of the sphere at height $z$, and $f(z) = \cos(\pi z)$, which represents an oscillating component.

1. Defining the $z$-coordinate and the Domain From the problem description, the particle starts at the south pole $(0, 0, -3)$ and ends at the north pole $(0, 0, 3)$. Therefore, we use $z$ as our parameter $t$, where: $z = t, \quad t \in [-3, 3]$ This sets the vertical progression of the particle from the bottom to the top of the sphere.

2. Determining the horizontal radius $r(z)$ Any point $(x, y, z)$ on a sphere of radius 3 must satisfy the equation $x^2 + y^2 + z^2 = 9$. Rearranging for the radius of the circle at a fixed height $z$, we get $r = \sqrt{x^2 + y^2} = \sqrt{9-z^2}$. $r(z) = \sqrt{9-z^2}$ This formula calculates the distance from the $z$-axis to the surface of the sphere for any given height $z$.

3. Establishing the angular frequency The particle must spiral around the $z$-axis exactly 3 times as $z$ moves from $-3$ to $3$ (a total distance of 6 units). A full rotation is $2\pi$ radians; thus, 3 rotations require a total change in angle $\Delta \theta = 6\pi$. $\theta(z) = k \cdot z$ To find $k$, we set the change in $\theta$ over the interval $\Delta z = 6$ to be $6\pi$, which gives $k = \pi$. $\theta(z) = \pi z$ This linear relationship ensures that as $z$ increases by 2 units, the particle completes exactly one full $2\pi$ rotation.

4. Constructing the Parametric Equations Using standard polar-to-rectangular conversion $x = r \cos \theta$ and $y = r \sin \theta$, and substituting our derived expressions for $r(z)$ and $\theta(z)$: $x(z) = \sqrt{9-z^2} \cos(\pi z)$ The x-coordinate oscillates with a frequency determined by $\pi$ and an amplitude determined by the sphere's width. $y(z) = \sqrt{9-z^2} \sin(\pi z)$ The y-coordinate provides the orthogonal oscillation required to create circular motion in the $xy$-plane.

Final Answer

A possible set of parametric equations for the curve $\mathcal{C}$ is: $\boxed{\begin{cases} x = \sqrt{9-z^2} \cos(\pi z) \\ y = \sqrt{9-z^2} \sin(\pi z) \\ z = t \end{cases} \text{ for } t \in [-3, 3]}$

Common Mistakes

• Rotation Count Error: Students often forget to verify that the coefficient of the angle (here, $\pi$) results in the correct number of turns. If you used $\theta = 3\pi z$, the curve would spiral 9 times instead of 3.
• Radius Neglect: Failing to scale the $x$ and $y$ components by $\sqrt{9-z^2}$ results in a cylinder rather than a sphere. The "envelope" of the oscillation must match the circular profile shown in the second graph.

Related Topics: Spherical coordinates, Parametric curves in $\mathbb{R}^3$, Surface of revolution.