2.1 Space Curves: the Helix

A vector-valued function packages three scalar functions into one moving point:

$\mathbf{r}(t) = f(t)\,\mathbf{i} + g(t)\,\mathbf{j} + h(t)\,\mathbf{k},$

and its graph is a space curve. The curve carries no information beyond its components — so the fastest way to understand one is to un-weave it. The tool for that is the curve's shadows: its projections onto the three coordinate planes. Each shadow keeps two components and throws the third away, which means each is a plane curve you already know how to read.

Our first specimen is the helix $\mathbf{r}(t) = (\cos t,\ \sin t,\ t/4)$ : a point running around a circle while climbing at a steady rate.

A red helix winding two full turns up around the z-axis, accompanied by three gray curves, its shadows: a circle lying flat in the xy-plane directly below it, a cosine-shaped wave standing in the xz-plane, and a sine-shaped wave standing in the yz-plane. Two sliders edit the helix and its shadows together. Slider q, starting at 0, bends the height component from t over 4 toward t squared over 16 pi, which reshapes the two standing waves and leaves the circle unchanged. Slider m, starting at 1, sets the first component to the cosine of m t; at m equals 2 the circle collapses into a parabolic arc and the xz wave changes, while the yz wave stays the same. — A red helix climbing two turns around the $z$ -axis, with its three shadows drawn in gray: a circle in the $xy$ -plane, a cosine wave standing in the $xz$ -plane, and a sine wave in the $yz$ -plane. Two sliders, $q$ and $m$ , edit a component of the helix and of every shadow that uses it. The opening camera is chosen so the helix appears to cross itself.

Explore

Before orbiting: where the red curve appears to cross itself on screen, decide which strand is in front. Commit. Then orbit a quarter turn — were you right? Why can a single flat view of a space curve always fool you this way?
Match each gray shadow to its coordinate plane and to the pair of component functions that build it. Which component does each shadow discard?
Orbit to look straight down the $z$ -axis. Which shadow does the helix appear to become, and what information does this view destroy?
Predict: if the third component $t/4$ were changed to $t^2/(16\pi)$ , which shadows would change and which would stay exactly the same? Test your prediction with the slider $q$ , which blends the third component from $t/4$ at $q = 0$ to $t^2/(16\pi)$ at $q = 1$ in the helix and in every shadow that keeps it. Drag it from 0 to 1, then back to 0.
Predict again for changing $\cos t$ to $\cos 2t$ : which shadow is untouched this time? The $xy$ -shadow stops being a circle — what shape does it collapse into? Test it by moving the slider $m$ from 1 to 2; the first component is $\cos(mt)$ .

Shadows are components, two at a time

Project the helix onto the $xy$ -plane and you get $(\cos t, \sin t, 0)$ — the pair $(f, g)$ with $h$ set to zero. The other two shadows are $(\cos t, 0, t/4)$ and $(0, \sin t, t/4)$ . That is the general rule: the shadow on a coordinate plane is the curve with the discarded component replaced by zero. It explains everything you tested above. Editing $h$ alone cannot touch the $xy$ -shadow, because that shadow never knew about $h$ ; it must change both vertical shadows, because each of them keeps $h$ . Every edit changes exactly the two shadows that use the edited component.

Read as graphs, the vertical shadows are old friends. In the $xz$ -plane the shadow satisfies $x = \cos(4z)$ — a cosine wave lying on its side, since $t = 4z$ — and the $yz$ -shadow is the matching sine wave. The helix is a circle and two sinusoids in a trench coat, and the shadows catch it.

Example 1 A shadow after surgery

Suppose the helix's first component is edited from $\cos t$ to $\cos 2t$ , giving $\mathbf{r}(t) = (\cos 2t,\ \sin t,\ t/4)$ . Find an $xy$ -equation for the new shadow in the $xy$ -plane.

Show solution

The new shadow is $(\cos 2t,\ \sin t,\ 0)$ , so $x = \cos 2t$ and $y = \sin t$ . The double-angle identity $\cos 2t = 1 - 2\sin^2 t$ eliminates the parameter at once:

$x = 1 - 2y^2.$

The circle has collapsed into a parabolic arc opening in the negative $x$ -direction, traced back and forth as $t$ runs — a simple Lissajous figure. The $yz$ -shadow, which never used the first component, is exactly the sine wave it was before the edit.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Aligned to OpenStax Calculus Volume 3 (Strang & Herman), © OpenStax (Rice University), licensed CC BY-NC-SA 4.0; no OpenStax content is reproduced, and this work is not affiliated with or endorsed by OpenStax or Rice University. License: CC-BY-NC-SA-4.0.