Parameterizing a circle of radius r: do we have to use trigonometric functions? Can't we just set $y=\sqrt{9-t^2}$, $z=t$?

Question

Let's consider a circle traced out on the $yz$-plane by $y=\sqrt{9-z^2}$ (note that $x=0$).

To parameterize this, wouldn't we simply set $z=t$ and $y=\sqrt{9-t^2}$ ?

For some reason my textbook takes the approach of using trigonometric functions:

Is there a difference between my solution and the one the textbook shows? If not, is there a reason it choose the less intuitive solution involving trigonometric equations instead of the traditional equation of a circle?

Furthermore, couldn't set $y=-3 $sin$t$ and $z=$cos$t$ and we'd get the same result? Just trying to gain some intuition behind this because I'm not used to circles being expressed using trignometric functions...

Answer 1

In general to parameterize a function you can choose $x=t, y=f(t)$. Of course, a circle fails the vertical line test in the plane so it's not a function and this means we must parameterize the circle using another method. This is typically the trig functions or stereographic projection.

Answer 2

Your parametric functions are not correct as $\sqrt c$ only returns positive values. You would only get half the circle with your equations.

Trigonometric functions $\sin$ and $\cos$ are used because of their simplicity and how easy it is to work with them. $\sin$ and $\cos$ functions are used to describe harmonic motion. A dot tracing a circular path is a superimposition of these two harmonic functions.

Putting $+/-$ in front of $\sin$ and $\cos$ wont change the shape drawn. It will only affect the starting point and direction(clockwise or anti clockwise) of the point.

I suggest you plot the trigonometric functions in something like desmos and play around a little. It will give you an intuitive sense of how they describe the circle. Plot $x={\cos t}$ and $y={\sin t}$ separately. Play around with the slider for $t$. Then plot the point $A({\cos t},{\sin t})$.