Sketch curve $r(t)=\cos t\hat{\bf i}+\sin t\hat{\bf j}+t\hat{\bf k}$

Question

This is from "Multivariable Calculus, Concepts and Contexts" by Stewart.

He says "The parametric equations for this curve are: $x=\cos t$, $y=\sin t$, $z=t$. This makes sense.

However, he then goes on to state that "Since $x^2+y^2=\cos^2t+\sin^2t=1$, the curve must lie on the circular cylinder $x^2+y^2=1$.

I understand the trigonometric identity, I just don't understand how we get to even consider $x^2+y^2=1$ since the vector function is defined as $r(t)=\cos t\hat{\bf i}+\sin t\hat{\bf j}+t\hat{\bf k}$.

Where does $x^2+y^2=1$ come from? How do we get to use that fact when $x=\cos t$ and $y=\sin t$?

Answer 1

It's a simple substitution. You know from trigonometry that $$\cos^2t+\sin^2t=1,$$ and you know from the parametric equation of the curve that $$x=\cos t, y=\sin t.$$ So for any point of the curve, it is true that $$x^2+y^2=1.$$ Just notice that you dropped the equality $$z=t,$$ meaning that you get more $(x,y,z)$ values that satisfy the equalities: you obtain a superset of the curve points. Indeed, the parametric equation corresponds to a linear entity (1D), while $x^2+y^2=1$ defines a surface (2D), which contains the curve.

COMPLEMENT

If you replace $z=t$ by $t=0$, which amounts to projecting the curve on the $xy$ plane, you get a circle centered at the origin, with radius $1$. Coming back to $z=t$, you will "unroll" the circle and extend it in the $z$ direction, while remaining on the cylinder, forming a spring (called an helix in geometry).

Answer 2

For parametrizing the cylindre you can parametrize a circle $(\cos(t),\sin(t),0)$ and then "move" along the axe passing through its center $(0,0,0)$ and orthogonal to the plane where the circle is contained (i.e. $(0,0,s)$). Thus the cylinder can be parametrized by $$c(s,t) = (\cos(t),\sin(t),s), \quad t,s \in \mathbb{R}$$ the curve you are considering is $$\gamma(t) = c(t,t),\quad t \in \mathbb{R}$$ and therefore must lie on the cylindre. The expression $\cos^2(t)+\sin^2(t)=1$, shows you that a point $(\cos(t),\sin(t),t)$ on your curve is always at a distance $1$ from the point $(0,0,t)$ (the axis of the cylindre).