#math/calculus An equation describes which points lie on a curve, but not the order or speed with which they are traced. A parametrization adds that missing motion: a curve in space is a moving point $ \mathbf r(t)=\langle x(t),y(t),z(t)\rangle, \qquad t\in I. $ The parameter determines not only the traced set, but also direction and speed. Different parametrizations can trace the same geometric curve differently. To recognize a curve, inspect projections and eliminate $t$. For $ \mathbf r(t)=\langle\cos t,\sin t,t\rangle, $ the $xy$ projection satisfies $x^2+y^2=1$, while $z$ rises steadily. The result is a helix on a cylinder. ![[tikz-b09ecf3c5488.svg]] %% tikz-source ```latex \usetikzlibrary{arrows.meta} \begin{document} \begin{tikzpicture}[>=Stealth,scale=1.0,line cap=round] \draw[->,black!45] (0,0)--(3.4,-1.0) node[right]{$x$}; \draw[->,black!45] (0,0)--(-2.4,-1.1) node[left]{$y$}; \draw[->,black!45] (0,0)--(0,5.2) node[above]{$z$}; \draw[blue!70!black,very thick,samples=180,domain=0:720,smooth,variable=\t] plot ({1.55*cos(\t)+.38*1.55*sin(\t)},{.0062*\t+.33*1.55*sin(\t)}); \coordinate (P) at ({1.55*cos(390)+.38*1.55*sin(390)},{.0062*390+.33*1.55*sin(390)}); \fill (P) circle (2.2pt) node[left]{$\mathbf r(t_0)$}; \draw[->,red!75!black,very thick] (P)--++(-1.2,1.0) node[above]{$\mathbf r'(t_0)$}; \node[align=left] at (5.0,3.3) {$\mathbf r(t)=(\cos t,\sin t,t)$\\velocity is tangent}; \end{tikzpicture} \end{document} ``` %% Straight lines have the form $\mathbf r(t)=\mathbf p+t\mathbf v$. Curves such as $\langle t,2t,\cos t\rangle$ are easiest to understand by first seeing the line $y=2x$ in projection and then adding the oscillating height. The same object can be viewed as a triple of scalar functions or as a [[14F2 - Vector Functions and Derivatives|vector-valued function]].