# Paramaterizing a curve

by fleazo
Tags: curve, paramaterizing
 P: 71 Sure, it's a simple procedure for straight lines. Your particle's velocity vector will be in line with the vector ##(1, 2)##, and will have a magnitude of ##e^t##. The normalization of ##(1, 2)## is ##(\frac{1}{\sqrt 5}, \frac{2}{\sqrt 5})##. The equation for your velocity vector will thus be \begin{align}v_x(t) &= \frac{e^t}{\sqrt 5}\\ \\ \\ \\ v_y(t) &= \frac{2e^t}{\sqrt 5}\end{align} You integrate to get your curve (or in this case "line") parameterization, which in this case comes out identical to the velocity because of the nature of the exponential function: \begin{align}r_x(t) &= \frac{e^t}{\sqrt 5}\\ \\ \\ \\ r_y(t) &= \frac{2e^t}{\sqrt 5}\end{align} Then you throw in a constant of integration to represent the fact that your particle doesn't go through the origin. We'll say that at ##t=0## the particle is at location ##(0, 1)##, so the equations become \begin{align}r_x(t) &= \frac{e^t}{\sqrt 5}\\ \\ \\ \\ r_y(t) &= 1 + \frac{2e^t}{\sqrt 5}\end{align} If you differentiate those ##r_x## and ##r_y## functions you get the velocity functions. And if you take those and pop them into ##\sqrt{v_x^2 + v_y^2}## you ought to get ##e^t##; that is, unless I made a mistake somewhere, because actually I rushed while writing this. Gota run!