
#1
Apr1612, 04:26 PM

P: 66

Hello. I had always been annoyed when dealing with parametric equations but it finally dawned on me how awesome the concept it. I want to make sure I understand things properly.
So say I have a regular straight line given by rectangular equation y=2x+1 Now I can graph this. Now say I want to consider a particle moving in motion along the path of that line. But say I want that particle to move with exponential speed, say e^t, so shouldn't I then be able to paramaterize the curve to represent that? Is there any simple algorithmic process to do so? Thanks! 



#2
Apr1612, 08:56 PM

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! 


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