Something seems to be wrong

  • Thread starter medwatt
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Hello,
Suppose y=x^2. So dy/dx=2x. If we let x=t then we have a vector valued function r(t)=<t,t^2>. Hence r'(t)=<1,2t>. As can be seen we will not have the gradient after we take the modules of r'(t). Where am I going wrong. Also suppose I went on plotting r'(t) I will get a vertical line which is pretty useless and not as informative as dy/dx=2x. What's happening?
 

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  • #2
Office_Shredder
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r'(t) gives a vector which points in the direction of the tangent line of the graph of r(t). The absolute value contains some information but is dependent on your specific parametrization of r(t) so won't give you information about the graph of y=x2.

Since we're in only 2 dimensions, you immediately get that a vector pointing in the direction (x,y) has slope y/x, which means the slope of the tangent line is 2t/1 = 2t which is what you expect
 
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I already know that dy/dx={dy/dt)/(dx/dt). I was trying to plot the velocity vectors which resulted from diff the parametric equations. instead of attaching these velocity vectors to points as on the plane curve in text books I decided to shift all of these vectors to the origin to see what curve they'd define. I was then going to plot acceleration vectors and see that they are indeed perpendicular to the original. What is wrong with the parametrization that I used and if I used gradient parametrization will that solve the vertical line of r'(t) I'm getting?
 
  • #4
Office_Shredder
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I don't understand what information you are expecting to get out of the graph of r'(t), and why the fact that it's a perpendicular line is supposed to be surprising.
 
  • #5
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Well I am practising maths in mathematica and I remember reading in a dynamics book that velocity vectors could be plotted to trace a velocity plane curve. In those days I simply acquiesced to the information. But the concept was never explored further in the book except when differentiating unit vectors. But now since I'm not the one doing the plotting I just wanted to investigate the velocity vectors. I know that the acceleration is 2m/s2 so seeing a vertical line although already indicated by the i component of r'(t) would make anyone think the acceleration is infinite and the vertical line is pretty useless. It makes using parametrization seem useless. So looking at projectile motion particularly velocity v(t)=<uCos(t),uSin(t)-gt> you will also see the same straight line.
 
  • #6
Office_Shredder
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You would calculate the acceleration by finding r''(t) and getting (0,2). The slope of that line in the x-y plane does not correspond to the second derivative of y=x2, and in general trying to mix and match parametrization derivatives and y=f(x) derivatives like that is not going to work.

When I see the vertical line for r'(t) it tells me that your parametrization is linear in the x coordinate.
 

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