Parameteric Curves: Partial Diff, Tangent Line?

  • Context: Graduate 
  • Thread starter Thread starter quietrain
  • Start date Start date
  • Tags Tags
    Curves
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
quietrain
Messages
648
Reaction score
2
lets say i have a parameterized curve r(t)

if i do r'(t), is it the same as if i were to do a partial differentiation of d/dx d/dy d/dz ?

so i get r'(x,y,z) = (dr/dx, dr/dy, dr/dz) ?

that means these all discribe the tangent line to the curve right?

thanks
 
Physics news on Phys.org
quietrain said:
lets say i have a parameterized curve r(t)
I presume that you mean that r(t)= (x(t), y(t), z(t)).

if i do r'(t), is it the same as if i were to do a partial differentiation of d/dx d/dy d/dz ?

so i get r'(x,y,z) = (dr/dx, dr/dy, dr/dz) ?
No. You get, instead, (dx/dt, dy/dt, dz/dt).

that means these all discribe the tangent line to the curve right?
it means that the tangent line, at [itex](x(t_0), y(t_0), z(t_0))= (x_0, y_0, z_0)[/itex] is given by [itex]((dx/dt)(t_0)t+ x_0, (dy/dt(t_0)t+ y_0, (dz/dt)(t_0)t+ z_0)[/tex]<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> thanks </div> </div> </blockquote>[/itex]
 
HallsofIvy said:
it means that the tangent line, at [itex](x(t_0), y(t_0), z(t_0))= (x_0, y_0, z_0)[/itex] is given by [itex]((dx/dt)(t_0)t+ x_0, (dy/dt(t_0)t+ y_0, (dz/dt)(t_0)t+ z_0)[/tex][/itex]
[itex] <br /> oh so it is like an equation of the line right? r<sub>x</sub> = x + tv where v is a vector parallel to the x-direction ? so in this case it is the dx/dt? <br /> also, i meant that if i have r(x,y,z) = (x<sup>2</sup>,y<sup>2</sup>,z<sup>2</sup>) , and i do <a href="https://www.physicsforums.com/insights/partial-differentiation-without-tears/" class="link link--internal">partial differentiation</a>, to get r'(x,y,z) = (2x,2y,2z) , then this is the tangent line equation right? <br /> <br /> so if i parameterize r(x(t),y(t),z(t)) , and do r'(t), i get the equation you wrote above right? which is the tangent line equation parameterized.<br /> <br /> so are they the same? just that one is parameterized one is not?[/itex]
 
quietrain said:
oh so it is like an equation of the line right? rx = x + tv where v is a vector parallel to the x-direction ? so in this case it is the dx/dt?
Yes, every vector can be written as a sum of vectors parallel to the axes.



also, i meant that if i have r(x,y,z) = (x2,y2,z2)
I have no idea at all what that means. Are (a, b, c) the x,y,z coordinates or not? If so, then you are saying that [itex]x= x^2[/itex], [itex]y= y^2[/itex], [itex]z= z^2[/itex] so that your set is not a curve at all but is 8 discrete points.

, and i do partial differentiation, to get r'(x,y,z) = (2x,2y,2z) , then this is the tangent line equation right?
NO.

so if i parameterize r(x(t),y(t),z(t)) , and do r'(t), i get the equation you wrote above right? which is the tangent line equation parameterized.
I don't know what you mean by that. Are you still saying that [itex]r(x,y,z)= (x^2, y^2, z^2)[/itex]? As I said before, points satifying that do NOT form a curve and it cannot be parameterized.

so are they the same? just that one is parameterized one is not?
A vector (or point) function in an xyz-coordinate system is always (x, y, z) by definition of "xyz-coordinate system". You have to have x, y, and z functions of some other variables in order to have a set. For example, if they are functions of one parameter, (x(t), y(t), z(t)), this is a one dimensional figure, a curve. If they are functions of two parameters, (x(u,v), y(u,v), z(u,v)) then it is a two dimensional figure, a surface. To write something like "r(x, y, z)= (f(x), g(y), h(z)) would mean that you are requiring the points (x,y,z) to satisfy x= f(x), y= g(y), z= h(z) which, typically, will reduce to a finite number of points.
 
HallsofIvy said:
A vector (or point) function in an xyz-coordinate system is always (x, y, z) by definition of "xyz-coordinate system". You have to have x, y, and z functions of some other variables in order to have a set. For example, if they are functions of one parameter, (x(t), y(t), z(t)), this is a one dimensional figure, a curve. If they are functions of two parameters, (x(u,v), y(u,v), z(u,v)) then it is a two dimensional figure, a surface. To write something like "r(x, y, z)= (f(x), g(y), h(z)) would mean that you are requiring the points (x,y,z) to satisfy x= f(x), y= g(y), z= h(z) which, typically, will reduce to a finite number of points.

oh... i see... so what about a straight line?

so 3 parameters = volume?

what if i write it this way r = x2+y2+z2? does this make sense? is this a curve or surface or? so if i do partial diff on this one, do i get the tangent line?btw: if i want to parameterize a vector (1,0,1), is it (t,0,t)?