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Tangent to parametric curves

  1. Aug 7, 2005 #1
    Stewart uses the chain rule to show how to find the tangent to parametric curves. Given:

    x=f(t) and y=g(t), and that y can be written in terms of t, in other words, y=h(x)

    then the chain rule gives us, dy/dx = (dy/dt)/(dx/dt).

    Thats fine. The same argument holds for polar coordinates where t is replaced by theta. Looking at polar coordinates for a second, an example he has is r=1+sin(theta). Here, I dont see how it is possible to write y=h(x) directly. Its not possible to write y as a function of x in all cases. I thought this method should work ONLY when that condition is met. However, from vector calculus, I already know that this is just a special 2d case, where you can write x(t)i+y(t)j+0k. And the derivative w.r.t time or theta will give you the tangent. So the slope is just going to be (dy/dt)/(dx/dt) IN ALL CASES. Here it is not necessary to use the chain rule or the condition y=h(x). Am I wrong? ( I usually am so it wouldent suprise me) But my gripe is that hes finding the tanget to a polar curve when one of the conditions in his proof, that y=h(x) is not met. Also, I thought that the reason why we used polar coordinates was becuase we can graph functions in a polar system that we cannot in a standard cartesian coordinate system, so it would seem rather odd to have the condition y=h(x) in the proof in the first place, since thats typically something we cant do, and so we have to use polar.
     
  2. jcsd
  3. Aug 8, 2005 #2
    Any thoughts?
     
  4. Aug 9, 2005 #3

    matt grime

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    Why should it matter whether you think you can rearrange the equations to get y as a fucntion on x (how do you know you can't? have you tried?)? i really don't see what the problem is.
    and incidentally a parametric equation doesn't allow you to write y as a f****ion of x. it allows you locally to have y as a function of x.
     
  5. Aug 9, 2005 #4
    The reason I say that is becasue the book says,

    let x=f(t), and y=g(t)

    and suppose we can write y as a function of x,

    then using the chain rule, we get:

    dy/dt = dy/dx (dx/dt)

    rearranging gives, dy/dx = (dy/dt)/(dx/dt)

    Saying that y is a funciton of x, doesnt that imply that we can elimate t betwen x=f(t) and y=g(t) in order to get y=h(x)?

    In the example he uses, r=1+sin(theta), Then x=(r)cos(theta) y=(r)sin(theta).

    I do not think its possible to write y=f(x) here, I tried but could not eliminate both r and theta.
     
    Last edited: Aug 9, 2005
  6. Aug 9, 2005 #5

    matt grime

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    just because you cannot do it doesn't make it impossible. I think i can do it remembering that r^2=x^2+y^2. But there is a delicate question, one that is routinely overlooked by such unmathematical books as are used in such courses, about what a function is. for instance take the parametrization of the circle


    x=cos(t) y=sin(t)

    this does not define y as a fucntion of x (when x=1 there are two choices for y).


    plus you seem to be assuming that when we say y is defined as a function of x that there is a nice formula y=h(x) that we can pbtain by rearranging and "solving for y", well, that isn't true in almost all cases. this si the notion of implicitly defining a function, ie given some relations between y and x we can hopefully uniquely choose for each x a y making the relations hold, this defines a function of y in terms of x but not necessarily in any nice way.
     
    Last edited: Aug 9, 2005
  7. Aug 9, 2005 #6
    Hmm, i might be wrong, but are you saying that it is not possible to write y=h(x) in all cases. If we cant write y=h(x), then are we in violation of using dy/dx = (dy/dt)/(dx/dt) becuase we have no way of performing dy/dx?

    I tried that, but could not get rid of either the r or the theta, depending on which variable you choose to leave. you will get (1+sin(theta))^2. After expanding it out, I couldent get rid of the theta.

    You dont mean x=0? I thought at x=1, y=0 and only 0, but at x=0, y = +/- 1


    At the very least, lets say we have x=f(t) and y=g(t). In order to express y as a function of x, wont we have to require that x be invertible, so we can substitute in for t in y=g(t), or is there a way to make y a function of x without being invertible? Because x=(1+sin(theta))cos(theta) is not invertible, which means it wont be possible to to solve y=h(x) by substitution.
     
    Last edited: Aug 9, 2005
  8. Aug 9, 2005 #7

    matt grime

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    If you wish to do this properly, ie mathematically, then you need to read the implicit function theorem.

    All of the thigns that are bothering you are addressed there in the correct statements of the results.


    Yes, i indeed meant x=0, not x=1 in that example.

    I suggest you try harder in you;re example since it cna be done (sin(theta) is y/r which is y/sqrt(x^2+y^2), it really isn't very diffcult.


    in anycase differentiaion is purely a local property. I do not need to invert f and g at all points to be able to work out the tangent to the curve at that point. I just restrict my attention to small area in the xy plane where i can invert f and/or g and and i can apply the chain rule there.


    Just read a decent defintion of the implicit function theorem.
     
  9. Aug 9, 2005 #8
    Ah, I see matt. It would be nice if stewart had at least bothered to mention that small but important fact, he left that detail out. Is it a very long/difficult theorem to read/understand?
     
  10. Aug 9, 2005 #9

    matt grime

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    no, it is straightforward and essentially puts on a firm footing what you are thinking already.


    take the circle example, obviously x=cost(t) and y=sin(t) doesn't define y as a fucntion of x in the *proper* meaning of function (if x=2 what is y? etc) but we are perfectly happy to work out the slope at any point using (dy/dt)(dt/dx). Why? well, if we restrict y to be in the upper half of the plane and x between -1 and 1 then y=sqrt(1-x^2) and we can do the opposite sign for the lower half plane. ie we just restrict attention to a small bit in the xy plane where it does define a function properly, and on that we are fine.


    It's not an important issue (as long as dt/dx isn't zero you're fine) and one they often safely gloss over.
     
  11. Aug 9, 2005 #10
    A minor detail, you wrote (dy/dt)(dt/dx), but can you state 1/(dx/dt)=(dt/dx)? Wouldnt it have to be (dy/dt)[(dx/dt)^-1] What you stated would hold only if x=f(t) is invertible.
     
  12. Aug 9, 2005 #11

    matt grime

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    i can state that and it has nothing to do with f being invertible and, as i told you, i was restricting to a place where i can invert f locally, that's the whole sodding point of the implicit function theorem as far as i can tell.
     
  13. Aug 9, 2005 #12
    Ah, I see.
     
  14. Aug 9, 2005 #13
    Im searching for it, but im finding a big variation of proofs. http://mathworld.wolfram.com/ImplicitFunctionTheorem.html
    is from mathworld. I dont know if this is what you were referring to. It does not seem like this proof from mathworld has to do with insuring that y can be written in terms of x. To generalize, does the implicit function theorem tell you, given a function x=f(t) and y=g(t), when it is and is not possible to write y as a function of x? It wont tell you how nor will it tell you if its going to be nice or not, but it will either say yes you can POSSIBLY do it, or no you cannot?


    As an example, lets say Im given some general function r=f(theta). And I want to find the slope dy/dx at some arbitrary point. Would I be able to just jump striaght into using (dy/dt)/(dx/dt), or would I have to use a check, such as th implicit function theorem you mentioned, before using (dy/dt)/(dx/dt)?
     
    Last edited: Aug 9, 2005
  15. Aug 9, 2005 #14

    matt grime

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    for the last time, y cannot be necessarily written in terms of x for ALL y and x, that is not what we're getting at.

    BUT

    if you want to find the slope of some parametrized curve then by looking close to that point we can do it and as derivatives are purely a local thing this is good enough. (and the criterion for say y=f(t) and x=g(t) is that both df/dt and dg/dt do not vanish simultaneously, ok?)

    do you get the difference between local behaviour and global behaviour?


    look, i'm losing patience here so i'm not going to respond any more, sorry. i don't have your book, i dont' know exactly what it says, and more importantly i really don't care.
     
    Last edited: Aug 9, 2005
  16. Aug 9, 2005 #15
    Yeah, I dont mean all x and y. I mean for ( some x and y), as you did in your example with the circle. Sorry I was not clear.

    You said earlier
    and

    Im a little bit confused as to how restricting the xy plane makes a function invertible. In your example, you restircted x between -1 and 1. And use the half circle above the x axis. But in this case, we were able to write x^2+y^2=1, becuase we could square the cos and sin term using the trig identity. So in a sense, we already had a way to write the equation in terms of x and y only, for certain intervals of x. We said x= -1,1 and y= sqrt(1-x^2). and x=-1,1 and y=-sqrt(1-x^2).

    Im a little bit confused on how this theorem works. If I have a function y=f(x) that is not invertible, how does looking at the function locally make it invertible? Even if I restrict the values of x, how would I go about re-writting y=f(x). For instance, y=t^3+t^(4/3)+sin(t) is not invertible. Lets say you look at it on a limited piece where it is invertible. I do not see how you could express the inverse of that equation on that domain while still using the same equation. Hopefully that makes my question clearer. Im not totally understanding how the invertable theorm works. (Sorry, Im not trying to anger you or not listen. This is the first time I have ever heard of it, and I was just trying to find out how it would apply and be used.)
     
    Last edited: Aug 9, 2005
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