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Smooth, parameterized, regular curve (diff geometry)

  1. Aug 11, 2009 #1
    1. The problem statement, all variables and given/known data

    α(t) = (sint, cost + ln tan t/2) for α: (0:π) -> R2

    Show that α is a smooth, parametrized curve, which is regular except for t = π/2

    3. The attempt at a solution

    I am familiar with the definitions of smooth and regular, which I have provided below, however I am unsure as to how to formally show what the question asks.

    Am I supposed to show that dα/dt at t=π/2 is zero and hence not regular?
    for what it's worth, I have computed dα/dt at t=π/2 and it is = 0!

    Smooth - a function α(t) = α1(t), α2(t)....αn(t) is smooth if each of its components α1, α2,...,αn of α is smooth, that is, all the derivatives dαi/dt, d2α/dt2.... exist for i = 1,2,...,n

    Regular - a curve is regular if all its points are regular, that is dα/dt is nonzero.
    Last edited: Aug 11, 2009
  2. jcsd
  3. Aug 11, 2009 #2


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    The derivative of (sin(t),cos(t)+log(tan(t/2))) at pi/2 isn't zero. Did you mean log(tan(t))/2? That's not zero either, but it is undefined.
    Last edited: Aug 11, 2009
  4. Aug 11, 2009 #3
    it isn't?

    I got:
    dα/dt = (cos(t),-sin(t)+sec2(t/2)/2tan(t/2))

    then subbing in t = pi/2,
    dα/dt = (cos(π/2),-sin(π/2)+sec2(π/4)/2tan(π/4))
    I got [0,0]

    you think I have made a mistake?
  5. Aug 11, 2009 #4


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    Nope. I did. Sorry. I substituted it into the original function without differentiating. Ok, since da/dt is zero at pi/2, isn't that enough to make it non-regular, or do you actually need to show the tangent vector is discontinuous?
  6. Aug 11, 2009 #5
    I personally beleive that proving the function is zero at pi/2 is sufficient!

    there is a part b) that I am stumped on....

    The range of α is called the tractrix, show that the length of the segment of the trangent to the tractrix between the point of tangency and the y-axis is constantly equal to 1.

    Well, to start, since α: (0/pi) -> R2, I wish to find the range so I sub in 0 and pi into α. However in trying to find the lowest point of the range (i.e. α evaluated at 0), I immediately run into problems, as ln(tan(0)) is not defined.

    Am I on the right track regarding firstly finding the range?
  7. Aug 12, 2009 #6


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    You don't have to put t=0. It's not in the domain. You are supposed to pick a value of t between 0 and pi, intersect the tangent through the point on the curve with the y-axis and show the length of the segment is always 1, regardless of the value of t. If you get around to plotting the resulting curve you will see it does have a discontinuous tangent at t=pi/2, where the derivative vanishes.
  8. Aug 12, 2009 #7
    oh ok, makes sense now. So the length of the tangent bound at the ends by the y-axis and the chosen point, at any point between t=0 and t=pi (except for t=pi/2) is 1.

    now mathematically, how am I to show this? I don't think choosing random t between 0 and pi and showing the above is sufficient as the question asks to show the length is constantly equal to 1. So is there a general proof that will satisfy the condition for all of the tractrix? Apart from graphically, as I do not have access to software that will plot multivariables.
  9. Aug 12, 2009 #8


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    You've already got the tangent vector, that gives you the slope of the segment. The equation of the tractrix gives you a point it passes through. Just use plain old analytic geometry to intersect that with the y axis. Now find the distance between the two points. I haven't actually worked this out for a while, but I have extreme faith that it will work. Just leave t unknown. Things should just cancel out and give you 1. Try it. If you have problems, give the solution up to where you've gotten and I'll try to help.
    Last edited: Aug 12, 2009
  10. Aug 12, 2009 #9
    righty oh!
    so the tangent vector = α' = dα(t)/dt = (cos(t),-sin(t)+sec2(t/2)/2tan(t/2))

    now to find the intersection with the y-axis, I set x=0 (?)
    I'm confused as how to proceed since the function has been parameterized. :/
  11. Aug 12, 2009 #10


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    The segment passes through the point P=(sint, cost + ln tan t/2)=(x(t),y(t)). If da/dt=(x'(t),y'(t)) then the usual slope m=dy/dx=y'(t)/x'(t). So yes, then find an equation for the tangent segment like y=mx+b and set x=0. It will probably help if you notice that sec^2(t/2)/(2*tan(t/2)) can be simplified a lot using a double angle formula. I just worked it out again. It's really not bad if you keep your head screwed on.
    Last edited: Aug 12, 2009
  12. Aug 12, 2009 #11
    I managed to simplify dy/dx=y'(t)/x'(t) down into dy/dx = -tan(t) + 0.5sin(2t) using trig identities and double angle formulas.

    So now I have the "gradient" (as shown above), and the a point P, that is on the line but I'm not sure how to put it all together and find the y-intercept since I have a function of x and y as independent variables. Is there an analogous equation to y=mx + c for the multivariable case?
  13. Aug 12, 2009 #12


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    I don't think your simplification of m=dy/dx came out right. I get something much simpler. There is no 'multivariable' version of y=mx+b. That's it. Consider t to be a constant for now. Put your m in there, put y=y(t) and x=x(t). Solve for b.
  14. Aug 12, 2009 #13
    damn, ill show my working for dy/dx, but it could get messy!

    dy/dx = y'(t)/x'(t) = [-sin(t) + sec^2(t/2)/2tan(t/2)]/[cos(t)]

    change the sec^2(t/2) to 1/cos^2(t/2) and change the tan(t/2) to sin(t/2)/cos(t/2)

    then dy/dx = [-sin(t) + 1/2cos(t/2)sin(t/2)]/[cos(t)]

    then used double angle formula to change 1/2cos(t/2)sin(t/2) to sin(t) such that

    dy/dx = [-sin(t) + 1/sin(t)]/[cos(t)]

    then -sin(t)/cos(t) = -tan(t) and 1/sin(t)/cos(t) = 0.5sin(2t)

    so that dy/dc = -tan(t) + 0.5sin(2t)

    Can you pick out the mistake?
  15. Aug 12, 2009 #14


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    That's right. I wouldn't use a double angle to simplify though. Just write it as (-sin^t+1)/(sin(t)*cos(t))=cos^2(t)/(sin(t)*cos(t))=cos(t)/sin(t)=cot(t).

    1/(sin(t)*cos(t))=2/sin(2t). But you really don't want to go that way.
  16. Aug 12, 2009 #15
    ok, that came out pretty nicely.

    so now y=mx+b, with m = cot(t), x = x(t) = sint, y = y(t) = cost + ln tan t/2
    subbing in leads to...

    cos(t) + ln tan t/2 = cot(t)sin(t) + b
    blah blah blah

    b = ln tan t/2

    So now I have the point P = (sint, cost + ln tan t/2) and the y-intercept (0,ln tan t/2) and I wish to find the length between the two.

    So I guess I shall revert to basics and use distance formula,

    d={[sin(t)-0]^2 + [cost + ln tan t/2 - ln tan t/2]^2}^1/2

    d={sin^2(t) + cos^2(t)}^1/2



    YAY!, thanks alot dick!
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