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Smooth, parameterized, regular curve (diff geometry) 
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#1
Aug1109, 03:06 AM

P: 581

1. The problem statement, all variables and given/known data
α(t) = (sint, cost + ln tan t/2) for α: (0:π) > R^{2} 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, d^{2}α/dt^{2}.... exist for i = 1,2,...,n Regular  a curve is regular if all its points are regular, that is dα/dt is nonzero. 


#2
Aug1109, 11:01 PM

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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.



#3
Aug1109, 11:32 PM

P: 581

I got: dα/dt = (cos(t),sin(t)+sec^{2}(t/2)/2tan(t/2)) then subbing in t = pi/2, dα/dt = (cos(π/2),sin(π/2)+sec^{2}(π/4)/2tan(π/4)) I got [0,0] you think I have made a mistake? 


#4
Aug1109, 11:42 PM

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Smooth, parameterized, regular curve (diff geometry)



#5
Aug1109, 11:49 PM

P: 581

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 yaxis is constantly equal to 1. Well, to start, since α: (0/pi) > R^{2}, 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? 


#6
Aug1209, 12:00 AM

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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 yaxis 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.



#7
Aug1209, 12:16 AM

P: 581

oh ok, makes sense now. So the length of the tangent bound at the ends by the yaxis 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. 


#8
Aug1209, 12:28 AM

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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.



#9
Aug1209, 12:38 AM

P: 581

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 yaxis, I set x=0 (?) I'm confused as how to proceed since the function has been parameterized. :/ 


#10
Aug1209, 08:04 AM

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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.



#11
Aug1209, 08:25 AM

P: 581

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 yintercept 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? 


#12
Aug1209, 08:31 AM

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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.



#13
Aug1209, 08:39 AM

P: 581

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? 


#14
Aug1209, 08:48 AM

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#15
Aug1209, 09:02 AM

P: 581

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 yintercept (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 d=1^1/2 d=1 YAY!, thanks alot dick! 


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