The behavior of the tracetrix( to verify properties)

[Cauchy1789]

The behavior of the tracetrix(need help to verify properties) :(

Homework Statement

Howdy

Given the parametric function [tex]\beta(t) = (sin(t), cos(t) + ln(tan(t/2)) [/tex]

where t is the angle between the tangent vector and the y-axis and where

[tex]\beta: (o,\pi) \rightarrow \mathbb{R}^2 [/tex] then show the following two properties of the tracetrix are true.

1) that's its a regular parameterized curve differentiable curve except at

t = [tex]\frac{\pi}{2} [/tex]

2) That the length of the line segment of tangent in a tangent point of the curve, then intersecting with y-axis, this length always will be 1.

The Attempt at a Solution

1) According to the definition from my textbook a curve is parameterized curve is said to be regular if [tex] \beta'(t) \neq 0[/tex] for all t in I

thusly since my tracetrix has the [tex]\beta'(t) = (cos(t), -sin(t) + \frac{1}{2 \cdot sin(t/2) \cdot cos(t/2)})[/tex] and thusly [tex]\beta'(\pi/2) = 0,0[/tex]. Therefore the derivative at t = [tex]\frac{\pi}{2} [/tex] is zero and according to the definition the curve is not regular at that point.
But what about the end points? From what I can see they don't have a corresponding (x,y) and thusly they tend to [tex]\pm \infty[/tex]. So what about the definition applying in those end points?

2) From I can understand here the point is that do to the geometrical behavior of the tracetrix then tangent is asymptotical to [tex]\beta(t)[/tex](do I need to show this) and thusly the distance of every tangent point on the interval I (except t = pi/2) will always be the same.
(A hint/idear on howto show this would be very much appricated :D)

Sincerely
Cauchy1789

 

[Dick]

The domain of the curve is (0,pi). t=0 and t=pi are not points on the curve that you have to worry about.

 

[Cauchy1789]

The domain of the curve is (0,pi). t=0 and t=pi are not points on the curve that you have to worry about.

Hi Dick and thank you for your reply :)

So anyway in question number one, have I covered all that needs to be covered?

I tried pretty much to verify the definition of the regular curve:)

Any hints for question 2? Howto verify the asymptotical behaviour of curve which supposedly corresponds to every possible tangent on our Interval I? Which then if I remember correctly should let me conclude that distance from any tangent always will be constant?
I am messed up in part 2?

Sincerely Cauchy1789

 

[Dick]

In the first part of the problem, you can write the tangent vector even more simply than you have. Once you have done that the second part should be pretty easy. From the tangent vector you should be able to figure out the slope of the tangent line in terms of t. Now pick a point p on the curve having parameter t. It's x coordinate is sin(t). Draw a right triangle with one leg the distance from p to the y-axis, the hypotenuse parallel to the tangent and the other leg along the y-axis. You know the horizontal side has length sin(t) and knowing the slope of the tangent you should be able to figure out the length of the vertical side. Then use Pythagoras to get the length of the hypotenuse. Which is what you want.

 

[Cauchy1789]

Hi again Dick an thanks for looking at my stuff again,

To simplify the tangent vector of [tex]\beta(t)[/tex] that can be written as

[tex]\frac{d\beta}{dt} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex] if [tex]\frac{dx}{dt}\neq 0[/tex].

Thus [tex]\frac{d\beta}{dt} = \frac{-sin(t) + \frac{1}{2 \cdot sin(t/2) \cdot cos(t/2)}}{cos(t)} = \frac{1}{2 \cdot cos(t) \cdot sin(t/2) \cdot cos(t/2)} - tan(t) = \frac{1}{sin(t) \cdot cos(t)} - tan(t) = \frac{1}{tan(t)} [/tex]

Since [tex]2 \cdot sin(t/2) \cdot cos(t/2) = sin 2(t/2)[/tex]

Before continueing to part 2. Isn't the above what you mean by simplifying (1)?? Because I still get the same properties as in the less simplyfied version of tangent vector where [tex]t = \frac{\pi}{2}[/tex] is the only place where [tex]\beta'(t)[/tex] is zero and thus doesn't "live up" to the defintion of a regular parameterized curve.

Sincerely
Cauchy

 

[Dick]

Yes, that's what I meant by simplifying beta'(t), but beta'(t) is supposed to be a vector, isn't it. I wasn't challenging your conclusion about regularity, that seems just fine. A simpler version of beta'(t) will make it easier to do the second part though.

 

[Cauchy1789]

Yes, that's what I meant by simplifying beta'(t), but beta'(t) is supposed to be a vector, isn't it. I wasn't challenging your conclusion about regularity, that seems just fine. A simpler version of beta'(t) will make it easier to do the second part though.

Hello again Dick,

I am not very good a drawing graph and figure. Have them in my head but anyway.

If the length tangent vector is

[tex](\frac{1}{tan(t)}) = cot t[/tex] (remembering that t is the angle of the angle tangent makes with the y-axis.

then we have the formula.

[tex]1+ (cot(t))^2 = csc(t)^2[/tex], and the since csc is defined as

[tex]csc(t) = \frac{hyp}{opp} = \frac{1}{sin(t)} = \frac{hyp}{sin(t)}[/tex]

thus [tex]hyp = \frac{sin(t)}{sin(t)} = 1[/tex].

thus the adj has the length of [tex]\frac{1}{sin(t)}[/tex]

and as long as t is defined on the interval mentioned above then the of line segment of tangent will have the length of 1.

How is that Dick?

Sincerely Cauchy...

p.s. Any chance you could illustrate this triangle. All mine look very fishy..

 

[Dick]

If it's clear to you, I guess it's ok. You keep talking about a tangent vector, but I don't see any vectors. It is true that the slope of the tangent vector is cot(t), if that's what you mean to say. From that if h is the horizontal leg and v is the vertical leg, I conclude v/h=cot(t). Now using h=sin(t), I figure out h=cos(t) and then hyp=sqrt(v^2+h^2). Same thing, I guess.

 

[Cauchy1789]

Hi again,

I understand my part 2 myself but understandable for others? Would I need to add something to present part 2 better?


Sincerely Cauchy...

 

[Dick]

It would probably be a lot clearer if you sketch the triangle diagram and label the parts which you can do when you turn the problems in.

 

[Cauchy1789]

It would probably be a lot clearer if you sketch the triangle diagram and label the parts which you can do when you turn the problems in.

Okay but besides from its cool? :D

and you say that tangent is parallel to distance of p to the y-axis? Doesn't that mean that they almost are on top of each other?

 

[Dick]

Okay but besides from its cool? :D

and you say that tangent is parallel to distance of p to the y-axis? Doesn't that mean that they almost are on top of each other?

I don't understand what "tangent is parallel to distance of p to the y-axis" means. I don't think I said that. I said the hypotenuse is parallel to the tangent vector.

 

[Cauchy1789]

I don't understand what "tangent is parallel to distance of p to the y-axis" means. I don't think I said that. I said the hypotenuse is parallel to the tangent vector.

Hi again Dick and thanks for replying.

I guess I read wrong.


Anyway here is my graph

http://img528.imageshack.us/img528/6731/tracetrixqk9.jpg

The blue is tangent and the red is hyp of the triangle(which is visable (a bit)).

Have illustrated it correctly?

Sincerely Cauchy

 

[Dick]

I drew the red segment as the horizontal connecting p to the y-axis. So it had length sin(t). That made the tangent the hypotenuse. Which sides in your picture have known length?

 

[Cauchy1789]

I drew the red segment as the horizontal connecting p to the y-axis. So it had length sin(t). That made the tangent the hypotenuse. Which sides in your picture have known length?

the red has the length sin(t) and the is the tangent which has tan(t), since cot(t) = 1/tan(t).

which hyp is the red lin between (some what parallel) to the tangent marked in blue.

Does that make my drawing understable ?

Sincerely Cauchy

 

[Dick]

The red segment in your picture doesn't have length sin(t). The x coordinate of p is sin(t). The distance from p to the y-axis is sin(t).

 

[Cauchy1789]

Hi Dick I was able to draw a figure which somewhat resembles what you surgested by hand, but thanks anyway for your patience and help..

Sincerely

Cauchy