Reparametrizing a space curve in terms of arclength:

PreDefined

Hello folks! I'd be happy for any help you could give me. Thanks! :)

First though I'd like to make clear that I want to solve this problem myself, even if something seems obvious to you and you feel there is no harm in making a certain logical leap - please take the time to consider whether or not it's a step which I should be working out myself. I have very little background in Calc. and it took me a while just to place this problem when digging through my old texts from uni, but I would like to understand what the concepts I'm dealing with are.

Here is the original question: (Everything else that follows are notes that I have tracked down on the net to help me understand it.)

Starting from the point (5, 5, -4), reparametrize the curve: r(t) = ( 5 -3 t )i + ( 5 -2 t )j + ( -4 -2 t )k in terms of arclength
This was all that was presented. Is this a matrix then? The spacing does not seem to indicate something as simple as f(t)=(5-3t); g(t)=(5-2t); h(t)=(-4-2t); but I don't want to make any assumptions.

I'm really quite at a loss as to where to start. In simple terms it seems that I really haven't gotten anywhere yet and must still:

1. Figure out what ( 5 -3 t )i + ( 5 -2 t )j + ( -4 -2 t )k represent in a way that I can understand what I'm working with.
2. Find the arclength.
3. Reparametrize.

I'm assuming the relationship between the point given and the curve is one where by the question is supposed to be that much similar. Something akin to t=0 or something?

Other notes I picked up and which I'm still digesting:
A vector function of a single variable is much like a parametrized curve,
f(t)=f1(t)i+f2(t)j+f3(t)k
(where i, j, k, are the vectors (1,0,0), (0,1,0), and (0,0,1), respectively)

The functions fi are the component functions and much of calculus of vector functions of one variable is done by components, for example,
f'(t)=f1'(t)i+f2'(t)j+f3'(t)k

A curve is simply defined as a vector function of a single variable:
r(t)=x(t)i+y(t)j+z(t)k

http://planetmath.org/encyclopedia/SpaceCurve.html [Broken]
http://rsp.math.brandeis.edu/GenericMirror/palais/public_html/3D-Filmstrip_html/Documentation/DocumentationPages/Curves.html (Documentation for 3D-XplorMath)
http://oregonstate.edu/dept/math/CalculusQuestStudyGuides/vcalc/arc/arc.html

Thanks for any help you can give! I'm quite in the dark. Last edited by a moderator:

Janitor

I'll take a whack at it.

Call arc length s. s is a function of the parameter t. Let s(t) = 0 at the point (5,5,-4), i.e. s(0)=0.

Then

ds = sqrt[(dx)^2 + (dy)^2 + (dz)^2]
= sqrt[(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2] dt
= sqrt[(-3)^2 + (-2)^2 + (-2)^2] dt

So s(t) = sqrt t.

Solving for t as a function of s, t(s) = s/sqrt.

Reparameterizing,

r(s)= (5-3s/sqrt)i + (5-2s/sqrt)j + (-4-2s/sqrt)k.

Am I even on the right track? I haven't tried to do this sort of thing before.

arildno

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You have the right answer, Janitor.
PreDefined:
1. Hint:
Try to find out why your curve is a straight line.

PreDefined

arildno said:
You have the right answer, Janitor.
PreDefined:
1. Hint:
Try to find out why your curve is a straight line.
Would I be crazy to try to attempt to do row-reduced form like this?

Code:
5 -3 t
5 -2 t
-4 -2 t
---------
0 -5  0     R1-R2
0  1  0     R2-R1
1 -5  2t    R3+R1
---------
1  0  2t    R3-R1,  R3 <-> R1
0  1  0
0  0  0     R1+5R3, R1 <-> R3

Your matrix doesn't have any meaning.

Why are you trying to use matrices to solve this problem?

Ebolamonk3y

? When are matrices used anyways? I dun see a real system of equations here...

PreDefined

That's what I'm trying to understand. I'm not a math student. I work as a receptionist/secretary in a physio clinic. Approaching this problem I'm trying not to let what little math I did learn all blend together. If you tell me I can drop what little I remember from linear, I will - but by talking about parameters and such it seemed to tie right in with it.

So is ( 5 -3 t ) actually (5-3t) ? Or is it a vector ( 5 -3 t)?

If so then where does i = <1,0,0>, j= <0,1,0>, k= <0,0,1> come in?

Does it become r(t) = (5) + (-2) + (t) ?

r is given in the problem statement.

r(t) = (5-3t)i + (5-2t)j + (-4-2t)k

Do you understand what a vector function is?

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arildno

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PreDefined:
I see now where you got lost in the first place:
The "spacing" between a number and "t" has absolutely no significance whatsoever!!
You are absolutely correct in your assumption that:
f(t)=5-3t, g(t)=(5-2t), h(t)=(-4-2t).

Good luck!

PreDefined

Thank you all for your time!

Ok, thanks a lot then! Sorry everyone for all that confusion. Please correct me anywhere I go wrong, and if there is a proper mathematical way of saying something please let me know. ;)

So following Janitor like he did it:

r(t) = (5-3t)i + (5-2t)j + (-4-2t)k

So given x, y, z in terms of parameter t; t= 0 as starting point; and t>0 as positive acrlength s; the formula is s(t)= int_0^t sqrt[x'^2+y'^2+z'^2]dt.

Is it because t can grow to be infinitely large that the numbers by themselves have no meaning? For example with (5-3t) in [5/10000] - [(-3*10000)/10000], [5/10000] has almost no meaning so we drop it. So far so good? Or is it because we are starting from the point (5, 5, -4)?

s(t)= int_0^t sqrt[(-3)^2 + (-2)^2 + (-2)^2] dt
s(t)= int_0^t sqrt[(-3)^2 + (-2)^2 + (-2)^2] dt
s(t)= int_0^t sqrt[(9) + (4) + (4)] dt
s(t)= int_0^t sqrt dt

So now with arclength s as a function of t, we solve for t as a function of s and substitute that formula into the parametric equation:

t(s) = s/sqrt
r(t(s))= (5-3s/sqrt)i + (5-2s/sqrt)j + (-4-2s/sqrt)k

Is that all?

I think I still need to understand that point above about whether starting from the point (5, 5, -4) affects things. Is there an upper limit? Are there any specific areas that you would suggest I read up on? :)

arildno

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Is it because t can grow to be infinitely large that the numbers by themselves have no meaning? For example with (5-3t) in [5/10000] - [(-3*10000)/10000], [5/10000] has almost no meaning so we drop it. So far so good? Or is it because we are starting from the point (5, 5, -4)?
No, absolutely not!!

First, let us regard the usual notation for an arbitrary point in 3-D:

An arbitrary point P in 3-D can be written as a 3-tuple: (X,Y,Z).
(X,Y,Z) is a code that allows us to locate the point P:
1)
We go a distance X along the x-axis from the origin (0,0,0); we have then reached the intermediate point (X,0,0)

2) We then travel from (X,0,0) a distance Y along an axis parallell to the y-axis, such that we reach the intermediate point (X,Y,0)

3) Finally, we reach P by going a distance Z along an axis parallell to the z-axis, such that we reach (X,Y,Z) (i.e., P itself!)

Now, in explicit vector notation, we have:
$$(1,0,0)=\vec{i}, (0,1,0)=\vec{j}, (0,0,1)=\vec{k}$$

Hence, the arbitrary point P can be represented as the vector $$\vec{P}$$:
$$\vec{P}=x\vec{i}+y\vec{j}+z\vec{k}$$
(The only reason why I used uppercase letters formerly, was that I there chose to use the lowercase letters in the axis descriptions)

Now, let us return to a vector representation of a curve C:
$$\vec{r}(t)=x(t)\vec{i}+y(t)\vec{j}+z(t)\vec{k}$$

That is, the x-coordinate of a point P on C is given as a function of the single variable "t", and we call that function x(t), for obvious reasons.
A similar argument holds for the y and z-coordinates of P.

Now, we will find an expression for the arclength of a curve segment:
Suppose we have 2 points on C, $$\vec{P}_{1},\vec{P}_{2}$$, where:
$$\vec{P}_{1}=x_{1}\vec{i}+y_{1}\vec{j}+z_{1}\vec{k}$$,
$$\vec{P}_{2}=x_{2}\vec{i}+y_{2}\vec{j}+z_{2}\vec{k}$$

Note in particular that I have not yet used the fact that I can represent the curve C as a function of "t" at all. This fact will enter a bit later!

Now, clearly, if we draw the straight line segment/secant L between $$\vec{P}_{1},\vec{P}_{2}$$, the length of L, $$\bigtriangleup{s}$$, is given by the Pythagorean theorem:
$$\bigtriangleup{s}=\sqrt{(\bigtriangleup{x})^{2}+(\bigtriangleup{y})^{2}+(\bigtriangleup{z})^{2}}$$,
where:
$$\bigtriangleup{x}=x_{2}-x_{1},\bigtriangleup{y}=y_{2}-y_{1},\bigtriangleup{z}=z_{2}-z_{1}$$

Let us now utilize that the curve C may be represented as a function of "t":
Let $$\vec{P}_{1}$$ be given at the evaluation $$t=t_{1}$$, whereas $$\vec{P}_{2}$$, be given at the evaluation $$t=t_{2}$$.
That is:
$$\vec{r}(t_{1})=x(t_{1})\vec{i}+y(t_{1})\vec{j}+z(t_{1})\vec{k}, x(t_{1})=x_{1}, y(t_{1})=y_{1}, z(t_{1})=z_{1}$$
$$\vec{r}(t_{2})=x(t_{2})\vec{i}+y(t_{2})\vec{j}+z(t_{2})\vec{k}, x(t_{2})=x_{2}, y(t_{2})=y_{2}, z(t_{2})=z_{2}$$

We now introduce:
$$\bigtriangleup{t}=t_{2}-t_{1}\rightarrow{t}_{2}=t_{1}+\bigtriangleup{t}$$,
and we assume that $$\bigtriangleup{t}$$ is small.
Then we have:
$$\bigtriangleup{x}=x_{2}-x_{1}=x(t_{1}+\bigtriangleup{t})-x(t_{1})\approx\frac{dx}{dt}\mid_{(t=t_{1})}\bigtriangleup{t}$$

Possibly, you are unfamiliar with the notation $$\frac{dx}{dt}\mid_{(t=t_{1})}$$.

This simply means that you are to evaluate the derivative of x(t) (that is the function $$\frac{dx}{dt}$$) at the t-value $$t_{1}$$.

For example, if $$x(t)=t^{2}$$, then $$\frac{dx}{dt}=2t$$ and
$$\frac{dx}{dt}\mid_{(t=t_{1})}=2t_{1}$$.

Clearly, similar expressions hold for $$\bigtriangleup{y},\bigtriangleup{z}$$

Therefore we have when we suppress the function evaluation notation $$\mid_{(t=t_{1})}$$:
$$\bigtriangleup{s}\approx\sqrt{ (\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}+(\frac{dz}{dt})^{2}}\bigtriangleup{t}$$

We now proceed to the limit $$\bigtriangleup{s}\rightarrow{0}$$, and sum together all line segments to find the total arclength S:
$$S\equiv\int_{C}{ds}=\int_{0}^{T}\sqrt{ (\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}+(\frac{dz}{dt})^{2}}dt$$,

where the curve C has been represented by t-values lying between 0 and T.

Hence, the reason why the constants disappears is that you worh with the derivatives of the functions (the derivative of a constant is zero)

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PreDefined

Wow, I really appreciate the work you put in to explaining that.

Thanks again for all your time and effort! :)

Calculus, by arildno, to be published in December. arildno to be made a millionaire by January.

Ebolamonk3y

man... so formal... cookie man could make a practical man edition out of such formality. Chrono

Calculus, by arildno, to be published in December. arildno to be made a millionaire by January.
Hey, I'd buy it. And after your first million, don't forget to show your love for the forum and contribute, ok? It's not like you need that much anyway.

Ebolamonk3y

Chrono said:
Hey, I'd buy it. And after your first million, don't forget to show your love for the forum and contribute, ok? It's not like you need that much anyway.
Should be a Schaum's guide thing... like arildno's guides... And have a through 3000page explanation for Calculus. :)

Chrono

Ebolamonk3y said:
Should be a Schaum's guide thing... like arildno's guides... And have a through 3000page explanation for Calculus. :)
It'll be better than the so called book we use.

Ebolamonk3y

What do you guys use? Clay tablets?

arildno

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Calculus, by arildno, to be published in December. arildno to be made a millionaire by January.

Just because I had another boring day, doesn't mean I'm getting rich..

arildno

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Ebolamonk3y said:
man... so formal... cookie man could make a practical man edition out of such formality. Well, since I didn't define the integral as the limit of Riemann sums, and didn't present $$\epsilon,\delta$$-proofs of my statements, I thought my argument was rather sloppy...

Hence, "formal," and not "rigorous." Physics Forums Values

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