Differential geometry : Tangent vector & reparameterization

In summary: AND complex numbers; where csgn = complex signum.The arc length function in your relevant equations gives the arc length in terms of a parameter t.IOW, ##\int_0^t 5 \sqrt{\cosh^2(w)} dw## is the reparameterization of the tangent vector.
  • #1
Schwarzschild90
113
1

Homework Statement


Problem statement uploaded as image.

Homework Equations


Arc-length function
eq0014M.gif

The Attempt at a Solution


Tangent vector:
r=-sinh(t), cosh(t), 3

Now, I just need to reparameterize it using arclength and verify my work is unit-speed. Will someone give me a hint? Should I use the arc-length function to accomplish this.
 

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  • #2
Schwarzschild90 said:

Homework Statement


Problem statement uploaded as image.

Homework Equations


Arc-length function
eq0014M.gif

The Attempt at a Solution


Tangent vector:
r=-sinh(t), cosh(t), 3

Now, I just need to reparameterize it using arclength and verify my work is unit-speed. Will someone give me a hint? Should I use the arc-length function to accomplish this.

What is preventing you from trying it for yourself?
 
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  • #3
It's that I have no means of checking the solution, so before I invest in it, I would like to know if my method is correct (assuming that I integrate correctly).
 
  • #4
Schwarzschild90 said:

Homework Statement


Problem statement uploaded as image.

Homework Equations


Arc-length function
eq0014M.gif

The Attempt at a Solution


Tangent vector:
r=-sinh(t), cosh(t), 3
This isn't the tangent vector.
Schwarzschild90 said:
Now, I just need to reparameterize it using arclength and verify my work is unit-speed. Will someone give me a hint? Should I use the arc-length function to accomplish this.
In future posts, please show more of your work. What you have here just barely qualifies as a problem attempt.
 
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  • #5
Tangent vector
upload_2015-12-13_18-41-41.png


Now, compute the norm of the tangent vector:
upload_2015-12-13_18-55-58.png

Using this, make the following substitution
upload_2015-12-13_19-1-35.png
 
  • #7
How do I compute the arclength, without knowing the range? For example [0 <= t <= 2pi]

Another shot at the arc length of the tangent vector

[itex]\sqrt{(9+9*sinh(t)^2+16*cosh(t)^2)}dt =^*
25 cosh^2(t) =
25 sinh^2(t)+25
[/itex]

* Using a trigonometric identity

PS: csgn is code used specifically by maple. It' not necessarily a mathematical function
 
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  • #8
Schwarzschild90 said:
How do I compute the arclength, without knowing the range? For example [0 <= t <= 2pi]
The arc length function in your relevant equations gives the arc length in terms of a parameter t.
Schwarzschild90 said:
Another shot at the arc length of the tangent vector

[itex]\sqrt{(9+9*sinh(t)^2+16*cosh(t)^2)}dt =^*
25 cosh^2(t) =
25 sinh^2(t)+25
[/itex]
The last expression above is not helpful, but the one before it is helpful. What happened to the square root?
Schwarzschild90 said:
* Using a trigonometric identity

PS: csgn is code used specifically by maple. It' not necessarily a mathematical function
Do you know what it means, though? I've never seen it, but I don't use Maple.
 
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  • #9
Right, the square root should've been preserved, in the above equation. Here it is, in all of its glory:

[itex]\sqrt{25cosh^2(t)}[/itex]
So, is this equation the reparameterization of the tangent vector?

csgn(x) is the sign function of real AND complex numbers; where csgn = complex signum.
 
  • #10
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  • #11
upload_2015-12-13_20-51-51.png

Plot of the 25cosh^2(t) function; the norm of the tangent vector
 
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  • #12
Schwarzschild90 said:
View attachment 93321
Plot of the 25cosh^2(t) function

This is not relevant. The question is what cosh(t) looks like, not its square. You should not even need to do an actual plot; just picture it in your mind.
 
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  • #13
upload_2015-12-13_21-9-33.png


Plot of [itex]5 \sqrt{cosh(t)}[/itex]

I can picture it in my mind. What am I supposed to "see"?
 
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  • #14
I get this for the parameterization by arclength of the tangent vector
[itex]int(5*sqrt(cosh(t)^2), t = 0 .. 1) = 5 sinh(1)[/itex]
 
  • #15
Schwarzschild90 said:
I get this for the parameterization by arclength of the tangent vector
[itex]int(5*sqrt(cosh(t)^2), t = 0 .. 1) = 5 sinh(1)[/itex]
This is not a parameterization -- it's a number.

As I said before...
Mark44 said:
The arc length function in your relevant equations gives the arc length in terms of a parameter t.
IOW, ##\int_0^t 5 \sqrt{\cosh^2(w)} dw##
 
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FAQ: Differential geometry : Tangent vector & reparameterization

1. What is a tangent vector in differential geometry?

A tangent vector in differential geometry is a vector that is tangent to a curve or surface at a specific point. It represents the direction and rate of change of the curve or surface at that point.

2. How is a tangent vector calculated?

A tangent vector can be calculated by taking the derivative of a curve or surface at a specific point. This can be done using calculus or other mathematical methods.

3. What is reparameterization in differential geometry?

Reparameterization in differential geometry is the process of changing the parameterization of a curve or surface. This can be done by redefining the parameterization function or by using a different set of parameters to describe the curve or surface.

4. Why is reparameterization important in differential geometry?

Reparameterization is important in differential geometry because it allows for the manipulation and transformation of curves and surfaces while preserving their geometric properties. It also allows for easier calculations and analysis of these objects.

5. What are some real-world applications of differential geometry and tangent vectors?

Differential geometry and tangent vectors have many applications in the fields of physics, engineering, and computer graphics. They can be used to model and analyze the motion of objects, calculate the curvature and torsion of curves and surfaces, and generate smooth and realistic animations in computer graphics.

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