# Differential geometry : Tangent vector & reparameterization

1. Dec 13, 2015

### Schwarzschild90

1. The problem statement, all variables and given/known data

2. Relevant equations
Arc-length function

3. 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. Dec 13, 2015

### Ray Vickson

What is preventing you from trying it for yourself?

3. Dec 13, 2015

### Schwarzschild90

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. Dec 13, 2015

### Staff: Mentor

This isn't the tangent vector.

5. Dec 13, 2015

### Schwarzschild90

Tangent vector

Now, compute the norm of the tangent vector:

Using this, make the following substitution

6. Dec 13, 2015

### Staff: Mentor

7. Dec 13, 2015

### Schwarzschild90

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

$\sqrt{(9+9*sinh(t)^2+16*cosh(t)^2)}dt =^* 25 cosh^2(t) = 25 sinh^2(t)+25$

* Using a trigonometric identity

PS: csgn is code used specifically by maple. It' not necessarily a mathematical function

Last edited: Dec 13, 2015
8. Dec 13, 2015

### Staff: Mentor

The arc length function in your relevant equations gives the arc length in terms of a parameter t.
The last expression above is not helpful, but the one before it is helpful. What happened to the square root?
Do you know what it means, though? I've never seen it, but I don't use Maple.

9. Dec 13, 2015

### Schwarzschild90

Right, the square root should've been preserved, in the above equation. Here it is, in all of its glory:

$\sqrt{25cosh^2(t)}$
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. Dec 13, 2015

### Ray Vickson

11. Dec 13, 2015

### Schwarzschild90

Plot of the 25cosh^2(t) function; the norm of the tangent vector

Last edited: Dec 13, 2015
12. Dec 13, 2015

### Ray Vickson

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.

13. Dec 13, 2015

### Schwarzschild90

Plot of $5 \sqrt{cosh(t)}$

I can picture it in my mind. What am I supposed to "see"?

Last edited: Dec 13, 2015
14. Dec 13, 2015

### Schwarzschild90

I get this for the parameterization by arclength of the tangent vector
$int(5*sqrt(cosh(t)^2), t = 0 .. 1) = 5 sinh(1)$

15. Dec 13, 2015

### Staff: Mentor

This is not a parameterization -- it's a number.

As I said before...
IOW, $\int_0^t 5 \sqrt{\cosh^2(w)} dw$