# Compute the curvature

1. Sep 10, 2014

### Shackleford

1. Compute the curvature.

$α(t) = (cos^3t, sin^3t)$

This is not a unit-speed curve. I want to use $κ(t) = \frac{||T'(t)||}{||σ'(t)||}$

When I find $α'(t)$ and then its norm, I run into an impasse. Am I supposed to use a trig identity?

2. Sep 10, 2014

### ehild

What was that impasse? Do you know how to get the norm of a vector?

ehild

3. Sep 10, 2014

### haruspex

Please post your working, or we cannot tell where you are stuck.

4. Sep 10, 2014

### Shackleford

Sorry. It was late last night. Maybe I'll get a different answer today. Heh.

$α(t) = (cos^3t, sin^3t)$

$α'(t) = (-3cos^2(t) sin(t), 3sin^2(t)cost(t)$

$||α'(t)|| = \sqrt{(-3cos^2(t) sin(t))^2 + (3sin^2(t)cos(t))^2}$

$||α'(t)|| = \sqrt{9cos^4(t) sin^2(t) + 9sin^4(t)cos^2(t)}$

Last edited: Sep 10, 2014
5. Sep 10, 2014

### haruspex

What common factor can you take outside the square root?

6. Sep 10, 2014

### Shackleford

I have no excuse. Thanks. I see it. Let me try to work the rest of the problem out.

$κ(t) = \frac{1}{3cos(t)sin(t)}$

Last edited: Sep 10, 2014
7. Sep 10, 2014

### Shackleford

3. Show that the curve is planar (i.e. contained in a plane). Hint: Show that its torsion is identically zero. Of course, I need to set the torsion $τ$ = 0.

$α(t) = (\frac{1+t^2}{t}, t+1, \frac{1-t}{t})$

$α'(t) = (1-t^{-2},1, -t^{-2})$

$||α'(t)|| = \sqrt{2(t^{-4}-t^{-2}+1})$

8. Sep 10, 2014

### haruspex

What's your equation for torsion? And how is $||α'(t)||$ useful?

9. Sep 10, 2014

### Shackleford

I see one entirely in terms of alpha up to its third derivative, a cross product, norm, etc.

The other one involves B'(t), N(t), and alpha.

I found alpha-prime because it's used to find T, T', which is then used for N, and then B, and so forth.

10. Sep 11, 2014

### haruspex

That's the one. But remember you are only interested in whether it is zero, which simplifies matters a little.
Yes, you need α', but I asked why its norm was interesting here.

11. Sep 11, 2014

### Shackleford

Oh, indeed. I suppose that I can simply show that the numerator $α^{'''}(t)\cdot (α^{'}(t) \times α^{''}(t))$ is zero.

12. Sep 11, 2014

### haruspex

Yes.

13. Sep 12, 2014

### Shackleford

Thanks. I have one more problem, and the HW is due today at 5 PM.

$Let σ be a unit-speed plane curve. Define a new curve (called parallel curve of σ) β by β(t) = σ(t) + 2N(t), where N(t) is the principal normal at α(t). Show that, if 2κ(t) < 1 for all t, where κ(t) is the curvature of σ at σ(t), then β(t) is a regular curve and that its curvature is κ(t)/(1-2κ(t)). I know that σ being a unit-speed plane curve implies that ||σ'(t)|| = 1 and that τ(t) = 0. Also, if β(t) is a regular curve, then ||β(t)|| = 1.$

Last edited: Sep 12, 2014
14. Sep 12, 2014

### haruspex

Sorry, only just saw this.
No, it is only required that it is differentiable and the derivative never changes sign.
Try to get an expression for the derivative in terms of that of the original curve.

15. Sep 12, 2014

### Shackleford

$β(t) = α(t) + 2N(t)$
$β'(t) = α'(t) + 2N'(t)$
$β'(t) = α'(t) + 2(-κT + τB)$
$β'(t) = α'(t) + 2(-κT)$
$β'(t) = α'(t) - 2κα'(t)$
$β'(t) = α'(t)(1 - 2κ)$

$||α'(t)|| = 1 \implies α'(t) ≠ 0$
If $κ(t) < \frac{1}{2},\ β'(t) > 0, \implies β'(t) \ is \ regular.$

Last edited: Sep 12, 2014
16. Sep 13, 2014

Looks right.