Calculating Curvature of Non-Unit-Speed Curve Using a Trig Identity

  • Thread starter Shackleford
  • Start date
  • Tags
    Curvature
In summary: Now find the curvature of β(t).Looks right. Now find the curvature of β(t).β'(t) = α'(t)(1 - 2κ)||β'(t)|| = |1 - 2κ|κ_β(t) = \frac{||β'(t)||}{||σ'(t)||} = \frac{|1 - 2κ|}{1} = |1 - 2κ|κ_β(t) = \frac{|1 - 2κ|}{1} = |1 - 2κ|In summary, the curvature of the curve β(t) is equal to |1-2κ|, where κ is the curvature of the original
  • #1
Shackleford
1,656
2
1. Compute the curvature.

[itex]α(t) = (cos^3t, sin^3t)[/itex]

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

When I find [itex]α'(t)[/itex] and then its norm, I run into an impasse. Am I supposed to use a trig identity?
 
Physics news on Phys.org
  • #2
What was that impasse? Do you know how to get the norm of a vector?

ehild
 
  • #3
Shackleford said:
1. Compute the curvature.

[itex]α(t) = (cos^3t, sin^3t)[/itex]

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

When I find [itex]α'(t)[/itex] and then its norm, I run into an impasse. Am I supposed to use a trig identity?
Please post your working, or we cannot tell where you are stuck.
 
  • #4
Sorry. It was late last night. Maybe I'll get a different answer today. Heh.

[itex]α(t) = (cos^3t, sin^3t)[/itex]

[itex]α'(t) = (-3cos^2(t) sin(t), 3sin^2(t)cost(t)[/itex]

[itex]||α'(t)|| = \sqrt{(-3cos^2(t) sin(t))^2 + (3sin^2(t)cos(t))^2} [/itex]

[itex]||α'(t)|| = \sqrt{9cos^4(t) sin^2(t) + 9sin^4(t)cos^2(t)} [/itex]
 
Last edited:
  • #5
Shackleford said:
Sorry. It was late last night. Maybe I'll get a different answer today. Heh.

[itex]α(t) = (cos^3t, sin^3t)[/itex]

[itex]α'(t) = (-3cos^2(t) sin(t), 3sin^2(t)cost(t)[/itex]

[itex]||α'(t)|| = \sqrt{(-3cos^2(t) sin(t))^2 + (3sin^2(t)cos(t))^2} [/itex]

[itex]||α'(t)|| = \sqrt{9cos^4(t) sin^2(t) + 9sin^4(t)cos^2(t)} [/itex]
What common factor can you take outside the square root?
 
  • #6
haruspex said:
What common factor can you take outside the square root?

:rolleyes:

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

[itex]κ(t) = \frac{1}{3cos(t)sin(t)}[/itex]
 
Last edited:
  • #7
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 [itex]τ[/itex] = 0.

[itex]α(t) = (\frac{1+t^2}{t}, t+1, \frac{1-t}{t})[/itex]

[itex]α'(t) = (1-t^{-2},1, -t^{-2})[/itex]

[itex]||α'(t)|| = \sqrt{2(t^{-4}-t^{-2}+1})[/itex]
 
  • #8
Shackleford said:
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 [itex]τ[/itex] = 0.

[itex]α(t) = (\frac{1+t^2}{t}, t+1, \frac{1-t}{t})[/itex]

[itex]α'(t) = (1-t^{-2},1, -t^{-2})[/itex]

[itex]||α'(t)|| = \sqrt{2(t^{-4}-t^{-2}+1})[/itex]

What's your equation for torsion? And how is [itex]||α'(t)||[/itex] useful?
 
  • #9
haruspex said:
What's your equation for torsion? And how is [itex]||α'(t)||[/itex] useful?

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
Shackleford said:
I see one entirely in terms of alpha up to its third derivative, a cross product, norm, etc.
That's the one. But remember you are only interested in whether it is zero, which simplifies matters a little.
I found alpha-prime because it's used to find T, T', which is then used for N, and then B, and so forth.
Yes, you need α', but I asked why its norm was interesting here.
 
  • #11
haruspex said:
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.

Oh, indeed. I suppose that I can simply show that the numerator [itex]α^{'''}(t)\cdot (α^{'}(t) \times α^{''}(t))[/itex] is zero.
 
  • #12
Shackleford said:
Oh, indeed. I suppose that I can simply show that the numerator [itex]α^{'''}(t)\cdot (α^{'}(t) \times α^{''}(t))[/itex] is zero.

Yes.
 
  • #13
haruspex said:
Yes.

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

[itex]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.[/itex]
 
Last edited:
  • #14
Sorry, only just saw this.
Shackleford said:
if β(t) is a regular curve, then ||β(t)|| = 1.
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
haruspex said:
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.

[itex]β(t) = α(t) + 2N(t)[/itex]
[itex]β'(t) = α'(t) + 2N'(t)[/itex]
[itex]β'(t) = α'(t) + 2(-κT + τB)[/itex]
[itex]β'(t) = α'(t) + 2(-κT)[/itex]
[itex]β'(t) = α'(t) - 2κα'(t)[/itex]
[itex]β'(t) = α'(t)(1 - 2κ)[/itex]

[itex]||α'(t)|| = 1 \implies α'(t) ≠ 0[/itex]
If [itex]κ(t) < \frac{1}{2},\ β'(t) > 0, \implies β'(t) \ is \ regular.[/itex]
 
Last edited:
  • #16
Shackleford said:
[itex]β(t) = α(t) + 2N(t)[/itex]
[itex]β'(t) = α'(t) + 2N'(t)[/itex]
[itex]β'(t) = α'(t) + 2(-κT + τB)[/itex]
[itex]β'(t) = α'(t) + 2(-κT)[/itex]
[itex]β'(t) = α'(t) - 2κα'(t)[/itex]
[itex]β'(t) = α'(t)(1 - 2κ)[/itex]

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

Looks right.
 

What is meant by "curvature"?

Curvature refers to the amount of bending or deviation from a straight line that is present in a curve or surface.

How is curvature calculated?

Curvature can be calculated using various mathematical formulas, depending on the type of curve or surface being analyzed. In general, it involves finding the rate of change of the tangent line or plane at a specific point on the curve or surface.

Why is it important to compute curvature?

Computing curvature is important in many fields of science, including physics, engineering, and mathematics. It allows us to analyze and understand the behavior of curves and surfaces, and can also help with designing and optimizing structures and systems.

What are some real-life applications of computing curvature?

Some examples of real-life applications of computing curvature include analyzing the shape of lenses in optics, designing roller coasters and other amusement park rides, and calculating the trajectory of objects in motion.

Are there different types of curvature?

Yes, there are several types of curvature, including Gaussian curvature, mean curvature, and principal curvature. Each type measures a different aspect of the bending or curvature of a curve or surface.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
1K
Replies
4
Views
874
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
119
  • Calculus and Beyond Homework Help
Replies
6
Views
275
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Replies
2
Views
1K
Back
Top