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

[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?

 

[ehild]

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

ehild

 

[haruspex]

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?

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

 

[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)}

 

[haruspex]

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)}

What common factor can you take outside the square root?

 

[Shackleford]

What common factor can you take outside the square root?

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)}

 

[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})

 

[haruspex]

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})

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

 

[Shackleford]

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

 

[haruspex]

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.

 

[Shackleford]

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 α^{'''}(t)\cdot (α^{'}(t) \times α^{''}(t)) is zero.

 

[haruspex]

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

Yes.

 

[Shackleford]

Yes.

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 <i>parallel curve </i>of σ) β by<br /> <br /> β(t) = σ(t) + 2<b>N</b>(t),<br /> <br /> where <b>N</b>(t) is the principal normal at α(t). Show that, if 2κ(t) &lt; 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)).<br /> <br /> I know that σ being a unit-speed plane curve implies that ||σ&#039;(t)|| = 1 and that τ(t) = 0. Also, if β(t) is a regular curve, then ||β(t)|| = 1.

 

[haruspex]

Sorry, only just saw this.

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.

 

[Shackleford]

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.

β(t) = α(t) + 2N(t)
β&#039;(t) = α&#039;(t) + 2N&#039;(t)
β&#039;(t) = α&#039;(t) + 2(-κT + τB)
β&#039;(t) = α&#039;(t) + 2(-κT)
β&#039;(t) = α&#039;(t) - 2κα&#039;(t)
β&#039;(t) = α&#039;(t)(1 - 2κ)

||α&#039;(t)|| = 1 \implies α&#039;(t) ≠ 0
If κ(t) &lt; \frac{1}{2},\ β&#039;(t) &gt; 0, \implies β&#039;(t) \ is \ regular.

 

[haruspex]

β(t) = α(t) + 2N(t)
β&#039;(t) = α&#039;(t) + 2N&#039;(t)
β&#039;(t) = α&#039;(t) + 2(-κT + τB)
β&#039;(t) = α&#039;(t) + 2(-κT)
β&#039;(t) = α&#039;(t) - 2κα&#039;(t)
β&#039;(t) = α&#039;(t)(1 - 2κ)

||α&#039;(t)|| = 1 \implies α&#039;(t) ≠ 0
If κ(t) &lt; \frac{1}{2},\ β&#039;(t) &gt; 0, \implies β&#039;(t) \ is \ regular.

Looks right.