Show reparameterized curvature equals curvature up to a sign

[MxwllsPersuasns]

Homework Statement

Let γ: I → ℝ² be a smooth regular curve and let λ = γ ο φ with φ: I_λ → I be a reparameterisation of γ. Show, by using the general formula for curvature of a regular curve that κ_λ = ±κ ο Φ where the ± depends on whether φ is orientation preserving (+) or reversing (-).

Homework Equations

κ = Det{γ',γ''}/{absval(γ')}³

The Attempt at a Solution

So, the way I would approach this problem I guess is by first taking the curvature of γ...
- Though the problem I immediately run into, and did on another problem I just posted, is whether this should be done in general or for a specific case of γ and λ
+ i.e., I would parameterize γ with t perhaps by <acos(t), bsin(t)> then take φ to be something orientation reversing (maybe like φ = -t) and again as something orientation preserving (like maybe φ = t²?) and then show that the curvature of each λ_± equals the curvature of the original curve (up to a sign) reparameterized with Φ.
+ Another question is, why are we using k composed with Φ for the reparameterised curvature when the reparamterization is done by composing with the function φ? Wouldn't the curvature then be something like κ_λ = ±κ ο φ?

Any help or guidance is extremely appreciated I'm having a bit of trouble grasping how to start and where to go with this.

 

[mighty2000]

Thanks!To start, we can use the given formula for curvature to find the curvature of γ:

κ = Det{γ',γ''}/{absval(γ')}3

Next, using the chain rule, we can find the curvature of λ:

κλ = Det{λ',λ''}/{absval(λ')}3

Now, we need to express λ in terms of γ and φ. Since λ = γ ο φ, we can substitute this into the above equation:

κλ = Det{(γ ο φ)',(γ ο φ)''}/{absval((γ ο φ)')}3

Using the chain rule again, we can expand (γ ο φ)' and (γ ο φ)'':

κλ = Det{(γ'φ' + γ''φ),(γ''φ' + γ'''φ + γ'φ'')}/{absval(γ'φ' + γ''φ)}3

Now, using the properties of determinants, we can simplify this expression to:

κλ = Det{γ',γ''}/{absval(γ')}3 * Det{φ',φ''}/{absval(φ')}3

Since κ = Det{γ',γ''}/{absval(γ')}3, we can substitute this into the above equation to get:

κλ = κ * Det{φ',φ''}/{absval(φ')}3

Now, we can consider the orientation of φ. If φ is orientation preserving, then φ' and φ'' will have the same sign. This means that Det{φ',φ''} will be positive. Similarly, if φ is orientation reversing, then φ' and φ'' will have opposite signs, and Det{φ',φ''} will be negative.

Therefore, if φ is orientation preserving, κλ = κ * Det{φ',φ''}/{absval(φ')}3 = κ * (+1) = κ. And if φ is orientation reversing, κλ = κ * Det{φ',φ''}/{absval(φ')}3 = κ * (-1) = -κ.

This shows that the curvature of the reparameterised curve λ is equal to the curvature of the original curve γ, up to a sign depending on the orientation of φ. This is what we wanted to prove.

Hope this helps! Let me know if you have any further questions.