Show plane curve can be described with graph @ tangent point

In summary, the homework statement provides a proof that a regular plane curve γ can near each point γ(t0) be written as a graph over the tangent line. There exists a smooth real valued map x → f(x) for small x with f(0) = 0 so that x → xT(t0) + f(x)JT(t0) parametrizes γ near γ(t0). Here T = γ'/||γ'|| is the unit length tangent vector.
  • #1
MxwllsPersuasns
101
0

Homework Statement



Provide a complete proof that a regular plane curve γ : I → R2 can near each point γ(t0) be written as a graph over the tangent line: more precisely, there exists a smooth real valued map x → f(x) for small x with f(0) = 0 so that x → xT(t0) + f(x)JT(t0) parametrizes γ near γ(t0). Here T = γ'/||γ'|| is the unit length tangent vector.

Homework Equations

The Attempt at a Solution


So I don't know what exactly they mean as a graph over the tangent line (like an xy-coordinate system where the tangent line is one of the axes of the graph?) but the further explanation kind of clears it up a little bit in that it seems like the map x → xT(t0) + f(x)JT(t0) basically lines up the x direction along the tangent line at the point t0 and since JT is orthogonal to T we can see that the map also lines up the y (or f(x)) direction along the normal to the tangent line, thus creating an orthonormal basis at every point t0 on the curve.

Can anyone tell me if this thinking is correct and if so, how I could potentially put this down in a mathematical argument?
 
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  • #2
MxwllsPersuasns said:

Homework Statement



Provide a complete proof that a regular plane curve γ : I → R2 can near each point γ(t0) be written as a graph over the tangent line: more precisely, there exists a smooth real valued map x → f(x) for small x with f(0) = 0 so that x → xT(t0) + f(x)JT(t0) parametrizes γ near γ(t0). Here T = γ'/||γ'|| is the unit length tangent vector.
I'm having a hard time trying to understand the notation.
In "regular plane curve γ : I → R2" I assume that 'I' here represents an interval on the x-axis, but I'm not sure.
It's clear from how T is defined that it's a unit vector, and the above also states this. What does J (or JT) mean in this notation?
"f(x)JT(t0)"
MxwllsPersuasns said:

Homework Equations

The Attempt at a Solution


So I don't know what exactly they mean as a graph over the tangent line (like an xy-coordinate system where the tangent line is one of the axes of the graph?) but the further explanation kind of clears it up a little bit in that it seems like the map x → xT(t0) + f(x)JT(t0) basically lines up the x direction along the tangent line at the point t0 and since JT is orthogonal to T we can see that the map also lines up the y (or f(x)) direction along the normal to the tangent line, thus creating an orthonormal basis at every point t0 on the curve.

Can anyone tell me if this thinking is correct and if so, how I could potentially put this down in a mathematical argument?
 
  • #3
Oh I do apologize Mark44. You are indeed correct that I signifies an interval. Essentially I is just a constrained subset of the real line, ℝ. I do believe J is the reflection matrix aka reflection across x-axis is diag(1, -1) and one across the y-axis is diag(-1, 1) one across the origin is diag(-1, -1). Now that I've said that though it doesn't make sense that JT(t0)f(x) would align f(x) direction along the normal to the curve.. that would require a rotation by 90 degrees.

Shoot, now I'm more confused.
 
  • #4
Now I'm trying to parse this statement:
MxwllsPersuasns said:
there exists a smooth real valued map x → f(x) for small x with f(0) = 0 so that x → xT(t0) + f(x)JT(t0) parametrizes γ near γ(t0)
Does x have anything to do with the interval I? I'm confused by this because t0 belongs to the interval.
The map referred to above takes x (from somewhere) and maps it to a sum of a multiple of the unit tangent at t0 and a different multiple of some reflection of the unit tangent vector, also evaluated at t0.

The notation is throwing me. Typically when someone talks about a map x → f(x), a formula for f is given. Here it looks like f(x) = x*T(t0) + f(x) * J * T(t0), which doesn't make sense to me, as f seems to be defined in terms of itself. Is what you showed in post #1 the exact wording of the problem?
 
  • #5
Yes I'm actually having a similar issue with this notation -- I've personally not seen mappings put like this before and it threw me through a major loop. So I'm glad to see I'm not the only one. I basically copy and pasted that straight from the homework set pdf (changing only the aesthetic issues which didn't transfer over well). My best guess is to try to go with whatever our intuition says and I guess I already laid out what my intuition was at the beginning, but like I said.. I'm very confused. Lol.
 
  • #6
My advice would be to contact the instructor to get clarification on what is being asked and what the notation means.
 
  • #7
Unfortunately the homework set is due tomorrow, I didn't think I'd have notational issues or anything like that so I didn't look ahead, assuming they'd be harder but not hard to even understand what's being asked. It's a shame.. definitely downsides to having overly idiosyncratic professors.
 

FAQ: Show plane curve can be described with graph @ tangent point

Can all plane curves be described with a graph at the tangent point?

No, not all plane curves can be described with a graph at the tangent point. Some curves, such as fractals, have an infinite number of tangent points and cannot be fully represented with a single graph.

How do you find the tangent point on a plane curve?

The tangent point on a plane curve can be found by taking the derivative of the curve's equation and solving for the point(s) where the derivative equals the slope of the tangent line at that point.

What is the significance of the tangent point on a plane curve?

The tangent point on a plane curve represents the point where the curve's slope is equal to the slope of the tangent line at that point. This is important because it allows us to understand the behavior of the curve at that specific point.

How does the graph at the tangent point help us understand the behavior of a plane curve?

The graph at the tangent point helps us understand the behavior of a plane curve by showing us the slope of the curve at that point. This can help us determine if the curve is increasing, decreasing, or changing direction at that specific point.

Can the graph at the tangent point be used to determine the curvature of a plane curve?

Yes, the graph at the tangent point can be used to determine the curvature of a plane curve. The curvature at a point on a curve is equal to the second derivative of the curve's equation at that point. By finding the second derivative and evaluating it at the tangent point, we can determine the curvature at that point.

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