If every tangent line of a curve passes through a point, it is a line

In summary, if every tangent line of a curve with unit speed parametrization passes through a fixed point, then the curve must be a line. Possible approaches to prove this statement include using the equation p = B(s) + r(s)B'(s) where r(s) is some function, using the fact that the acceleration of a unit speed curve must be perpendicular to the curve, and using Frenet-Serret frames to show that the curvature of the curve is identically zero. However, in characteristic p algebraic geometry, there can be counterexamples with concurrent tangents.
  • #1
demonelite123
219
0
If every tangent line of some curve B(s) which has the unit speed parametrization passes through a fixed point P, then the curve B(s) must be a line.

As a hint, my book says p = B(s) + r(s)B'(s) where r(s) is some function.

so i have that any tangent line L(t) = B(s) + t B'(s) and for some t, L(t) = p = B(s) + r(s)B'(s). i am having trouble continuing with this problem. i cannot think of anything else to do. could someone give me a hint or two on how to proceed? thanks!
 
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  • #2
demonelite123 said:
If every tangent line of some curve B(s) which has the unit speed parametrization passes through a fixed point P, then the curve B(s) must be a line.

As a hint, my book says p = B(s) + r(s)B'(s) where r(s) is some function.

so i have that any tangent line L(t) = B(s) + t B'(s) and for some t, L(t) = p = B(s) + r(s)B'(s). i am having trouble continuing with this problem. i cannot think of anything else to do. could someone give me a hint or two on how to proceed? thanks!

Here are two ideas.

- To use your equation recall that the acceleration of a unit speed curve must be perpendicular to the curve.

- Pick two points on the curve and draw the two lines from these two points to the fixed point,p. These lines are tangent to the curve and form a cone. Show that the entire curve must lie inside this cone.
 
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  • #3
hi, using your suggestion i differentiated the equation p = B(s) + r(s)B'(s) on both sides to get:
0 = B'(s) + r'(s)B'(s) + r(s)B''(s) and since |B'(s)| = 1, B'(s) * B''(s) = 0 so i take the dot product of B''(s) on both sides to get: 0 = r(s)|B''(s)|2. now if r(s) is equal to 0 for all s, then it is the trivial case of a line p + sv where v = 0. if r(s) is not equal to 0 for any s then i can divide both sides by it and i end up with |B''(s)| = 0 which means B''(s) is the zero vector or that B'(s) is a constant and that B(s) = p + sv where v = B'(s).

the only trouble I'm having is the case when r(s) is 0 for only some values of s. Then p must be some point on the curve itself. I argue that in order for every tangent line to pass through p, they must all be parallel to the tangent line that passes through p itself. however this is more of an intuitive thought and i don't really have a rigorous justification for it. what can be used to prove the statement in this case when r(s) is 0 for only some values of s?
 
  • #4
I think you can also use the Frenet-Serret frames to conclude that the curvature of the
curve is identically zero. May be a good idea to translate/shift the curve so that it goes
thru the origin.

EDIT: I know that to use Frenet-Serret frames we need to have non-trivial curvature;
the idea was to get a contradiction by assuming we have a curve with non-trivial torsion
and curvature (a line having both equal to zero), and then arriving at a contradiction, but
I have not been able to get much from this yet; I'll try for a while longer.
 
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  • #5
for a counter example in characteristic p algebraic geometry google "strange curves".
 
  • #6
mathwonk said:
for a counter example in characteristic p algebraic geometry google "strange curves".

i.e. in characteristic 2, a conic can have concurrent tangents.
 

1. What does the statement "If every tangent line of a curve passes through a point, it is a line" mean?

The statement means that if every line that touches a curve at any point also passes through a specific point, then that curve is actually a straight line passing through that point.

2. How is this statement related to calculus?

This statement is related to calculus because it involves the concept of tangent lines, which are a fundamental part of calculus. Tangent lines are used to find the instantaneous rate of change of a curve at a specific point.

3. What is the significance of this statement in mathematics?

This statement is significant in mathematics because it helps us determine whether a curve is actually a straight line passing through a point or not. It also helps us understand the relationship between a curve and its tangent lines.

4. Can you give an example to illustrate this statement?

Yes, for example, consider the curve y = x^2. At any point on this curve, we can draw a tangent line that passes through the point (1,1). This shows that the curve is actually a straight line passing through (1,1).

5. Is this statement always true for any curve and point?

No, this statement is not always true. It only applies to certain curves and points. For example, if a curve is a circle, then every tangent line will pass through the center of the circle, but the curve itself is not a straight line passing through the center point.

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