How is T the tangent to a vector

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The tangent vector T is defined as T = v/||v||, where v is a vector, making T a unit vector in the direction of v. In the context of motion, if an object moves in a straight line, T remains parallel to the path, confirming that it is indeed a tangent vector. The confusion arises from equating tangent vectors with normal vectors; tangent vectors are parallel to the path, not perpendicular. A tangent line at a point on the path has the same slope as the path at that point, reinforcing that a line is its own tangent line. This clarification resolves the misunderstanding about the nature of tangent vectors in relation to motion.
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The tangent vector is defined as :
T=v/||v||
Where v is some vector.
Then how is T the tangent vector to v? It's the unit vector in the direction of v right?
 
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My bad.
v is the velocity of a distance vector.
 
But even so, if v is the velocity vector I understand how T will be perpendicular to the path of the object as long as its direction is changing like if its moving in a circle. But suppose the object was moving in a straight line. Then would'nt the tangent vector given by the above equations be a unit vector in the same straight line. In other words it would not be a tangent vector. Help?
 
I think you are a bit confused about what it means for something to be tangent. A tangent vector is NOT a normal vector, it should not be perpendicular to the path; it should be "parallel". Recall that a tangent line to a point t=x on the path β(t) is a line that passes through the point (x,β(x)) and has slope β'(x). Hence a line is its own tangent line, so there is no problem with your definition of a tangent vector. Hope that helps
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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