Let the endpoint of a differentiable, vector-valued function a ( t ) : ( t a . . t b ) → R move on a circle in an Euclidean plane. Let t 0 ∈ ( t a . . t b ). Let s ( t ) be the length of the arc between t 0 and point a ( t ). Let there be a positive real number δ such that ∀ t : 0 &lt; | t − t 0 | &lt; δ : s ( t ) − s ( t 0 ) ≠ 0.My (physics) book uses a proof which seems to imply that s( t), where t is time, is differentiable in respect to t. How to prove its differentiability?

Question

Let the endpoint of a differentiable, vector-valued function a  (  t  )  :  (      t    a      .  .      t    b    )  →      R   move on a circle in an Euclidean plane. Let       t    0    ∈  (      t    a      .  .      t    b    ). Let   s  (  t  ) be the length of the arc between       t    0   and point a  (  t  ). Let there be a positive real number   δ such that   ∀  t  :  0  &amp;lt;      |    t  −      t    0        |    &amp;lt;  δ  :  s  (  t  )  −  s  (      t    0    )  ≠  0.My (physics) book uses a proof which seems to imply that   s(  t), where   t is time, is differentiable in respect to   t. How to prove its differentiability?

lorienoldf7 · Accepted Answer

The arclength is   s  (  t  )  =      ∫                  t        0              t    |            a        ′    (  τ  )  |      d    τSo,   s  (  t  ) is differentiable because       s    ′    (  t  )  =  |            a        ′    (  t  )  | exists: composition of continuous functions and the derivative of       a    (  t  ), that exists.

Let the endpoint of a differentiable, vector-valued function a ( t ) : ( t

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