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Compute $T_t(S^1)$ in $(\cos(t),\sin(t))$ with parametrization $f(t)=(\cos(t),\sin(t)).$

I have this:

I know, $T_tf:T_t\mathbb{R}\to T_{f(t)}S^1$ now $T_tf(\left.\frac{d}{ds}\right|_{s=0}(t+sK))=\left.\frac{d}{ds}\right|_{s=0} f(t+sK)=K(-\sin(t),\cos(t))$.

Therefore $T_tf(K)=K(-\sin(t),\cos(t))$.

Now that I have the tangent function. How do I get $T_t(S^1)$?

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    $\begingroup$ There are more rigorous ways to do this, but as you can see from drawing a picture, the tangent space is simply spanned by $(-\sin(t),\cos(t))$. $\endgroup$ – Cave Johnson Jul 30 '18 at 1:35
  • $\begingroup$ From what I did, we can conclude that: $T_{\cos(t),\sin(t)}S^1=\left\{K(-\sin(t),\cos(t):K\in T_t\mathbb{R}\right\}$ ? $\endgroup$ – eraldcoil Jul 30 '18 at 1:44

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