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Please provide suggestions for improvement! I think I may have made the first major revision to this article. I reorganized the old material into the intro and added a definition and properties as well as reference. I may add more later. Gray's definition of a tangent vector is actually prior to his definition of a directional derivative. I may add this material which talks about a vector part and a point of application as comprising a tangent vector which is just these two pair of points in Euclidean space. I don't like this definition however. I need a little more mathematical sophistication, but I intend to cover tangent vectors as elements of a tangent space [in] a differentiable manifold. I do not think that this material is too technical. A tangent is part of elementary calculus, but a tangent vector is part of vector calculus. Therefore, our readers will have some very small bit of mathematical sophistication (like me).Stewart.M.Nash (talk) 02:34, 15 May 2015 (UTC)Reply[reply]

${\displaystyle \mathbf {T} (0)={\frac {\mathbf {r} ^{\prime }(0)}{|\mathbf {r} ^{\prime }(0)|))=\left.{\frac {(2t,2e^{2t},\sin {t})}{\sqrt {4t^{2}+e^{2t}+\sin ^{2}{t))))\right|_{t=0}=(0,1,0)\,.}$

Here the bottom part is not equal to the vector length. AFAIK it must be

${\displaystyle \mathbf {T} (0)={\frac {\mathbf {r} ^{\prime }(0)}{|\mathbf {r} ^{\prime }(0)|))=\left.{\frac {(2t,2e^{2t},\sin {t})}{\sqrt {4t^{2}+4e^{4t}+\sin ^{2}{t))))\right|_{t=0}=(0,1,0)\,.}$ Arkadi kagan (talk) 17:34, 12 December 2015 (UTC)Reply[reply]

The example only makes sense if the denominator evaluates to the value 2 at t=0; e.g., (4t² + 4e^4t + sin²t )^0.5 = 2. Arkadi is correct.

Krakengreen (talk) 08:37, 23 December 2015 (UTC)Reply[reply]