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I am studying basic vector calculus and am on tangential and normal vectors. I understand why the derivative of a vector is tangential, and I also understand why the second derivative of a vector points perpendicularly to the tangent. But does this not apply to vectors defining linear functions? Because I can have a vector function $s(t)=(2t^3)i+(3t^3)j$ that defines a linear function. The second derivative is $s''(t)=(12t)i+(18t)j$ so neither the first nor second derivative are constant, yet the second derivative of vector $s(t)$ does not point perpendicularly to the tangent. Can someone explain why? This is my first post, so sorry for any formatting errors.

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The second derivative is perpendicular to the first derivative if the latter has constant norm. This is because $$\langle s'(t),s'(t)\rangle = {\rm const.} \implies 2\langle s''(t),s'(t)\rangle = 0. $$ Your $s'(t)$ doesn't have constant norm. You'll be sure that the derivatives come out perpendicular if, for example, the curve is parametrized by arc-length (and you can always take such a reparametrization, it is an early result in most differential geometry book).

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  • $\begingroup$ Is this why we can use the tangential unit vector to find the perpendicular one? Since the tangent unit vector has constant norm? $\endgroup$ – B. Sandoval May 31 '16 at 0:07
  • $\begingroup$ Yes, precisely. $\endgroup$ – Ivo Terek May 31 '16 at 0:08

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