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If I have the curve $$r(t)=\left\langle \cos t+t\sin t, \sin t-t\cos t,1 \right\rangle$$

How would I find the unit tangent vector as a function of t? Here is a result that I ended up with after taking the derivative of the vector function and finding magnitude.

$$T(t)={{\left\langle -\sin t+t\cos t+\sin t, \cos t+t\sin t+\cos t,0 \right\rangle}\over{\sqrt{(-\sin t+t\cos t+\sin t)^2+(\cos t+t\sin t+\cos t)^2}}}$$

Would that be the correct answer? Is there any way to make it look nicer? As in, simplify it a bit more. I can't see any identities that would work.

Thanks for any help!

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For a start, $-\sin t +t\cos t+\sin t=t\cos t$, which helps. Also $\cos t+t\sin t-\cos t=t\sin t$. Using these, your final expression simplifies a lot.

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That would be a correct solution, yes. Generally with these problems the magnitude of the initial vector tends to look very ugly, but can simplify down to something far more workable. In this case, there's a very simple way to reduce the denominator: the first polynomial contains $-\sin{t}$ and $\sin{t}$, and the second contains $-\cos{t}$ and $\cos{t}$. This leaves you with $\sqrt{(t\cos{t})^2+(t\sin{t})^2}$ as the denominator. From there, distribute the powers. You should be able to work out the rest from there.

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The tangent vector simplifies as others mentioned .. I am not repeating the same.

The tangent to the curve and originating base circle radius ( of the involute curve that these parametrizations describe) are parallel, where the taut string of length $L = a t $ is the common normal. The situation should be grasped comprehensively.

Angular parameter $t$ of circle radius rotation links the generating relation

$$ r^2 = a^2 + L^2 = a^2(1+t^2) $$

The radius of curvature of the involute is simply $a t$.

Involute Relations

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