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OpenStudy (anonymous):

Find T when r'(t)= (t^2, 2t, 2) for any time t

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OpenStudy (anonymous):

I got as far as \[<t^2, 2t, 2>/\sqrt(t^4+4t^2+4)\]

OpenStudy (anonymous):

If its the unit tangent vector your want then it's \[\frac{\vec{r}\ '(t)}{|\vec{r}\ '(t)|}\]

OpenStudy (anonymous):

Do you know how to simplify the bottom half?

OpenStudy (anonymous):

I don't think that's a square. But it shouldn't matter.

OpenStudy (anonymous):

The bottom is finding the norm of r'(t), which is equivalent to \[\sqrt(x^2 + y^2 + z^2)\]

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OpenStudy (anonymous):

Yup. I meant the thing under the square root doesn't seem to be a perfect square so it's probably simplified enough

OpenStudy (anonymous):

Ohhhhh gotcha. Haha thanks!

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