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Mathematics 17 Online
OpenStudy (anonymous):

If T(t)=(2ti+2j)/(2sqrt(t^2+1)) what is N(t)?

OpenStudy (anonymous):

what do u mean by N

OpenStudy (anonymous):

principle unit normal vector

OpenStudy (anonymous):

I got you. Let me type it :P

OpenStudy (anonymous):

K

OpenStudy (anonymous):

\[N(t)=\frac{T'(t)}{\left|| T'(t) \right||}\].

OpenStudy (anonymous):

As long as you can differentiate it :P

OpenStudy (anonymous):

I can work it all the way out if you need help with that.

OpenStudy (anonymous):

\[(i / (\sqrt{t ^{2}+1})^{3} -jt/(\sqrt{t ^{2}+1})^{3} )/\sqrt{(1)/(t ^{2}+1)^{2}}\]) i think

OpenStudy (anonymous):

I got \[N(t)=\frac{1}{t}i-j\].

OpenStudy (anonymous):

For my derivative:

OpenStudy (anonymous):

lololol im wrong

OpenStudy (anonymous):

\[T'(t)=\frac{(2)(2\sqrt{t^2+1})-(2t)(2t)(t^2+1)^{-\frac{1}{2}}}{4(t^2+1)}i+(-1/2)(2t)(t^2+1)^{-3/2}j\].

OpenStudy (anonymous):

What I wrote above is what I got after dividing by the magnitude and everything^^ (The N(t) one)

OpenStudy (anonymous):

wow i just got schooled

OpenStudy (anonymous):

You want a walk through of my differentiation?

OpenStudy (anonymous):

me? it's not my question lol. I probably just made a stupid mistake somewhere

OpenStudy (anonymous):

Its a large quotient rule then a chain rule (you can rewrite it as (t^2+1)^(-1/2)

OpenStudy (anonymous):

Haha, its okay :P All about bookkeeping and not losing negatives and what not.

OpenStudy (anonymous):

yeah, plus it's super annoying to type out the answer

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