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Can someone please help!!! Find the unit tangent vector T(t) to the curve r(t) at the point with the given value of the parameter t. r(t)= , t=−2 T(-2)=

Mathematics
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Ummm maybe you could refresh my memory, I can't remember how the Tangent vector comes about.\[\large \vec T=\frac{\vec r(t)}{\vec r\;'(t)}\qquad \qquad \qquad \vec T=\frac{\vec r(t)}{|\vec r\;'(t)|}\] I think it's one of these equations, yes? :o
\[\large \vec r(t)=\]Taking the derivative of each component gives us,\[\large \vec r\;'(t)=<-2e^{-2t},\quad 2t,\quad \frac{1}{2}>\]
Ok ok I looked it up, I was being silly. The equation is this,\[\large T=\frac{\vec r\;'(t)}{|\vec r\;'(t)|} \qquad \rightarrow \qquad T=\frac{v(t)}{|v(t)|}\]

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Other answers:

\[\large \vec v(t)=<-2e^{-2t},\quad 2t,\quad \frac{1}{2}>\]The magnitude will give us,\[\large |\vec v(t)|=\sqrt{\left(-2e^{-2t}\right)^2+ \left(2t\right)^2+ \left(\frac{1}{2}\right)^2}\]
You'll save yourself a lot of trouble at this point, if you skip trying to simplify it from here. Just plug in the t=-2 at this point.\[\large |\vec v(-2)|=\sqrt{\left(-2e^{4}\right)^2+ \left(-4\right)^2+ \left(\frac{1}{2}\right)^2}\]Hmm I guess we'll still end up with a weird number because of that exponential :3
\[\large |\vec v(-2)|=\sqrt{4e^{8}+ 16+ \frac{1}{4}}\]This is probably a perfect square in disguise, gimme one sec to think :d
hmm
i dont think thats a vector
I think its the second equation you wrte in the beginning
r(t) is the `position vector`. r'(t) is the `velocity vector`. The velocity vector is `tangent` to the position vector. But it's not a unit vector! That's why we divide by the magnitude of itself ~ to turn it into a unit vector. \[\large T=\frac{r'}{|r'|} \qquad = \qquad \frac{v}{|v|} \qquad = \qquad \frac{v}{s}\](Where s is speed) Yah I'm pretty sure it's an r' on top. Do you have notes from class saying otherwise? :o
\[\large \hat T(-2)=\frac{r\;'(-2)}{|r\;'(-2)|}\] \[\large =\frac{<-2e^4,\quad -4,\quad \dfrac{1}{2}>}{\sqrt{4e^{8}+ 16+ \frac{1}{4}}}\]
It feels like that bottom part should simplify down.. But I'm sorry, I can't seem to figure that out :\
its ok thanks
@zepdrix it says that the third coordinate is incorrect
is it \[r(t)=\left\] or \[r(t)=\left\]
yes it is
@Zarkon sorry t took me long ot respond I was trying to solve another problem
"yes it is" what?
oh sorry i misread what you wrote it its the first one
that is why zepdrix's third component is not correct.
you should have corrected their 2nd post. Then he could have derived the answer you need
if \[r(t)=\left\] then \[r'(t)=\left\]
oh sorry i didnt catch that

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