anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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zepdrix
  • zepdrix
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
zepdrix
  • zepdrix
\[\large \vec r(t)=\]Taking the derivative of each component gives us,\[\large \vec r\;'(t)=<-2e^{-2t},\quad 2t,\quad \frac{1}{2}>\]
zepdrix
  • zepdrix
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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zepdrix
  • zepdrix
\[\large \vec v(t)=<-2e^{-2t},\quad 2t,\quad \frac{1}{2}>\]The magnitude will give us,\[\large |\vec v(t)|=\sqrt{\left(-2e^{-2t}\right)^2+ \left(2t\right)^2+ \left(\frac{1}{2}\right)^2}\]
zepdrix
  • zepdrix
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
zepdrix
  • zepdrix
\[\large |\vec v(-2)|=\sqrt{4e^{8}+ 16+ \frac{1}{4}}\]This is probably a perfect square in disguise, gimme one sec to think :d
zepdrix
  • zepdrix
hmm
anonymous
  • anonymous
i dont think thats a vector
anonymous
  • anonymous
I think its the second equation you wrte in the beginning
anonymous
  • anonymous
@zepdrix?
zepdrix
  • zepdrix
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
zepdrix
  • zepdrix
\[\large \hat T(-2)=\frac{r\;'(-2)}{|r\;'(-2)|}\] \[\large =\frac{<-2e^4,\quad -4,\quad \dfrac{1}{2}>}{\sqrt{4e^{8}+ 16+ \frac{1}{4}}}\]
zepdrix
  • zepdrix
It feels like that bottom part should simplify down.. But I'm sorry, I can't seem to figure that out :\
anonymous
  • anonymous
its ok thanks
anonymous
  • anonymous
@zepdrix it says that the third coordinate is incorrect
Zarkon
  • Zarkon
is it \[r(t)=\left\] or \[r(t)=\left\]
anonymous
  • anonymous
yes it is
anonymous
  • anonymous
@Zarkon sorry t took me long ot respond I was trying to solve another problem
Zarkon
  • Zarkon
"yes it is" what?
anonymous
  • anonymous
oh sorry i misread what you wrote it its the first one
Zarkon
  • Zarkon
that is why zepdrix's third component is not correct.
Zarkon
  • Zarkon
you should have corrected their 2nd post. Then he could have derived the answer you need
Zarkon
  • Zarkon
if \[r(t)=\left\] then \[r'(t)=\left\]
anonymous
  • anonymous
oh sorry i didnt catch that

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