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ranyai12

  • one year ago

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)=<e^(−2t), t^2, 1/2t> , t=−2 T(-2)=

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  1. zepdrix
    • one year ago
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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

  2. zepdrix
    • one year ago
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    \[\large \vec r(t)=<e^{-2t},\quad t^2,\quad \frac{1}{2}t>\]Taking the derivative of each component gives us,\[\large \vec r\;'(t)=<-2e^{-2t},\quad 2t,\quad \frac{1}{2}>\]

  3. zepdrix
    • one year ago
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    Ok ok I looked it up, I was being silly. The equation is this,\[\large T=\frac{\vec r\;'(t)}{|\vec r\;'(t)|} \qquad \rightarrow \qquad T=\frac{v(t)}{|v(t)|}\]

  4. zepdrix
    • one year ago
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    \[\large \vec v(t)=<-2e^{-2t},\quad 2t,\quad \frac{1}{2}>\]The magnitude will give us,\[\large |\vec v(t)|=\sqrt{\left(-2e^{-2t}\right)^2+ \left(2t\right)^2+ \left(\frac{1}{2}\right)^2}\]

  5. zepdrix
    • one year ago
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    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

  6. zepdrix
    • one year ago
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    \[\large |\vec v(-2)|=\sqrt{4e^{8}+ 16+ \frac{1}{4}}\]This is probably a perfect square in disguise, gimme one sec to think :d

  7. zepdrix
    • one year ago
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    hmm

  8. ranyai12
    • one year ago
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    i dont think thats a vector

  9. ranyai12
    • one year ago
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    I think its the second equation you wrte in the beginning

  10. ranyai12
    • one year ago
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    @zepdrix?

  11. zepdrix
    • one year ago
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    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

  12. zepdrix
    • one year ago
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    \[\large \hat T(-2)=\frac{r\;'(-2)}{|r\;'(-2)|}\] \[\large =\frac{<-2e^4,\quad -4,\quad \dfrac{1}{2}>}{\sqrt{4e^{8}+ 16+ \frac{1}{4}}}\]

  13. zepdrix
    • one year ago
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    It feels like that bottom part should simplify down.. But I'm sorry, I can't seem to figure that out :\

  14. ranyai12
    • one year ago
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    its ok thanks

  15. ranyai12
    • one year ago
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    @zepdrix it says that the third coordinate is incorrect

  16. Zarkon
    • one year ago
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    is it \[r(t)=\left<e^{−2t}, t^2, \frac{1}{2t}\right>\] or \[r(t)=\left<e^{−2t}, t^2, \frac{1}{2}t\right>\]

  17. ranyai12
    • one year ago
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    yes it is

  18. ranyai12
    • one year ago
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    @Zarkon sorry t took me long ot respond I was trying to solve another problem

  19. Zarkon
    • one year ago
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    "yes it is" what?

  20. ranyai12
    • one year ago
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    oh sorry i misread what you wrote it its the first one

  21. Zarkon
    • one year ago
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    that is why zepdrix's third component is not correct.

  22. Zarkon
    • one year ago
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    you should have corrected their 2nd post. Then he could have derived the answer you need

  23. Zarkon
    • one year ago
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    if \[r(t)=\left<e^{−2t}, t^2, \frac{1}{2t}\right>\] then \[r'(t)=\left<e^{−2t}(-2), 2t, \frac{-1}{2t^2}\right>\]

  24. ranyai12
    • one year ago
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    oh sorry i didnt catch that

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