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ranyai12
Group Title
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)=
 one year ago
 one year ago
ranyai12 Group Title
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)=
 one year ago
 one year ago

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zepdrix Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
\[\large \vec r(t)=<e^{2t},\quad t^2,\quad \frac{1}{2}t>\]Taking the derivative of each component gives us,\[\large \vec r\;'(t)=<2e^{2t},\quad 2t,\quad \frac{1}{2}>\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
Ok ok I looked it up, I was being silly. The equation is this,\[\large T=\frac{\vec r\;'(t)}{\vec r\;'(t)} \qquad \rightarrow \qquad T=\frac{v(t)}{v(t)}\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
\[\large \vec v(t)=<2e^{2t},\quad 2t,\quad \frac{1}{2}>\]The magnitude will give us,\[\large \vec v(t)=\sqrt{\left(2e^{2t}\right)^2+ \left(2t\right)^2+ \left(\frac{1}{2}\right)^2}\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.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
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
\[\large \vec v(2)=\sqrt{4e^{8}+ 16+ \frac{1}{4}}\]This is probably a perfect square in disguise, gimme one sec to think :d
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
i dont think thats a vector
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
I think its the second equation you wrte in the beginning
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
\[\large \hat T(2)=\frac{r\;'(2)}{r\;'(2)}\] \[\large =\frac{<2e^4,\quad 4,\quad \dfrac{1}{2}>}{\sqrt{4e^{8}+ 16+ \frac{1}{4}}}\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.2
It feels like that bottom part should simplify down.. But I'm sorry, I can't seem to figure that out :\
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
its ok thanks
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix it says that the third coordinate is incorrect
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
is it \[r(t)=\left<e^{−2t}, t^2, \frac{1}{2t}\right>\] or \[r(t)=\left<e^{−2t}, t^2, \frac{1}{2}t\right>\]
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
yes it is
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
@Zarkon sorry t took me long ot respond I was trying to solve another problem
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
"yes it is" what?
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
oh sorry i misread what you wrote it its the first one
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
that is why zepdrix's third component is not correct.
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
you should have corrected their 2nd post. Then he could have derived the answer you need
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
if \[r(t)=\left<e^{−2t}, t^2, \frac{1}{2t}\right>\] then \[r'(t)=\left<e^{−2t}(2), 2t, \frac{1}{2t^2}\right>\]
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
oh sorry i didnt catch that
 one year ago
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