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 2 years 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)=
 2 years 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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zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.2Ummm 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
 2 years ago
Best 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,\[\large \vec r\;'(t)=<2e^{2t},\quad 2t,\quad \frac{1}{2}>\]

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

zepdrix
 2 years ago
Best 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,\[\large \vec v(t)=\sqrt{\left(2e^{2t}\right)^2+ \left(2t\right)^2+ \left(\frac{1}{2}\right)^2}\]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.2You'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
 2 years ago
Best 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

ranyai12
 2 years ago
Best ResponseYou've already chosen the best response.0i dont think thats a vector

ranyai12
 2 years ago
Best ResponseYou've already chosen the best response.0I think its the second equation you wrte in the beginning

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

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

ranyai12
 2 years ago
Best ResponseYou've already chosen the best response.0@zepdrix it says that the third coordinate is incorrect

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

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

ranyai12
 2 years ago
Best ResponseYou've already chosen the best response.0oh sorry i misread what you wrote it its the first one

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.2that is why zepdrix's third component is not correct.

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

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

ranyai12
 2 years ago
Best ResponseYou've already chosen the best response.0oh sorry i didnt catch that
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