A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 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)=
 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)=

This Question is Closed

zepdrix
 one year 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
 one year 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
 one year 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
 one year 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
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0i dont think thats a vector

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

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

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

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

Zarkon
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0oh sorry i didnt catch that
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.