## 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

1. 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

2. zepdrix

$\large \vec r(t)=<e^{-2t},\quad t^2,\quad \frac{1}{2}t>$Taking the derivative of each component gives us,$\large \vec r\;'(t)=<-2e^{-2t},\quad 2t,\quad \frac{1}{2}>$

3. zepdrix

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

$\large \vec v(t)=<-2e^{-2t},\quad 2t,\quad \frac{1}{2}>$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

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

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

7. zepdrix

hmm

8. ranyai12

i dont think thats a vector

9. ranyai12

I think its the second equation you wrte in the beginning

10. ranyai12

@zepdrix?

11. 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

12. zepdrix

$\large \hat T(-2)=\frac{r\;'(-2)}{|r\;'(-2)|}$ $\large =\frac{<-2e^4,\quad -4,\quad \dfrac{1}{2}>}{\sqrt{4e^{8}+ 16+ \frac{1}{4}}}$

13. zepdrix

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

14. ranyai12

its ok thanks

15. ranyai12

@zepdrix it says that the third coordinate is incorrect

16. Zarkon

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

yes it is

18. ranyai12

@Zarkon sorry t took me long ot respond I was trying to solve another problem

19. Zarkon

"yes it is" what?

20. ranyai12

oh sorry i misread what you wrote it its the first one

21. Zarkon

that is why zepdrix's third component is not correct.

22. Zarkon

you should have corrected their 2nd post. Then he could have derived the answer you need

23. Zarkon

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

oh sorry i didnt catch that