Question

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)= , t=−2 T(-2)=

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

hmm

Answer 8

i dont think thats a vector

Answer 9

I think its the second equation you wrte in the beginning

Answer 10

@zepdrix?

Answer 11

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

Answer 12

\[\large \hat T(-2)=\frac{r\;'(-2)}{|r\;'(-2)|}\]

\[\large =\frac{<-2e^4,\quad -4,\quad \dfrac{1}{2}>}{\sqrt{4e^{8}+ 16+ \frac{1}{4}}}\]

Answer 13

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

Answer 14

its ok thanks

Answer 15

@zepdrix it says that the third coordinate is incorrect

Answer 16

is it \[r(t)=\left<e^{−2t}, t^2, \frac{1}{2t}\right>\]
or
\[r(t)=\left<e^{−2t}, t^2, \frac{1}{2}t\right>\]

Answer 17

yes it is

Answer 18

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

Answer 19

"yes it is" what?

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

if \[r(t)=\left<e^{−2t}, t^2, \frac{1}{2t}\right>\]

then

\[r'(t)=\left<e^{−2t}(-2), 2t, \frac{-1}{2t^2}\right>\]

Answer 24

oh sorry i didnt catch that