Question

Please help me with unit tangent and normal vectors. I am asked to find T(t) and N(t) at the given point t=1.

(click to see equation)

Answer 1

\[r(t)=(t^2-1)i+tj\]

Answer 2

\[r'(t)=2ti+j\]

Answer 3

\[T(1)=\frac{ 2 }{ \sqrt5 }i+\frac{ 1 }{ \sqrt5 }j\]

Answer 4

I tried to find N(t) a t=1 but i am not getting the right answer

Answer 5

\[N(t)=\frac{ T'(t) }{ ||T'(t)|| }\]

Answer 6

\[T(t)=\frac{ 2 }{ \sqrt5 }ti+\frac{ 1 }{ \sqrt5 }j\]

Answer 7

\[T'(t)=\frac{ 2 }{ \sqrt5 }\]

Answer 8

\[\large \hat T(1)=\frac{<2,1>}{\sqrt5} \\ \large \hat T(1)=\left<\frac{2}{\sqrt5},\frac{1}{\sqrt5}\right>\]

Bahhh I was doing N wrong, lemme try this again..
Your Tangent vector looks good though.

Answer 9

Ya i checked my t-vector in the back of the book but my N is way off for some reason....

Answer 10

I'm a little confused by your T(t).
Shouldn't we get something like this?
Remember we're not plugging in t=1 at this point.
\[\large \hat T(t)=\frac{<2t,1>}{\sqrt{4t^2+1}} \qquad = \qquad \left<\frac{2t}{\sqrt{4t^2+1}},\;\frac{1}{\sqrt{4t^2+1}}\right>\]

And then we'd have to differentiate this ... I guess :P

Answer 11

Oh ok, I see my mistake now....i wasn't taking t into account when finding the magnitude of r'(t)

Answer 12

i think i closed this too soon lol
how do you take the derivative of T(t)?

Answer 13

Oh boy, it's a doozy.

Quotient rule on the \(\large \hat i\) component.
And for the \(\large \hat j\) component we can just use the power rule by changing the square root to a fractional exponent.

Answer 14

Do you happen to have an answer key handy?

I worked it out, I wanna know if I'm on the right track here.

Answer 15

the answer to N(1)

Answer 16

ya

Answer 17

\[<\frac{ 1 }{ \sqrt5 },-\frac{ 2 }{ \sqrt5 }>\]

Answer 18

Ah ok, I messed up somewhere then lol.
I'll have to try again.

Answer 19

Yesssss finally got it!

Answer 20

Getting stuck at any particular spot?

Answer 21

ya im stuck simplifying the derivative

Answer 22

That's where I was making a mistake also.
Don't both simplifying it.
If you've taken the derivative, simply plug in t=1 from that point.

Answer 23

It will be much easier to simplify down after you've plugged in t=1.

Answer 24

ok and what about when im finding the magnitude of T'(t)? it doesnt need to be simplified?

Answer 25

Once you have found, \(\large \hat T'(t)\), plug in t=1.

From there you can simplify \(\large \hat T'(1)\) down to 2 nice looking components.

Then finding \(\large ||\hat T'(1)||\) from there will be fairly straight forward.

Answer 26

You'll want to use the simplified \(\large \hat T'(1)\) to find your \(\large ||\hat T'(1)||\).

Don't try to find the magnitude until you've simplified the thing down! :O

Answer 27

Hope that makes sense :3

Answer 28

ya ill just try and work it out..

this is pretty insane because this is the first problem out of 15 and I imagine each one is a bit more difficult

Answer 29

Are they all the same type of problem?

Yah the vector problems can take a lot of steps to get through :\

Answer 30

i have 12 problems like this and more on binormal vectors

Answer 31

Just to help lead you in the right direction,
Once you plug in t=1, and simplify your derivative, you should get this,

\[\large \hat T'(1)=\left<\frac{2}{\sqrt{125}},\;-\frac{4}{\sqrt{125}}\right>\]

Answer 32

ok thanks