Finding unit normal vectors and normal/tangent components of accelerat

1. Sep 24, 2013

icesalmon

1. The problem statement, all variables and given/known data
given r(t) = <t, 1/t,0> find T(t) N(t) aT and aN at t = 1

2. Relevant equations
T(t) = r'/||r'||
N(t) = T'/||T'||
aT = a . T = (v . a)/||T||
aN = a . N = ||v x a||/||v|| = sqrt(||a||2 - aT2)

3. The attempt at a solution
for my T(t) I get <t2, -1 , 0>/(sqrt(1+t4) (I like keeping things in 3 dimensions even if there is no contribution in the z direction)
and I am not calculating the normal vector if there isn't some algebra I can use to simplify this greatly , which I am not seeing, t4 + 1 I don't believe I can factor and I can't think of any other way to simplify this one so i'm just moving around through the problem set looking for some r(t) = <cost, sint, t> type of vector that I can simply differentiate and use identities with, desperately trying to avoid those other problems

Last edited: Sep 24, 2013
2. Sep 24, 2013

jbunniii

How did you get that? Your first equation in the "relevant equations" section is $T(t) = r' / \|r\|$. What is $r'$ in this case?

3. Sep 25, 2013

icesalmon

r' = <1,-1/t2>
|r'| = sqrt(1 + 1/t4)
did I mess something up when differentiating/doing algebra/both?

4. Sep 25, 2013

icesalmon

|r'| = sqrt((t4+1)/t4) = (1/t2)sqrt(t4+1)
which should be t2r'/(|r'|)
which I believe is actually <t2,-1>/(sqrt(t4+1)
which is the same thing...oops i'm not sure what i'm messing up here.

5. Sep 25, 2013

jbunniii

OK, that looks fine. So what is $T'$? Just use the quotient rule, it shouldn't be too horrible.

6. Sep 25, 2013

icesalmon

well that's what i'm saying, are there any ways I can simplify this particular unit tangent vector before differentiating? I know how to differentiate it, but on an exam I will chew up a ton of time making sure I keep everything straight with the differentiation in terms of signs and cancellations.