Homework Help: Reparamterizing a curve

1. Oct 14, 2007

1. The problem statement, all variables and given/known data

"Reparameterize the curve with respect to the arc length measured from the point where t=0 in the direction of increasing t. r(t)=e^(2t)*cos2t i +2 j + e^(t2)*sin2t k"

2. Relevant equations

3. The attempt at a solution

the method we were taught is take r'(t) and then find |r'(t)| and then integrate that.

but in every example we were given, the derivative was a real number, not a function (IE: 12, sqrt(23) etc)

i took the derivative and got [-2e^(2t)sin(2t)+2te^(2t)cos(2t)]i +0j + [2e^(2t)cos(2t)+2te^(2t)sin(2t)] k

then |r'(t)|=(after simplifying it i got) =2e^(2t)sqrt(1+t^2) which seems nice and simple now....but how to i integrate that...

the form they used was s=s(t)=integral from 0-t of |r'(u)|du
for every other problem of this nature

2. Oct 14, 2007

Dick

Where did the t's in front of the exponentials in your derivative come from? They definitely shouldn't be there.

3. Oct 14, 2007

this derivative? [-2e^(2t)sin(2t)+2te^(2t)cos(2t)]i +0j + [2e^(2t)cos(2t)+2te^(2t)sin(2t)] k

r(t)=e^(2t)*cos2t i +2 j + e^(t2)*sin2t k

ill just do the firts part as "k" is the same as "i"

f'(t) "e^(2t)*cos2t"

u'v*v'u corrrect?

2t*e^t*cos2t + e(2t)*-sin2t

or did i do that wrong?
i thought the derivative of e^(2x)=2xe(2x) but maybe its juts 2e^(2x)? (this is calc 3 and i took calc 1 and 2 2 years ago so i kinda forget them! :(

4. Oct 14, 2007

Dick

Yes, it's just 2e^(2x). It's a pretty easy chain rule.

5. Oct 14, 2007

so doing it the RIGHT way i get the derivative is

[-2e^(2t)sin(2t)+2e^(2t)cos(2t)]i +0j + [2e^(2t)cos(2t)+2e^(2t)sin(2t)] k

which gives me a simplified |r'(t)| of 2e^(2t)sqrt2?
that look right now

6. Oct 14, 2007

and assuming thats right i would just integrate 2e^(2t)sqrt2? and get e^(2t)sqrt2? and then evaluate from o to t?
which would give me e^(2t)sqrt2-sqrt(2)

Last edited: Oct 14, 2007
7. Oct 14, 2007

Dick

So what's your new parameter t? So far, you only have arc length.

8. Oct 14, 2007

e^(2t)sqrt2-sqrt(2) would be the arc legnth

so the new parameter is since s=e^(2t)sqrt2-sqrt(2)
s+sqrt2=e^(2t)sqrt2
(s+sqrt2)/sqrt2=e^(2t)
ln((s+sqrt2)/sqrt2)=2tlne
ln((s+sqrt2)/sqrt2) /2 =t

original e^(2t)*cos2t i +2 j + e^(t2)*sin2t

final e^(ln((s+sqrt2)/sqrt2))*cos(ln((s+sqrt2)/sqrt2)) i +2 j + e^((ln((s+sqrt2)/sqrt2)))*sin(ln((s+sqrt2)/sqrt2))

9. Oct 14, 2007

Dick

Gotta confess, I'm not checking all the details for you. But it looks about right. You can check yourself. Take d/ds and see if you get a unit vector. BTW you could have skipped the integration constant. Why?

Last edited: Oct 14, 2007
10. Oct 14, 2007

what do you mean by the integration constant? the 2?

11. Oct 14, 2007

Dick

No, the sqrt(2). The lower limit in your integral. It doesn't hurt you to include it though. Don't worry about it.

12. Oct 14, 2007