# 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