Answer check: Arc length x = e^t - e^-t , y = 5-2t [0,3]?

Question

Here are my steps:

dx = e^t + e^-t
dy = -2

then

L = ∫ √ [ (e^t + e^-t)² - (-2)² ] dt

= ∫ √ [ (e^2t) +( 2*e^t *e^-t) + (e^-2t) - 4 ] dt

2*e^t *e^-t = 2

so

L = ∫ √ [ (e^2t) - 2+ (e^-2t) ] dt

Which is the same as

L = ∫ √ [ (e^2t) - (2*e^t *e^-t)+ (e^-2t) ] dt

This is a perfect square

L = ∫ √ [ (e^2t ) - (e^-2t) ]² dt

L = ∫ (e^2t ) - (e^-2t)

= (1/2)e^2t + (1/2)e^-2t

The integral is from 0 to 3

the answer is (1/2) (e^6 + e^-6)

Is this right? I don't want the answer in decimals. To keep it in this format is okay.

Supastremph: you're right,

I shoudl have:

e^t + e^-t from 0 to 3

thus my answer is
(e^3 + e^-3) - (1+1)

= e^3 + e^-3 - 2

Is this the correct answer then?

Answer 1

Yes, this looks correct. The fact that you find the radical is a perfect square is an indicator that you did the problem right (it was engineered to be easy) as in general the integrals to find arc lengths are horrendous.

I found one error:

you want (e^t - e^t)^2 not (e^2t - e^-2t)^2

Answer 2

You were almost completely correct. When you evaluate the integral from 0 to 3, you will get (1/2)[(e^3 + e^-3 - 2)]--you didn't plug in t=0.


edit:
You updated yourself while I was posting.

Very nice work!

Answer 3

You're on the right track but be careful,

e^(2t) -2 + e^(-2t) = ( e^(t) - e^(-t) )^2

and don't forget to evaluate at the lower limit.

Good Luck!

UPDATE

Yes, you got the correct answer!!!