So, I am in Calculus 3 right now and we are covering vector calculus. This week we are covering arc lengths.

While I understand the formula for arc length, I have a specific problem that I do not understand the answer which the book computed. It is not so much a question of the application of the formula as something the book did in its solutions manual without any obvious reason.

The problem is formated as:

While I understand the formula for arc length, I have a specific problem that I do not understand the answer which the book computed. It is not so much a question of the application of the formula as something the book did in its solutions manual without any obvious reason.

The problem is formated as:

r(t) = 2

r'(t) = 2

^{1/2}t**i**+ e^{t}**j**+ e^{-t}**k**r'(t) = 2

^{1/2}+ e^{t}-e^{-t}|r'(t)| = ( (2

^{1/2})

^{2}+ (e

^{t})

^{2}+ (-e

^{-t})

^{2})

^{(1/2)}

This is where my answer diverges from the books...According to my understanding of math, 2

After arbitrarilly dropping the '2', the book continues on its merry way. with:

^{1/221/2}=2^{1/2}. According to the book, the '2' simply disapears.After arbitrarilly dropping the '2', the book continues on its merry way. with:

|r'(t)| = (e

|r'(t)| = ((e

|r'(t)| = (e

Integrate |r'(t)| ... and so on and so forth...

^{2t}+ e^{-2t})^{1/2}|r'(t)| = ((e

^{t}+ e^{-t})^{2})^{1/2}|r'(t)| = (e

^{t}+ e^{-t})Integrate |r'(t)| ... and so on and so forth...

What I fail to understand is how and why the '2' is arbitrarilly dropped. In my understanding, the 2 should stay in and be integrated to a value of 2

Can anyone explain what happens to the 2?

^{1/2}t . Resultantly my answer and the books answer differ by the value of 2^{1/2}.Can anyone explain what happens to the 2?

## Error