Question

Question 13 of the final exam regarding arclength is giving me trouble. Does anyone know how to solve this?

Answer 1

I think that the hard part of this question is interpreting the question. The curves are defined as integrals. To calculate the arclength you need the derivatives. I believe that your integrands are essentially the derivatives. Next you sum the squares and take the square root. I think that you will essentially wind up with the square root of cosine squared plus sine squared... which is one. Taking the integral from zero to t0 of 1 will give you t0. It seems a little freaky though. I would appreciate some back-up.

Answer 2

That's what I recorded, so perhaps it is correct. I didn't have a lot of confidence in my answer though, so I wanted to see if I was completely off the mark.

Answer 3

I think it is a mistake and the correct equation is shown in the solution, corresponding to the ellipse (x2/a2 + y2/b2 = 1, a, b> 0) and not x2*a2 + y2*b2 = 1.
Because a,b are constants you can consider a=(1/a) and b=(1/b).