Question

Find the arc length of the curve from t = 0 to t = 1 whose derivatives in parametric form are dx/dt=2-cos(t) and dy/dt=ln(t^2). Type your answer in the space below and give 2 decimal places

My work:
2-cost(t) = sin(t) = sin^2(t)
ln(t^2)= 2/t= 4/t^2
integral from 0 to 1 √(sin^2(t)+4/t^2) dt
I got stuck, please finish this. Show me your work

Answer 1

part of your problem is that you didn't use what they gave you.
You don't have to take the derivative, since they gave you the derivatives.
That is, you already know dx/dt and dy/dt. Just plug them into the formula for ds.

And your sloppy use of the "=" signs makes things even more confusing.

That said, good luck with the integral; you will have to use some numeric method to evaluate it.

Answer 2

I got 1.98 oobleck

Answer 3

Hmmm. wolframalpha got 2.52

https://www.wolframalpha.com/input/?i=%E2%88%AB%5B0,1%5D+sqrt((2-cost)%5E2%2Bln(t%5E2)%5E2)+dt