Find the arclength of the curve r(t) = <9sqrt(2t), e^(9t), e^(-9t)>, 0<_t<_1

Answer 1

You have to use the standard formula:

L = ∫ sqrt( (dx/dt)² + (dy/dt)² + (dz/dt)² ) dt, the integral is from 0 to 1.

For the curve you gave, the integral can be taken numerically. I don't think you can solve it to a formula, unless there is a typo in the given curve.

Answer 2

I believe that the first component is 9√2 t, not 9√(2t).

In that case, Arc length = ∫_{(0 to 1)} √[((9√2 t)')² + ((e^9t)')² + ((e^-9t)')²] dt

= ∫_{(0 to 1)} √(162 + (9e^9t)² + (-9e^-9t)²) dt

= 9∫_{(0 to 1)} √(e^18t + 2 + e^-18t)dt = 9∫_{(0 to 1)} √(e^9t + e^-9t)² dt

= 9∫_{(0 to 1)} (e^9t + e^-9t)dt

Evaluate the integral.