1. ## Arc length question

I understand how to calculate arc length given parametric equations. However, this one has me completely stumped.

Find the arc length of the curve $r(t)= <2\sqrt{2t}, e^{2t}, e^{-2t}>$. The interval given is [0,1]

Following the same procedure I use for other arc length calculations, I arrive at:

$\int_0^1 2\sqrt{.5t+e^{4t}-e^{-4t}}$

I have no idea how to even approach that integral, and have spent the last two hours just on this problem.

Any help would be appreciated.

Thanks!

2. Originally Posted by nicknc
I understand how to calculate arc length given parametric equations. However, this one has me completely stumped.

Find the arc length of the curve $r(t)= <2\sqrt{2t}, e^{2t}, e^{-2t}>$. The interval given is [0,1]

Following the same procedure I use for other arc length calculations, I arrive at:

$\int_0^1 2\sqrt{.5t+e^{4t}-e^{-4t}}$

I have no idea how to even approach that integral, and have spent the last two hours just on this problem.

Any help would be appreciated.

Thanks!
Once the derivatives are correctly found and substituted into the formula you will get $2 \int_0^1 \sqrt{\frac{1}{2t} + e^{4t} + e^{-4t}} \, dt$. An exact answer is not possible using a finite number of elementary functions. To get an approximate answer, use technology.

However ....... I suspect there's a small but fatal typo in the question and that the definition of the curve should be $r(t)= <{\color{red}(2\sqrt{2}) t}, e^{2t}, e^{-2t}>$, in which case the integrand becomes a perfect square ....

3. If you gave accurate information and the first component of the r(t) function is indeed 2 sqrt(2t), then I can't help...

But if the component is actually 2t sqrt(2), then it's much simpler to evaluate.
r'(t) = 2 < sqrt(2) , e^(2t) , -e^(-2t)>

Just find |r'(t)|, and you should have:
2 sqrt( 2 + e^(4t) + e^(-4t))

Try simplifying this by realizing that e^(4t) + 2 + e^(-4t) is in the form a^2 + ab + b^2. Then the integral should be easy to evaluate.

But once again, if the information you first provided is accurate, then I'm stuck too.

Edit: Mr. Fantastic beat me to the punch!