Calculus 2 test question that even the professor could not solve (Parametric Arc length) everyone got this question wrong from my calculus 3 midterm.
" find the arc length of the curve on the given interval: $x=\sqrt(t),  \space y=8t-6,\space on\space 0 \leq t \leq 3$"
I set the problem up just fine, however at that point I got stuck.
$$ \int_{0}^{3}\sqrt{\left ( \frac{1}{2\sqrt{t}} \right )^{2}+64} dt$$
we asked the teacher for a review, he got the problem set up then quit and said he would not count it against us. but it's killing me I need to know how to do it!
thanks

First simplify. $\left( \frac{1}{2\sqrt{t}} \right) ^2 = \frac{1}{4t}$.
Next, factor out the $4t$: $$f(t) = \sqrt{\frac{1}{4t} + 64} = \frac{1}{2\sqrt{t}}\cdot \sqrt{1 + 256t}$$
Now let $u = \sqrt{t}$, so $du = \frac{1}{2\sqrt{t}} dt$ and
$$f(t) dt = \sqrt{1+256u^2}du$$
From here you can use a trigonometric substitution where $256u^2 = \tan^2 \theta$, so $16u = \tan \theta$ and $du = \frac{1}{16}\sec^2 \theta d\theta$, resulting in
$$f(t)dt = \sqrt{1 + \tan^2 \theta}\frac{1}{16}\sec^2 d\theta = \frac{1}{16}\sec^3 \theta d\theta$$
From there you can use a reduction of order formula to relate that integral to the integral of $\sec \theta$, which is doable.
