I'm working on an arclength question right, now, and I'm consistently landing very close to the answer but not on it. I'm trying to find the arclength of the following:
$x = 6e^tcos(t) ; y = 6e^tsin(t) ; z = 17e^t$ from t = 0 to t = 1
I know that the arc length is the integral from 0 to 1 of v(t)
So here's what I've done:
$x' = -6(e^t)sin(t)$
$y' = 6(e^t)cos(t)$
$z' = 17e^t$
And then finding the squares of these values:
$x'^2 = 36(e^{2t})sin(t)^2$
$y'^2 = 36(e^{2t})cos(t)^2$
$z'^2 = 289e^{2t}$
And now that I have those individually, I add them under a square root. This is where things go wrong.
$\sqrt{36(e^{2t})sin(t)^2 + 36(e^{2t})cos(t)^2 + 289e^{2t}}$
$\sqrt{e^{2t}(36sin(t)^2 + 36cos(t)^2 + 289)}$
$\sqrt{e^{2t}(36(sin(t)^2 + cos(t)^2) + 289)}$
$\sqrt{e^{2t}(36 + 289)}$
$e^{t}\sqrt{36 + 289}$
$e^{t}\sqrt{325}$
So my method tells me to take the integral of this. But according to the book, I should actually be taking the integral of $e^{t}\sqrt{361}$, or $e^{t}\sqrt{289 + 36 + 36}$.
In other words, somewhere in my method I'm losing an additional 6$^2$, but for the life of me I can't figure out where I'm going wrong. Can someone point me in the right direction?
The computation of the derivatives $x'$ and $y'$ are incorrect. You should have $$x'(t) = (6 e^t \cos t)' = 6 e^t (\cos t - \sin t),$$ so $$x'(t)^2 = 6 e^{2t} (1 - 2 \sin t \cos t).$$ This and a similar computation for $y(t)$ give $$x'(t)^2 + y'(t)^2 = 12 e^{2t},$$ which accounts exactly for the deviation from the given solution.