Answer:
Step-by-step explanation:
Consider the following curve:
[tex]r(t)=10\cos(t)i + 10\sin(t)j+9tk[/tex] where [tex]0\leqt\leq\frac{\pi}{6}[/tex]
Need to find the arc length parameter and legth of the indicated portion of the curve
Difference [tex]r(t)[/tex] with respect to [tex]t[/tex]
[tex]r'(t) = -10\sin(t)i + 10\cos (t)j +9k\\\\|r'(t)|=\sqt{(-10\sin t)^2+(10\cos t)^2+(9)^2}\\\\=\sqrt{100\sin^2t + 100cos^2t +81}\\\\=\sqrt{100(\sin^2t+\cos^2t)+81}\\\\=\sqrt{100(1)+81}=\sqrt{181}[/tex]
The formular for arc length of a curve is [tex]L=\int\limits^b_a|r'(t)|dt[/tex]
So,
[tex]s(t)=\int\limits^b_a|v(\tau)|dt\\\\=\int\limits^{\frac{\pi}{6}}_a\sqrt{181}dt\, since\, |v|=|r'(t)|\, or\, |v|=\sqrt{181}\\\\=\sqrt{181}[t]\limits^{\frac{\pi}{6}}_0\\\\=\sqrt{181}[\frac{\pi}{6}-0]\\\\=\frac{\sqrt{181\times \pi}}{6}=7.0443[/tex]
hence the arc length is 7.0443
