Find the arc length parameter along the given curve from the point where t=0 by evaluating the integral s(t)=lv(t)ldt The

Question

Find the arc length parameter along the given curve from the point where t=0 by evaluating the integral s(t)=lv(t)ldt

Then find the length of indicated portion of the curve r(t)= 5costi+5sintj+8tk, where 0<_t<_pi

The arc length parameter along the curve, starting at t=0 is s(t)=____. (Type exact answers, using radicals as needed)

The length of the indicated portion of the curve is L=___. (Type exact answers, using radicals and Pi as needed)

Find the arc length parameter along the given curve from the point where t=0 by evaluating the integral s(t)=lv(t)ldt

Then find the length of indicated portion of the curve r(t)=(2e^tcost)i+(2e^tsint)j-2e^tk, -ln4<_t<_0 s(t)=__ (Type exact answers, using radicals as needed)

The length of rage indicated portion of the curve is __ units. (Type exact answers, using radicals as needed)

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Harper 4 weeks 2021-09-19T03:17:02+00:00 1 Answer 0

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    2021-09-19T03:18:16+00:00

    Answer:

    Step-by-step explanation:

    Consider the following curve:

    r(t)=10\cos(t)i + 10\sin(t)j+9tk where 0\leqt\leq\frac{\pi}{6}

    Need to find the arc length parameter and legth of the indicated portion of the curve

    Difference r(t) with respect to t

    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}

    The formular for arc length of a curve is L=\int\limits^b_a|r'(t)|dt

    So,

    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

    hence the arc length is 7.0443

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