Solución

Apartado a

Usando la ecuación 3.11, r′(t)=3i+4j\bold{r'}(t) = 3\bold{i} + 4\bold{j}, entonces

s=∫ab∥r′(t)∥dt=∫1532+42dt=∫155dt=5t∣15=20\begin{aligned} s &= \int_a^b\|\bold{r'}(t)\|dt\\ &= \int_1^5\sqrt{3^2+4^2}dt\\ &= \int_1^5 5dt = 5t\bigg|_1^5 = 20 \end{aligned}

Apartado b

Usando la ecuación 3.12, r′(t)=⟨cost−tsent,sent+tcost,2⟩\bold{r'}(t) = \lang cost − tsent, sent + tcost, 2\rang, entonces

s=∫ab∥r′(t)∥dt=∫02π(cost−tsent)2+(sent+tcost)2+22dt=∫02π(cos2t−2tsentcost+t2sen2t)+(sen2t+2tsentcost+t2cos2t)+4dt=∫02πcos2t+sen2t+t2(cos2t+sin2t)+4dt=∫02πt2+5dt\begin{aligned} s &= \int_a^b\|\bold{r'}(t)\|dt\\ &= \int_0^{2\pi}\sqrt{(cost−tsent)^2+(sent+tcost)^2+2^2}dt\\ &= \int_0^{2\pi}\sqrt{(cos^2t−2tsentcost+t^2sen^2t)+ (sen^2t+2tsentcost+t^2cos^2t)+4}dt\\ &= \int_0^{2\pi}\sqrt{cos^2t+sen^2t+t^2(cos^2t+sin^2t)+4}dt\\ &= \int_0^{2\pi}\sqrt{t^2+5}dt \end{aligned}

Aquí podemos usar una fórmula de una tabla de integrales

∫u2+a2du=u2u2+a2+a22ln∣u+u2+a2∣+C\int \sqrt{u^2+a^2}du = \frac{u}{2}\sqrt{u^2+a^2}+\frac{a^2}{2}ln\bigg|u+\sqrt{u^2+a^2}\bigg|+C

entonces obtenemos

∫02πt2+5dt=12(tt2+5+5ln∣t+t2+5∣)o2π=12(2π4π2+5+5ln(2π+4π2+5))−525≈25.343\begin{aligned} \int_0^{2\pi}\sqrt{t^2+5}dt &= \frac{1}{2}\bigg(t\sqrt{t^2+5} + 5ln|t+\sqrt{t^2+5}|\bigg)_o^{2\pi}\\ &= \frac{1}{2}\bigg(2\pi\sqrt{4\pi^2+5} + 5ln\bigg(2\pi+\sqrt{4\pi^2+5}\bigg)\bigg)-\frac{5}{2}\sqrt{5}\\ &\approx 25.343 \end{aligned}