# Solución

Apartado a

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

$$\begin{aligned} s &= \int_a^b\|\bold{r'}(t)\|dt\\ &= \int_1^5\sqrt{3^2+4^2}dt\\ &= \int_1^5 5dt = 5t\bigg|_1^5 = 20 \end{aligned}$$

Apartado b

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

$$\begin{aligned} s &= \int_a^b\|\bold{r'}(t)\|dt\\ &= \int_0^{2\pi}\sqrt{(cost−tsent)^2+(sent+tcost)^2+2^2}dt\\ &= \int_0^{2\pi}\sqrt{(cos^2t−2tsentcost+t^2sen^2t)+ (sen^2t+2tsentcost+t^2cos^2t)+4}dt\\ &= \int_0^{2\pi}\sqrt{cos^2t+sen^2t+t^2(cos^2t+sin^2t)+4}dt\\ &= \int_0^{2\pi}\sqrt{t^2+5}dt \end{aligned}$$

Aquí podemos usar una fórmula de una tabla de integrales

$$\int \sqrt{u^2+a^2}du = \frac{u}{2}\sqrt{u^2+a^2}+\frac{a^2}{2}ln\bigg|u+\sqrt{u^2+a^2}\bigg|+C$$

entonces obtenemos

$$\begin{aligned} \int_0^{2\pi}\sqrt{t^2+5}dt &= \frac{1}{2}\bigg(t\sqrt{t^2+5} + 5ln|t+\sqrt{t^2+5}|\bigg)_o^{2\pi}\\ &= \frac{1}{2}\bigg(2\pi\sqrt{4\pi^2+5} + 5ln\bigg(2\pi+\sqrt{4\pi^2+5}\bigg)\bigg)-\frac{5}{2}\sqrt{5}\\ &\approx 25.343 \end{aligned}$$