Question

# Find the arc length of the curve r(t) on the interval [−π/3].

Given $$\displaystyle{r}′{\left({t}\right)}=⟨{\sec{{2}}}{t},−{\sin{{t}}}⟩$$, find the arc length of the curve r(t) on the interval $$\displaystyle{\left[-\frac{\pi}{{3}}\right]}$$

2021-03-10

Let $$\displaystyle{y}={f{{\left({x}\right)}}},{a}\leq{x}\leq{b}$$ be the given curve. The arc length LL of such curve is given by the definite integral
$$\displaystyle={\int_{{{a}}}^{{{b}}}}\sqrt{{{1}+{\left[\int'{\left({x}\right)}\right]}^{{{2}}}}}{\left.{d}{x}\right.}$$
Let $$\displaystyle{x}={g{{\left({t}\right)}}},{y}={h}{\left({t}\right)}$$ where c≤x≤d be the parametric equations of the curve y=f(x).
Then the arc length of the curve is given by
$$\displaystyle{L}={\int_{{{c}}}^{{{d}}}}\sqrt{{{\left({\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}\right)}^{{2}}+{\left({\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}\right)}^{{2}}}}{\left.{d}{t}\right.}$$
Here $$x(t)=\sec2t,y(t)=−\sin t$$ where $$−π/3≤x≤π/3$$ be the parametric equations of the curve y=f(x).

Then the arc length of the curve is given by

$$L=\int_{-π/3}^{π/3}\sqrt{\sec2t)^2+-\sin t)^2}dt =\int_{-π/3}^{π/3}\sqrt{\sec^22t+\sin^2}tdt$$