Given r′(t)=⟨sec2t,−sint⟩, find the arc length of the curve r(t) on the interval [−π/3].

Question
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[−

Answers (1)

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 \(\displaystyle{x}{\left({t}\right)}={\sec{{2}}}{t},{y}{\left({t}\right)}=−{\sin{{t}}}\) where \(\displaystyle−
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