Question

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

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

Answers (1)

2020-12-31

Let \(y=f(x),a\leq x\leq b\) be the given curve. The arc length LL of such curve is given by the definite integral
\(=\int_{a}^{b} \sqrt{1+[\int '(x)]^{2}}dx\)
Let \(x=g(t),y=h(t)\) where c≤x≤d be the parametric equations of the curve y=f(x).
Then the arc length of the curve is given by
\(L=\int_{c}^{d}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt\)
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
\(L=\int_{-π/3}^{π/3}\sqrt{\sec(2t)^2+-\sin (t)^2}dt =\int_{-π/3}^{π/3}\sqrt{\sec^22t+\sin^2}tdt.\)

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