Specialist2024 NHT Exam 2 Section A · Q9CAS active1 mark

Length of a parametric curve

The length of the curve given by $x = 3\left(1 - \tan^2(t)\right)$ , $y = 4\sec^2(t)$ where $t \in \left[\frac{\pi}{6}, \frac{\pi}{3}\right]$ , is given by

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\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 10\tan(t)\sec^2(t)\,dt

\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 100\tan^2(t)\sec^4(t)\,dt

\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 2\tan(t)\sec^2(t)\,dt

\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 28\tan^2(t)\sec^4(t)\,dt

\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{2}\sqrt{\tan(t)}\sec^2(t)\,dt

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