smc-1196-45-Curves

Which of the following gives the same curve as \(\left(\begin{array}{c}\cos (t) \\ -t \\ \sin (t)\end{array}\right)\) for \(t \in \mathbb{R}\) ?

\(\left(\begin{array}{c}\cos (2 t) \\ 2 t \\ \sin (2 t)\end{array}\right)\)
\(\left(\begin{array}{c}\cos \left(t^2+\dfrac{\pi}{2}\right) \\ t^2+\dfrac{\pi}{2} \\ \sin \left(t^2+\dfrac{\pi}{2}\right)\end{array}\right)\)
\(\left(\begin{array}{c}\cos \left(t^2\right) \\ -t^2 \\ \sin \left(t^2\right)\end{array}\right)\)
\(\left(\begin{array}{c}\cos \left(2 t+\dfrac{\pi}{2}\right) \\ 2 t+\dfrac{\pi}{2} \\ -\sin \left(2 t+\dfrac{\pi}{2}\right)\end{array}\right)\)

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\(D\)

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\(\text{Find which option is a re-parametrisation of the given curve.}\)

\(\text{Consider option D:}\)

\(\text{Let}\ \ u=- \left(2t + \dfrac{\pi}{2}\right)\)

\(\text{Since}\ t \in \mathbb{R}\ \ \Rightarrow \ \ u \in \mathbb{R}\)

\(\cos \left(2 t+\dfrac{\pi}{2}\right) = \cos\left(- \left(2 t+\dfrac{\pi}{2}\right) \right) = \cos\,u\)

\(2 t+\dfrac{\pi}{2} = -\left( -\left(2 t+\dfrac{\pi}{2} \right) \right) = -u\)

\(-\sin \left(2 t+\dfrac{\pi}{2}\right) = \sin\left(- \left(2 t+\dfrac{\pi}{2}\right) \right) = \sin\,u\)

\(\Rightarrow D\)

Vectors, EXT2 V1 2025 HSC 10 MC