Let $\mathbf { r } ( t ) = \cos 2 t \mathbf { i } + 2 t \mathbf { j } + \sin 2 t \mathbf { k }$ . Show that the acceleration vector is parallel to the normal vector N(t).

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The Answer of Let $\mathbf { r } ( t ) = \cos 2 t \mathbf { i } + 2 t \mathbf { j } + \sin 2 t \mathbf { k }$ . Show that the acceleration vector is parallel to the normal vector N(t).

Let r(t)=cos⁡2ti+2tj+sin⁡2tk\mathbf { r } ( t ) = \cos 2 t \mathbf { i } + 2 t \mathbf { j } + \sin 2 t \mathbf { k }r(t)=cos2ti+2tj+sin2tk

Let $\mathbf { r } ( t ) = \cos 2 t \mathbf { i } + 2 t \mathbf { j } + \sin 2 t \mathbf { k }$