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Question 27
Let r(t) =cos(2t) i−sin(2t) j+4tk\mathbf { r } ( t ) = \cos ( 2 t ) \mathbf { i } - \sin ( 2 t ) \mathbf { j } + 4 t \mathbf { k }r(t) =cos(2t) i−sin(2t) j+4tk Then the unit tangent vector T(π4) \mathbf { T } \left( \frac { \pi } { 4 } \right) T(4π) is
A) −15i+25k- \frac { 1 } { \sqrt { 5 } } \mathbf { i } + \frac { 2 } { \sqrt { 5 } } \mathbf { k }−51i+52k B) 15i+25k\frac { 1 } { \sqrt { 5 } } \mathbf { i } + \frac { 2 } { \sqrt { 5 } } \mathbf { k }51i+52k C) −15i−25k- \frac { 1 } { \sqrt { 5 } } \mathbf { i } - \frac { 2 } { \sqrt { 5 } } \mathbf { k }−51i−52k D) 15i−25k\frac { 1 } { \sqrt { 5 } } \mathbf { i } - \frac { 2 } { \sqrt { 5 } } \mathbf { k }51i−52k E) −25i+15k- \frac { 2 } { \sqrt { 5 } } \mathbf { i } + \frac { 1 } { \sqrt { 5 } } \mathbf { k }−52i+51k
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