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Question 9
Let r(t) =cos(2t+1) i+t2+1j+t22k\mathbf { r } ( t ) = \cos ( 2 t + 1 ) \mathbf { i } + \sqrt { t ^ { 2 } + 1 } \mathbf { j } + \frac { t ^ { 2 } } { 2 } \mathbf { k }r(t) =cos(2t+1) i+t2+1j+2t2k Then r′′(t) \mathbf { r } ^ { \prime \prime } ( t ) r′′(t) is
A) 4cos(2t+1) i+1(t2+1) 3j+k4 \cos ( 2 t + 1 ) \mathbf { i } + \frac { 1 } { \left( \sqrt { t ^ { 2 } + 1 } \right) ^ { 3 } } \mathbf { j } + \mathbf { k }4cos(2t+1) i+(t2+1) 31j+k B) 4cos(2t+1) i−1(t2+1) 3j+k4 \cos ( 2 t + 1 ) \mathbf { i } - \frac { 1 } { \left( \sqrt { t ^ { 2 } + 1 } \right) ^ { 3 } } \mathbf { j } + \mathbf { k }4cos(2t+1) i−(t2+1) 31j+k C) −4cos(2t+1) i+1(t2+1) 3j+k- 4 \cos ( 2 t + 1 ) \mathbf { i } + \frac { 1 } { \left( \sqrt { t ^ { 2 } + 1 } \right) ^ { 3 } } \mathbf { j } + \mathbf { k }−4cos(2t+1) i+(t2+1) 31j+k D) −4cos(2t+1) i−1(t2+1) 3j+k- 4 \cos ( 2 t + 1 ) \mathbf { i } - \frac { 1 } { \left( \sqrt { t ^ { 2 } + 1 } \right) ^ { 3 } } \mathbf { j } + \mathbf { k }−4cos(2t+1) i−(t2+1) 31j+k E) −4cos(2t+1) i+1(t2+1) 3j−k- 4 \cos ( 2 t + 1 ) \mathbf { i } + \frac { 1 } { \left( \sqrt { t ^ { 2 } + 1 } \right) ^ { 3 } } \mathbf { j } - \mathbf { k }−4cos(2t+1) i+(t2+1) 31j−k
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