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Let r(t)=esinti+costj+ln(5t)k\mathbf { r } ( t ) = e ^ { \sin t } \mathbf { i } + \cos t \mathbf { j } + \ln ( 5 - t ) \mathbf { k }

Question 75

Multiple Choice

Let r(t) =esinti+costj+ln(5t) k\mathbf { r } ( t ) = e ^ { \sin t } \mathbf { i } + \cos t \mathbf { j } + \ln ( 5 - t ) \mathbf { k } Then r(t) \mathbf { r } ^ { \prime \prime } ( t ) is


A) esint(cos2tsint) i+costj1(t5) 2ke ^ { \sin t } \left( \cos ^ { 2 } t - \sin t \right) \mathbf { i } + \cos t \mathbf { j } - \frac { 1 } { ( t - 5 ) ^ { 2 } } \mathbf { k }
B) etint(cos2sint) icostj+1(t5) 2ke ^ { \operatorname { tin } t } \left( \cos ^ { 2 } - \sin t \right) \mathbf { i } - \cos t \mathbf { j } + \frac { 1 } { ( t - 5 ) ^ { 2 } } \mathbf { k }
C) esint(cos2tsint) i+costj+1(t5) 2ke ^ { \sin t } \left( \cos ^ { 2 } t - \sin t \right) \mathbf { i } + \cos t \mathbf { j } + \frac { 1 } { ( t - 5 ) ^ { 2 } } \mathbf { k }
D) esint(cos2tsint) icostj1(t5) 2ke ^ { \sin t } \left( \cos ^ { 2 } t - \sin t \right) \mathbf { i } - \cos t \mathbf { j } - \frac { 1 } { ( t - 5 ) ^ { 2 } } \mathbf { k }
E) esint(cos2tsint) icostj+1(t5) 2k- e ^ { \sin t } \left( \cos ^ { 2 } t - \sin t \right) \mathbf { i } - \cos t \mathbf { j } + \frac { 1 } { ( t - 5 ) ^ { 2 } } \mathbf { k }

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