Solve the Initial Value Problem $\frac { d \mathbf { r } } { d t } = \mathbf { i } + \left( 4 t ^ { 3 } - 10 t \right) j + \frac { 1 } { t + 2 } \mathbf { k }$

question 52

Multiple Choice

Solve the initial value problem.
-Differential Equation: $\frac { d \mathbf { r } } { d t } = \mathbf { i } + \left( 4 t ^ { 3 } - 10 t \right) j + \frac { 1 } { t + 2 } \mathbf { k }$
Initial Condition: $\mathbf { r } ( 0 ) = 2 \mathrm { j } + ( \ln 2 ) \mathbf { k }$

• A) \(r ( t ) = t i + \left( \frac { t ^ { 2 } } { 4 } \left( 4 t ^ { 2 } - 20 \right) + 1 \right) j + ( \ln 2 ) \mathbf { k }\)
• B) \(r ( t ) = t i + \left( \frac { t ^ { 2 } } { 2 } \left( 4 t ^ { 2 } - 20 \right) + 2 \right) j + ( \ln t ) k\)
• C) \(\mathbf { r } ( \mathrm { t } ) = \mathrm { i } + \left( \frac { \mathrm { t } ^ { 2 } } { 4 } \left( 4 \mathrm { t } ^ { 2 } - 20 \right) + 2 \right) \mathrm { j } + ( \ln 2 ) \mathbf { k }\)
• D) \(r ( t ) = t i + \left( \frac { t ^ { 2 } } { 4 } \left( 4 t ^ { 2 } - 20 \right) + 2 \right) j + \ln ( t + 2 ) k\)

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