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) $r ( t ) = t i + \left( \frac { t ^ { 2 } } { 4 } \left( 4 t ^ { 2 } - 20 \right) + 2 \right) j + \ln ( t + 2 ) k$

Question

Quizplus · Accepted Answer

We can solve the differential equation by integrating each component separately:
\begin{align*}
\frac{dx}{dt} &= 1 \
\Rightarrow x &= t + C_1 \
\frac{dy}{dt} &= 4t^3 - 10t \
\Rightarrow y &= \int (4t^3 - 10t)dt = \frac{1}{2}t^4 - 5t^2 + C_2 \
\frac{dz}{dt} &= \frac{1}{t+2} \
\Rightarrow z &= \int \frac{1}{t+2} dt = \ln(t+2) + C_3
\end{align*}
Using the initial condition $\mathbf{r}(0) = 2\mathrm{j} + (\ln 2)\mathbf{k}$, we get:
\begin{align*}
x(0) &= C_1 = 0 \
y(0) &= -5(0)^2 + C_2 = 2 \
z(0) &= \ln(0+2) + C_3 = \ln 2 \
\Rightarrow C_2 &= 2, \quad C_3 &= \ln 2
\end{align*}
Therefore, the solution to the initial value problem is $\mathbf{r}(t) = ti + \left(\frac{t^2}{4}(4t^2 - 20) + 2ight)j + \ln(t+2)\mathbf{k}$, which matches choice D.

Solve the Initial Value Problem drdt=i+(4t3−10t)j+1t+2k\frac { d \mathbf { r } } { d t } = \mathbf { i } + \left( 4 t ^ { 3 } - 10 t \right) j + \frac { 1 } { t + 2 } \mathbf { k }dtdr​=i+(4t3−10t)j+t+21​k

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