Solve the initial value problem.
-Differential Equation: $\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm { dt } ^ { 2 } } = 3 \mathrm { t } ^ { 2 } \mathbf { i } - \mathrm { tj }$
Initial Conditions: $\left. \frac { \mathrm { d } r } { \mathrm { dt } } \right| _ { \mathrm { r } = 0 } = - 4 \mathbf { i } , \mathrm { r } ( 0 ) = 10 \mathbf { i } - 3 \mathbf { k }$
A) $\mathbf { r } ( \mathrm { t } ) = \left( \frac { 1 } { 4 } \mathrm { t } ^ { 4 } - 4 \mathrm { t } + 10 \right) \mathrm { i } - \frac { \mathrm { t } ^ { 3 } } { 2 } \mathrm { j } - 3 \mathrm { k }$
B) $\mathbf { r } ( \mathrm { t } ) = \left( \frac { 3 } { 10 } \mathrm { t } ^ { 4 } - 4 \mathrm { t } + 10 \right) \mathrm { i } - \frac { \mathrm { t } ^ { 3 } } { 6 } \mathrm { j } - 3 \mathrm { k }$
C) $\mathbf { r } ( \mathrm { t } ) = \left( \frac { 1 } { 4 } \mathrm { t } ^ { 4 } + 4 \mathrm { t } + 10 \right) \mathrm { i } + \frac { \mathrm { t } ^ { 3 } } { 6 } \mathrm { j } - 3 \mathbf { k }$
D) $\mathbf { r } ( \mathrm { t } ) = \left( \frac { 1 } { 4 } \mathrm { t } ^ { 4 } - 4 \mathrm { t } + 10 \right) \mathrm { i } - \frac { \mathrm { t } ^ { 3 } } { 6 } \mathrm { j } - 3 \mathbf { k }$

Question

Quizplus · Accepted Answer

We can solve the differential equation by integrating twice with respect to time:
\begin{align*}
\frac{\mathrm{d}^2 \mathbf{r}}{\mathrm{dt}^2} &= 3\mathrm{t}^2\mathbf{i} - \mathrm{tj} \
\int \frac{\mathrm{d}^2 \mathbf{r}}{\mathrm{dt}^2} \mathrm{d}t &= \int (3\mathrm{t}^2\mathbf{i} - \mathrm{tj}) \mathrm{d}t \
\frac{\mathrm{d} \mathbf{r}}{\mathrm{dt}} &= \mathbf{a} \mathrm{t}^3 + \mathbf{b} \mathrm{t} + \mathbf{c} \
\int \frac{\mathrm{d} \mathbf{r}}{\mathrm{dt}} \mathrm{d}t &= \int (\mathbf{a} \mathrm{t}^3 + \mathbf{b} \mathrm{t} + \mathbf{c}) \mathrm{d}t \
\mathbf{r}(\mathrm{t}) &= \frac{\mathbf{a}}{4}\mathrm{t}^4 + \frac{\mathbf{b}}{2}\mathrm{t}^2 + \mathbf{c}\mathrm{t} + \mathbf{d}
\end{align*}
where $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, and $\mathbf{d}$ are constants of integration that we will determine using the initial conditions.

From the initial condition $\left.\frac{\mathrm{d} \mathbf{r}}{\mathrm{dt}}\right|_{\mathbf{r}=0}=-4\mathbf{i}$, we have $\mathbf{b}=-4\mathbf{i}$.

From the initial condition $\mathbf{r}(0) = 10\mathbf{i} - 3\mathbf{k}$, we have $\mathbf{d} = 10\mathbf{i} - 3\mathbf{k}$.

To find $\mathbf{a}$ and $\mathbf{c}$, we need to differentiate $\mathbf{r}(\mathrm{t})$ once and evaluate it at $\mathrm{t}=0$:
\begin{align*}
\frac{\mathrm{d} \mathbf{r}}{\mathrm{dt}} &= \mathbf{a} \mathrm{t}^3 - 4\mathbf{i} + \mathbf{c} \
\left.\frac{\mathrm{d} \mathbf{r}}{\mathrm{dt}}\right|_{\mathrm{t}=0} &= \mathbf{c} - 4\mathbf{i} = -4\mathbf{i} \
\Rightarrow \mathbf{c} &= 0
\end{align*}

Therefore, the solution to the initial value problem is:
$$\mathbf{r}(\mathrm{t}) = \left(\frac{1}{4}\mathrm{t}^4 - 4\mathrm{t} + 10\right)\mathbf{i} - \frac{1}{6}\mathrm{t}^3\mathbf{j} - 3\mathbf{k}$$
which matches choice D.

Solve the Initial Value Problem d2rdt2=3t2i−tj\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm { dt } ^ { 2 } } = 3 \mathrm { t } ^ { 2 } \mathbf { i } - \mathrm { tj }dt2d2r​=3t2i−tj

Solve the Initial Value Problem $\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm { dt } ^ { 2 } } = 3 \mathrm { t } ^ { 2 } \mathbf { i } - \mathrm { tj }$