If $\frac{d^2\vec{r}}{dt^2}=6t\hat{i}-24t^2\hat{j}+4\sin t \hat{k}$, then prove that $\vec{r}=(t^3-t+2)\hat{i}+(1-2t^4)\hat{j}+(t-4\sin t)\hat{k}$

Question

Problem : If $\dfrac{d^2\vec{r}}{dt^2}=6t\hat{i}-24t^2\hat{j}+4\sin t \hat{k}$ and if $\vec{r}=2\hat{i}+\hat{j}$ and $\dfrac{d\vec{r}}{dt}=-\hat{i}-3\hat{k}$ when $t=0$, then show that $\vec{r}=(t^3-t+2)\hat{i}+(1-2t^4)\hat{j}+(t-4\sin t)\hat{k}$.

Try: We have \begin{align*} \dfrac{d^2\vec{r}}{dt^2}= 6t\hat{i}-24t^2\hat{j}+4\sin t \hat{k}\implies & d\left(\dfrac{d\vec{r}}{dt}\right)=(6t\hat{i}-24t^2\hat{j}+4\sin t \hat{k})\, dt \\ \implies & \dfrac{d\vec{r}}{dt}=3t^2\hat{i}-8t^3\hat{j}-4\cos t \hat{k}+ \vec{C}\\ \implies & \dfrac{d\vec{r}}{dt}=(3t^2-1)\hat{i}-8t^3\hat{j}-(4\cos t+7) \hat{k}\\ \implies & \vec{r}=(t^3-t)\hat{i}-2t^4\hat{j}-(4\sin t+7t) \hat{k}+\vec{D}\\ \implies & \vec{r}=(t^3-t)\hat{i}-2t^4\hat{j}-(4\sin t+7t) \hat{k} \quad [\because \vec{D}=\vec 0] \end{align*} which is different from the original result. But the given result is true. My derived result is also satisfies the given diff equn. Where is the problem?

Answer 1

It seems you made $2$ mistakes. First, with

$$\dfrac{d\vec{r}}{dt}=3t^2\hat{i}-8t^3\hat{j}-4\cos t \hat{k}+ \vec{C} \tag{1}\label{eq1A}$$

From $\dfrac{d\vec{r}}{dt}=-\hat{i}-3\hat{k}$ when $t=0$, you get

$$\begin{equation}\begin{aligned} -4\hat{k} + \vec{C} & = -\hat{i}-3\hat{k} \\ \vec{C} & = -\hat{i}+\hat{k} \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Thus, you would get

$$\dfrac{d\vec{r}}{dt}=(3t^2-1)\hat{i}-8t^3\hat{j}-(4\cos t-1) \hat{k} \tag{3}\label{eq3A}$$

It seems you subtracting $4\hat{k}$ on the right instead of adding it. Anyway, this is basically what the answer also states.

Near the end, using the correct derivative expression, you would get

$$\vec{r}=(t^3-t)\hat{i}-2t^4\hat{j}-(4\sin t-t) \hat{k}+\vec{D} \tag{4}\label{eq4A}$$

Using the initial condition $\vec{r}=2\hat{i}+\hat{j}$ when $t = 0$ gives

$$2\hat{i}+\hat{j} = \vec{D} \tag{5}\label{eq5A}$$

I don't know why you think $\vec{D} = \vec 0$. Using \eqref{eq5A} in \eqref{eq4A}, you would get

$$\begin{equation}\begin{aligned} \vec{r} & =(t^3-t+2)\hat{i}-(2t^4-1)\hat{j}-(4\sin t-t) \hat{k} \\ & = (t^3-t+2)\hat{i}+(1-2t^4)\hat{j}+(t-4\sin t)\hat{k} \end{aligned}\end{equation}\tag{6}\label{eq6A}$$

This matches what you're asked to show.

Answer 2

It looks like you might have an error in the following line:

$\dfrac{d\vec{r}}{dt}=(3t^2-1)\hat{i}-8t^3\hat{j}-(4\cos t+7)\hat{k}$

I believe you should be subtracting 1 in the $\hat{k}$ term instead of adding 7, to match the initial condition given for $t=0$. I believe it should read:

$\dfrac{d\vec{r}}{dt}=(3t^2-1)\hat{i}-8t^3\hat{j}-(4\cos t-1)\hat{k}$

I hope this helps.