# Explanation for integration of Dr/Dt

1. Mar 7, 2017

### harmyder

1. The problem statement, all variables and given/known data

Why integration of $$\frac{D^2\mathbf r}{Dt^2}=−2\mathbf w \times \frac{D\mathbf r}{Dt}−g\mathbf R$$ gives us
$$\frac{D\mathbf r}{Dt}= \mathbf v_0 −2\mathbf w×(\mathbf r−\mathbf r_0)−gt\mathbf R$$

2. Relevant equations

Consider a time-varying vector written in the body coordinate system, $\xi(t) = R(t)\mathbf s(t).$
$$\frac{d\xi}{dt} = R\frac{d\mathbf s}{dt}+\mathbf w \times \xi = \frac{D\xi}{Dt} +\mathbf w \times \xi.$$

3. The attempt at a solution
It looks for me like they incorporated $-\mathbf v_0$ for LHS and $-\mathbf r_0$ for $\frac{D\mathbf r}{Dt}.$

Ah! Probably they took definite integral $\int_0^t$!

Last edited: Mar 7, 2017
2. Mar 12, 2017

### Ssnow

Hi, if you take $\int_{0}^{t} \cdot ds$ (as you said at the end) of both sides of $\frac{D^{2} \mathbf{r}}{Ds^2}=-2\mathbf{w}\times \frac{D\mathbf{r}}{Ds}-g\mathbf{R}$ you obtain in the first side $\frac{D \mathbf{r}}{Ds}(t)-\frac{D \mathbf{r}}{Ds}(0)$ that is problably $\frac{D \mathbf{r}}{Ds}(t)-\mathbf{v}_{0}$, the second side is a simple integration and calling $\mathbf{r}=\mathbf{r}(t)$ and $\mathbf{r}_{0}=\mathbf{r}(0)$ you have the result ...

Ssnow