- #1

Lambda96

- 189

- 65

- Homework Statement
- I should show the following ##\bigl\langle \dot{\textbf{r}}'(t) ,\ddot{\textbf{r}}'(t) \bigr\rangle=0##

- Relevant Equations
- none

Hi,

I am having problems with task b

I then defined the velocity vector and the acceleration vector as follows

##dot{\textbf{r}}'(t) = \frac{1}{||\dot{\textbf{r}}(t)||} \left(\begin{array}{c} \dot{r_1}(t) \\ \dot{r_2}(t) \end{array}\right)##

and

##ddot{\textbf{r}}'(t) = \frac{1}{||\dot{\textbf{r}}(t)||} \frac{d}{dt}\left(\begin{array}{c} \dot{r_1}(t) \\ \dot{r_2}(t) \end{array}\right)##

Then I calculated the following:

##bigl\langle \dot{\textbf{r}}'(t) ,\ddot{\textbf{r}}'(t) \bigr\rangle=\dot{\textbf{r}}'(t) = \frac{1}{||\dot{\textbf{r}}(t)||^2} \left(\begin{array}{c} \dot{r_1}(t) \\ \dot{r_2}(t) \end{array}\right) \cdot \frac{d}{dt}\left(\begin{array}{c} \dot{r_1}(t) \\ \dot{r_2}(t) \end{array}\right)= \frac{1}{||\dot{\textbf{r}}(t)||^2} \Bigl( \dot{r_1}(t) \frac{d}{dt} \dot{r_1}(t) + \dot{r_2}(t) \frac{d}{dt} \dot{r_2}(t) \Bigr)=\frac{1}{||\dot{\textbf{r}}(t)||^2} \Bigl( \dot{r_1}(t) \ddot{r_1}(t) + \dot{r_2}(t) \ddot{r_2}(t) \Bigr)##

Unfortunately, I can't get any further. The following must apply ##\bigl\langle \dot{\textbf{r}}'(t) ,\ddot{\textbf{r}}'(t) \bigr\rangle=0## so that this is the case, the expression ##\dot{r_1}(t) \ddot{r_1}(t) + \dot{r_2}(t) \ddot{r_2}(t)=0##, but why is this the case?

My derivation is probably already wrong, but I don't know how else to solve the problem.

I am having problems with task b

I then defined the velocity vector and the acceleration vector as follows

##dot{\textbf{r}}'(t) = \frac{1}{||\dot{\textbf{r}}(t)||} \left(\begin{array}{c} \dot{r_1}(t) \\ \dot{r_2}(t) \end{array}\right)##

and

##ddot{\textbf{r}}'(t) = \frac{1}{||\dot{\textbf{r}}(t)||} \frac{d}{dt}\left(\begin{array}{c} \dot{r_1}(t) \\ \dot{r_2}(t) \end{array}\right)##

Then I calculated the following:

##bigl\langle \dot{\textbf{r}}'(t) ,\ddot{\textbf{r}}'(t) \bigr\rangle=\dot{\textbf{r}}'(t) = \frac{1}{||\dot{\textbf{r}}(t)||^2} \left(\begin{array}{c} \dot{r_1}(t) \\ \dot{r_2}(t) \end{array}\right) \cdot \frac{d}{dt}\left(\begin{array}{c} \dot{r_1}(t) \\ \dot{r_2}(t) \end{array}\right)= \frac{1}{||\dot{\textbf{r}}(t)||^2} \Bigl( \dot{r_1}(t) \frac{d}{dt} \dot{r_1}(t) + \dot{r_2}(t) \frac{d}{dt} \dot{r_2}(t) \Bigr)=\frac{1}{||\dot{\textbf{r}}(t)||^2} \Bigl( \dot{r_1}(t) \ddot{r_1}(t) + \dot{r_2}(t) \ddot{r_2}(t) \Bigr)##

Unfortunately, I can't get any further. The following must apply ##\bigl\langle \dot{\textbf{r}}'(t) ,\ddot{\textbf{r}}'(t) \bigr\rangle=0## so that this is the case, the expression ##\dot{r_1}(t) \ddot{r_1}(t) + \dot{r_2}(t) \ddot{r_2}(t)=0##, but why is this the case?

My derivation is probably already wrong, but I don't know how else to solve the problem.