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[SOLVED] taking the dot product

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
\frac{(\dot{\mathbf{r}}\times\ddot{\mathbf{r}}) \times\dot{\mathbf{r}}}{\lvert\dot{\mathbf{r}}
\rvert\lvert\dot{\mathbf{r}}\times\ddot{\mathbf{r}}\rvert}
$$
How do I take that dot product of the expression of above with itself?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
$$
\frac{(\dot{\mathbf{r}}\times\ddot{\mathbf{r}}) \times\dot{\mathbf{r}}}{\lvert\dot{\mathbf{r}}
\rvert\lvert\dot{\mathbf{r}}\times\ddot{\mathbf{r}}\rvert}
$$
How do I take that dot product of the expression of above with itself?
Let's define $\mathbf a = \dot{\mathbf r},\ \mathbf b = \ddot{\mathbf r}$.

Then, if I understand you correctly, you're asking for:
$$((\mathbf a \times \mathbf b) \times \mathbf a)^2$$
According to the vector triple product we have:
$$(\mathbf a \times \mathbf b) \times \mathbf a = \mathbf b(\mathbf a \cdot \mathbf a) - \mathbf a (\mathbf a \cdot \mathbf b)$$
So:
$$((\mathbf a \times \mathbf b) \times \mathbf a)^2 = a^2b^2 - a^2 (\mathbf a \cdot \mathbf b)^2$$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Let's define $\mathbf a = \dot{\mathbf r},\ \mathbf b = \ddot{\mathbf r}$.

Then, if I understand you correctly, you're asking for:
$$((\mathbf a \times \mathbf b) \times \mathbf a)^2$$
According to the vector triple product we have:
$$(\mathbf a \times \mathbf b) \times \mathbf a = \mathbf b(\mathbf a \cdot \mathbf a) - \mathbf a (\mathbf a \cdot \mathbf b)$$
So:
$$((\mathbf a \times \mathbf b) \times \mathbf a)^2 = a^2b^2 - a^2 (\mathbf a \cdot \mathbf b)^2$$
I really need to get back
$$
\frac{(\dot{\mathbf{r}}\times\ddot{\mathbf{r}})
\cdot\dddot{\mathbf{r}}}{ \lvert\dot{\mathbf{r}}\times
\ddot{\mathbf{r}}\rvert^2}
$$
afterwards though. I don't see how I can arrange your solution to do the job.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
I really need to get back
$$
\frac{(\dot{\mathbf{r}}\times\ddot{\mathbf{r}})
\cdot\dddot{\mathbf{r}}}{ \lvert\dot{\mathbf{r}}\times
\ddot{\mathbf{r}}\rvert^2}
$$
afterwards though. I don't see how I can arrange your solution to do the job.
It appears I simply do not understand your question.
Perhaps you can clarify.
 

dwsmith

Well-known member
Feb 1, 2012
1,673

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780

dwsmith

Well-known member
Feb 1, 2012
1,673
I see that document also provides the proof for the last theorem...
Is there a step that you do not understand?
I am not using there method. I was just showing you intent.

How is MIT going from step 2 to 3?

 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
I am not using there method. I was just showing you intent.

How is MIT going from step 2 to 3?

What is MIT?

Assuming you mean the 2nd equality, they are using that:
\begin{aligned}
\mathbf T(s) &= \mathbf r'(s) \\
\mathbf T'(s) &= \kappa(s) \mathbf n(s) \\
\mathbf n(s) &= \frac{\mathbf r''(s)}{\kappa(s)} \\
\mathbf b(s) &= \mathbf T(s) \times \mathbf n(s) \\
\end{aligned}

Or if you meant the 3rd equality, it is explained on page 2 in your pdf.

Btw, can you please be more specific?
I dislike guessing what someone means.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
What is MIT?

Assuming you mean the 2nd equality, they are using that:
\begin{aligned}
\mathbf T(s) &= \mathbf r'(s) \\
\mathbf T'(s) &= \kappa(s) \mathbf n(s) \\
\mathbf n(s) &= \frac{\mathbf r''(s)}{\kappa(s)} \\
\mathbf b(s) &= \mathbf T(s) \times \mathbf n(s) \\
\end{aligned}

Or if you meant the 3rd equality, it is explained on page 2 in your pdf.

Btw, can you please be more specific?
I dislike guessing what someone means.
MIT = Massachusetts Institute of Technology.

As a note, I have shown that \(\frac{1}{\rho} = \frac{\lvert\dot{\mathbf{r}}\times \ddot{\mathbf{r}}\rvert} {\lvert\dot{\mathbf{r}}\rvert^3}\)

Here is what I worked out.
\begin{align}
\frac{1}{\tau} &= -\hat{\mathbf{n}}\cdot \frac{d\hat{\mathbf{b}}}{ds}\\
&= -\rho\frac{d^2\mathbf{r}}{ds^2}\cdot \left(\hat{\mathbf{u}}\times
\hat{\mathbf{n}}\right)_t
\end{align}
Now let's write \(\frac{d^2\mathbf{r}} {ds^2}\) as
\begin{align}
\frac{d^2\mathbf{r}}{ds^2} &= \frac{d^2\mathbf{r}} {dt^2}
\left(\frac{dt} {ds}\right)^2\\
&= \frac{1} {v^2}\ddot{\mathbf{r}}
\end{align}
and substitute back in to equation above.
\begin{align}
&= -\frac{\rho} {v^2}\ddot{\mathbf{r}} \cdot \frac{\rho} {v^3}\left(\dot{\mathbf{r}}\times
\ddot{\mathbf{r}}\right)_t\\
&= \frac{\lvert\dot{\mathbf{r}}\rvert} {\lvert\dot{\mathbf{r}}\times \ddot{\mathbf{r}}\rvert}\dddot{\mathbf{r}} \cdot \frac{1}{\lvert\dot{\mathbf{r}}\times \ddot{\mathbf{r}}\rvert} (\dot{\mathbf{r}}\times\ddot{\mathbf{r}})\\
&= ???\\
&= \frac{(\dot{\mathbf{r}} \times\ddot{\mathbf{r}}) \cdot
\dddot{\mathbf{r}}}
{\lvert \dot{\mathbf{r}} \times\ddot{\mathbf{r}} \rvert^2}
\end{align}

So I still have a \(\lvert\dot{\mathbf{r}}\rvert\)
 
Last edited: