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- Thread starter dwsmith
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- Mar 5, 2012

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Let's define $\mathbf a = \dot{\mathbf r},\ \mathbf b = \ddot{\mathbf r}$.$$

\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?

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:

$$((\mathbf a \times \mathbf b) \times \mathbf a)^2 = a^2b^2 - a^2 (\mathbf a \cdot \mathbf b)^2$$

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I really need to get backLet'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:

$$((\mathbf a \times \mathbf b) \times \mathbf a)^2 = a^2b^2 - a^2 (\mathbf a \cdot \mathbf b)^2$$

$$

\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.

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It appears I simply do not understand your question.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.

Perhaps you can clarify.

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- #5

It appears I simply do not understand your question.

Perhaps you can clarify.

It is related to proving the last theorem here:

http://www.math.wisc.edu/~seeger/234/frenet.pdf

which has to do with this questions here:

http://mathhelpboards.com/advanced-applied-mathematics-16/frenet-equation-torsion-6229.html

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I see that document also provides the proof for the last theorem...It is related to proving the last theorem here:

http://www.math.wisc.edu/~seeger/234/frenet.pdf

which has to do with this questions here:

http://mathhelpboards.com/advanced-applied-mathematics-16/frenet-equation-torsion-6229.html

Is there a step that you do not understand?

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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.

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- #9

MIT = Massachusetts Institute of Technology.

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.

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 \(\lvert\dot{\mathbf{r}}\rvert\)

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