# Thread: 243.13.01.18 motion of a particle

1. $$\tiny{243.13.01.18}$$
$\textsf{The following equations each describe the motion of a particle.}\\$
$\textsf{For which path is the particle's velocity vector always orthogonal to its acceleration vector?}$

\begin{align*} \displaystyle (1) r(t)&=t^8i+t^5j\\ (2) r(t)&=\cos(8t)i+\sin(2t)j\\ (3) r(t)&=ti+t^3j\\ (4) r(t)&=\cos{(10t)}i+\sin(10t)j\\ \end{align*}

I presume this would mean a constant value?

2.

3. How do we know if two vectors are orthogonal?

I would suppose perpendicular to each other
But how would that be determined from the selection.

5. Originally Posted by karush
I would suppose perpendicular to each other
But how would that be determined from the selection.
Yes, perpendicular, normal, orthogonal all mean the same thing. Is there a vector formula involving the angle between two vectors? If two vectors in the plane are orthogonal, what is the projection of one onto the other?

6. More simply, if two vectors are orthogonal, what does that say about the dot product of one vector with the other? With velocity vector t2i+ t8j, the acceleration vector is 2ti+ 8t7j. Now, how do you determine if they are orthogonal?

7. Originally Posted by karush
$\textit{In the notation of the dot product,}\\$
$\textit{the angle between two vectors$u$and$v$is:}$
\begin{align*}\displaystyle
\theta&=\cos^{-1}\left[\frac{u\cdot v}{|u||v|} \right]
\end{align*}
So then if $u=t^2i+ t^8j$ and $v=2ti+ 8t7j$ then
\begin{align*}\displaystyle
\theta&=\cos^{-1}\left[\frac{(t^2i+ t^8j)\cdot (2ti+ 8t^7j)}{|t^2i+ t^8j||2ti+ 8t^7j|} \right]
\end{align*}
We know:

$\displaystyle \vec{a}\cdot\vec{b}=\left|\vec{a}\right|\left|\vec{b}\right|\cos(\theta)$

If $\displaystyle \theta=\frac{\pi}{2}$, then:

$\displaystyle \vec{a}\cdot\vec{b}=0$

Or (in the case of two-dimensional vectors):

$\displaystyle a_1b_1+a_2b_2=0$

The book answer to this was (4)

\begin{align*} \displaystyle
(1) r(t)&=t^8i+t^5j\\
(2) r(t)&=\cos(8t)i+\sin(2t)j\\
(3) r(t)&=ti+t^3j\\
(4) r(t)&=\cos{(10t)}i+\sin(10t)j\\
\end{align*}

so
\begin{align*} \displaystyle
r(t)&=\cos{(10t)}\textbf{i}+\sin(10t)\textbf{j}\\
r'(t)&=10\cos(10t)\textbf{j}-10\sin(10t)\textbf{i}
\end{align*}

9. Originally Posted by karush
The book answer to this was (4)

\begin{align*} \displaystyle
(1) r(t)&=t^8i+t^5j\\
(2) r(t)&=\cos(8t)i+\sin(2t)j\\
(3) r(t)&=ti+t^3j\\
(4) r(t)&=\cos{(10t)}i+\sin(10t)j\\
\end{align*}
Yes, that's what I got as well, but can you demonstrate this is the case?

\begin{align*}\displaystyle
\theta&=\cos^{-1}\left[\frac{u\cdot v}{|u||v|} \right]\\
&=\cos^{-1}\left[
\frac{(\cos{(10t)}\textbf{i}+\sin(10t)\textbf{j}\cdot (-10\sin(10t)\textbf{i}+10\cos(10t)\textbf{j})}
{|\cos{(10t)}\textbf{i}+\sin(10t)\textbf{j}||-10\sin(10t)\textbf{i}+10\cos(10t)\textbf{j}|}
\right]
\end{align*}
hopefully before I plow into it

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ok I don't see how you do a dot product on this?

11. I would just set:

$\displaystyle \vec{r}'(t)\cdot\vec{r}''(t)=0$

Using the dot-product formula:

$\displaystyle \vec{a}\cdot\vec{b}=\sum_{k=1}^{n}\left(a_kb_k\right)$