# 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?  Reply With Quote

2.

3.  Reply With Quote

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

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?  Reply With Quote

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?  Reply With Quote

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):

$\displaystyle a_1b_1+a_2b_2=0$  Reply With Quote

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*}  Reply With Quote

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?  Reply With Quote

\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?  Reply With Quote

11. I would just set:

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

Using the dot-product formula:

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

displaystyle, motion, rt&=cos8ti+sin2tj, rt&=ti+t^3j, rt&=t^8i+t^5j #### Posting Permissions 