
MHB Master
#1
November 27th, 2017,
22:21
$$\tiny{243.13.01.18}$$
$\textsf{The following equations each describe the motion of a particle.}\\$
$\textsf{For which path is the particle's velocity vector always orthogonal to its acceleration vector?}$
$\begin{align*} \displaystyle
(1) r(t)&=t^8i+t^5j\\
(2) r(t)&=\cos(8t)i+\sin(2t)j\\
(3) r(t)&=ti+t^3j\\
(4) r(t)&=\cos{(10t)}i+\sin(10t)j\\
\end{align*}$
I presume this would mean a constant value?

November 27th, 2017 22:21
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Pessimist Singularitarian
#2
November 27th, 2017,
22:34
How do we know if two vectors are orthogonal?

MHB Master
#3
November 28th, 2017,
05:07
Thread Author
I would suppose perpendicular to each other
But how would that be determined from the selection.

Pessimist Singularitarian
#4
November 28th, 2017,
05:24
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?

#5
November 28th, 2017,
09:37
More simply, if two vectors are orthogonal, what does that say about the dot product of one vector with the other? With velocity vector t^{2}i+ t^{8}j, the acceleration vector is 2ti+ 8t^{7}j. Now, how do you determine if they are orthogonal?

Pessimist Singularitarian
#6
November 28th, 2017,
15:11
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}{uv} \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^8j2ti+ 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 twodimensional vectors):
$ \displaystyle a_1b_1+a_2b_2=0$

MHB Master
#7
November 28th, 2017,
20:14
Thread Author
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*}
Last edited by karush; November 28th, 2017 at 20:47.

Pessimist Singularitarian
#8
November 28th, 2017,
20:19
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?

MHB Master
#9
November 28th, 2017,
21:36
Thread Author
\begin{align*}\displaystyle
\theta&=\cos^{1}\left[\frac{u\cdot v}{uv} \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
   Updated   
ok I don't see how you do a dot product on this?

Pessimist Singularitarian
#10
November 28th, 2017,
22:34
I would just set:
$ \displaystyle \vec{r}'(t)\cdot\vec{r}''(t)=0$
Using the dotproduct formula:
$ \displaystyle \vec{a}\cdot\vec{b}=\sum_{k=1}^{n}\left(a_kb_k\right)$