243.13.01.18 motion of a particle

[karush]

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

$\begin{align*} \displaystyle
(1) r(t)&=t^8i+t^5j\\
(2) r(t)&=\cos(8t)i+\sin(2t)j\\
(3) r(t)&=ti+t^3j\\
(4) r(t)&=\cos{(10t)}i+\sin(10t)j\\
\end{align*}$I presume this would mean a constant value?

 

[MarkFL]

How do we know if two vectors are orthogonal?

 

[karush]

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

 

[MarkFL]

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?

 

[HallsofIvy]

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?

 

[MarkFL]

$\textit{In the notation of the dot product,}\\$
$\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):

\(\displaystyle a_1b_1+a_2b_2=0\)

 

[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*}

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*}

 

[MarkFL]

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?

 

[karush]

\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

- - - Updated - - -

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

 

[MarkFL]

I would just set:

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

Using the dot-product formula:

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

 

[HallsofIvy]

Before you do all that, since the cosine of 90 degrees is 0, and a fraction is 0 if and only if the numerator is 0, the point we have been trying to make is that two vectors are orthogonal if and only if their dot product is 0!

 

[karush]

$\cos{(10t)}=0$
$\displaystyle 10t=\frac{\pi}{2}$
$\displaystyle t=\frac{\pi}{20}$

\begin{align*}\displaystyle
\theta&=\cos^{-1}\left[\frac{u\cdot v}{|u||v|} \right]\\
&=\cos^{-1}\left[
\frac{(\cos{\left(\frac{\pi}{2}\right)}\textbf{i}+\sin(\left(\frac{\pi}{2}\right))\textbf{j}
\cdot (-10\sin(\left(\frac{\pi}{2}\right))\textbf{i}+10\cos(\left(\frac{\pi}{2}\right))\textbf{j})}
{|\cos{(\left(\frac{\pi}{2}\right))}\textbf{i}+\sin(\left(\frac{\pi}{2}\right))\textbf{j}||-10\sin(\left(\frac{\pi}{2}\right))\textbf{i}+10\cos(\left(\frac{\pi}{2}\right))\textbf{j}|} \right]_{\displaystyle t=\frac{\pi}{20}} \\
&=\cos^{-1}\left[
\frac{(0\textbf{i}+1\textbf{j})
\cdot (-10(1)\textbf{i}+10(0)\textbf{j})}
{|0\textbf{i}+1)\textbf{j}||-10(1)\textbf{i}+10(0)\textbf{j}|} \right]\\
&=\cos^{-1}\left[
\frac{0}
{|1||200|} \right]\\
&=0^o
\end{align*}

suggestions?

 

[MarkFL]

$\cos{(10t)}=0$
$\displaystyle 10t=\frac{\pi}{2}$
$\displaystyle t=\frac{\pi}{20}$

We're not concerned with the argument of the trig functions describing the position, velocity and acceleration being any particular value. It is the angle between the velocity and acceleration vectors we are conerned about, specifically that it is always \(\displaystyle \frac{\pi}{2}\) which will be the case when the dot product of the velocity and acceleration is zero.

\begin{align*}\displaystyle
\theta&=\cos^{-1}\left[\frac{u\cdot v}{|u||v|} \right]\\
&=\cos^{-1}\left[
\frac{(\cos{\left(\frac{\pi}{2}\right)}\textbf{i}+\sin(\left(\frac{\pi}{2}\right))\textbf{j}
\cdot (-10\sin(\left(\frac{\pi}{2}\right))\textbf{i}+10\cos(\left(\frac{\pi}{2}\right))\textbf{j})}
{|\cos{(\left(\frac{\pi}{2}\right))}\textbf{i}+\sin(\left(\frac{\pi}{2}\right))\textbf{j}||-10\sin(\left(\frac{\pi}{2}\right))\textbf{i}+10\cos(\left(\frac{\pi}{2}\right))\textbf{j}|} \right]_{\displaystyle t=\frac{\pi}{20}} \\
&=\cos^{-1}\left[
\frac{(0\textbf{i}+1\textbf{j})
\cdot (-10(1)\textbf{i}+10(0)\textbf{j})}
{|0\textbf{i}+1)\textbf{j}||-10(1)\textbf{i}+10(0)\textbf{j}|} \right]\\
&=\cos^{-1}\left[
\frac{0}
{|1||200|} \right]
$=0^o
\end{align*}

suggestions?

If testing option (4), I would write (using differentiation):

\(\displaystyle \vec{v}(t)=-10\sin(10t)\hat{\imath}+10\cos(10t)\hat{\jmath}\)

\(\displaystyle \vec{a}(t)=-100\cos(10t)\hat{\imath}-100\sin(10t)\hat{\jmath}\)

And so:

\(\displaystyle \vec{v}\cdot\vec{a}=1000\sin(10t)\cos(10t)-1000\sin(10t)\cos(10t)=0\)

Since this dot product is zero for all values of $t$, we know they are always orthogonal. If I was going to work this problem, I would want to derive the type of motion that results from setting the velocity and accleration orthogonal to one another.

Spoiler

Let:

\(\displaystyle \vec{r}(t)=x(t)\hat{\imath}+y(t)\hat{\jmath}\)

Hence:

\(\displaystyle \vec{v}(t)=\vec{r'}(t)=x'(t)\hat{\imath}+y'(t)\hat{\jmath}\)

\(\displaystyle \vec{a}(t)=\vec{v'}(t)=\vec{r''}(t)=x''(t)\hat{\imath}+y''(t)\hat{\jmath}\)

Now, we require:

\(\displaystyle \vec{v}(t)\cdot\vec{a}(t)=0\)

\(\displaystyle x'(t)x''(t)+y'(t)y''(t)=0\)

Integrating w.r.t $t$, there results:

\(\displaystyle \left[x'(t)\right]^2+\left[y'(t)\right]^2=C\)

This means the magnitude of the velocity (the speed) is constant, and in order for the acceleration to be non-zero, the direction of the velocity must be changing over time. If we interpret the above geometrically, we could say that $x'(t)$ and $y'(t)$ are the legs of a right triangle, allowing us to state (where $af'(t)=\sqrt{C}\implies f(t)=bt+c$) :

\(\displaystyle x'(t)=\pm af'(t)\sin(f(t))=\pm ab\sin(bt+c)\implies x(t)=\pm a\cos(bt+c)+x_0\)

\(\displaystyle y'(t)=\pm af'(t)\cos(f(t))=\pm ab\cos(bt+c)\implies y(t)=\pm a\sin(bt+c)+y_0\)

 

[karush]

Much Mahalo
Very Helpful

 

[HOI]

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

$\begin{align*} \displaystyle
(1) r(t)&=t^8i+t^5j\\
(2) r(t)&=\cos(8t)i+\sin(2t)j\\
(3) r(t)&=ti+t^3j\\
(4) r(t)&=\cos{(10t)}i+\sin(10t)j\\
\end{align*}$I presume this would mean a constant value?

(1) $r(t)= t^8i+ t^5j$
so $v(t)= r'(t)= 8t^7i+ 5t^4j$ and $a(t)= v'(t)= 42t^6i+ 20t^3j$
$v(t)\cdot a(t)= 336t^{13}i+ 140t^7j$

(2) $r(t)&=\cos(8t)i+\sin(2t)j$
so $v(t)= r'(t)= -8sin(8t)I+ 2cos(2t)j$ and $a(t)= v'(t)= -64 cos(8t)- 4 sin(2t)j$
$v(t)\cdot a(t)= 512 sin(8t)cos(8t)- 8sin(2t)cos(2t)$

(3) $r(t)&=ti+t^3j$
so $v(t)= r'(t)= i+ 3t^2j$ and $a(t)= v'(t)= 6tj$
$v(t)\cdot a(t)= 18t^4$

(4) $r(t)&=\cos{(10t)}i+\sin(10t)j$
so $v(t)= r'(t)= -10sin(10t)i+ 10cos(10t)j$ and $a(t)= v'(t)= -100cos(10t)i- 100sin(10t)$
$v(t)\cdot a(t)= 1000 sin(10t)cos(10t)- 1000 cos(10t)sin(10t)= 0$!

This last one is, in fact, motion about a circle with center at (0, 0) and radius 1.