# 243.13.01.18 motion of a particle

• MHB
• karush
In summary, the particle's velocity vector will always be orthogonal to its acceleration vector for the path described by option (4): $r(t)=\cos{(10t)}i+\sin(10t)j$. This can be determined by calculating the dot product of the velocity and acceleration vectors, which will always equal zero, indicating that the vectors are always orthogonal.
karush
Gold Member
MHB
$$\tiny{243.13.01.18}$$
$\textsf{The following equations each describe the motion of a particle.}\\$
$\textsf{For which path is the particle's velocity vector always orthogonal to its acceleration vector?}$

\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?

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.

karush said:
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?

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?

karush said:
$\textit{In the notation of the dot product,}\\$
$\textit{the angle between two vectors$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*}

Last edited:
karush said:
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

- - - Updated - - -

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

I would just set:

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

Using the dot-product formula:

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

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!

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

\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?

Last edited:
karush said:
$\cos{(10t)}=0$
$\displaystyle 10t=\frac{\pi}{2}$
$\displaystyle t=\frac{\pi}{20}$

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.

karush said:
\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 oft$, 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. 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$af'(t)=\sqrt{C}\implies f(t)=bt+c$) : $$\displaystyle x'(t)=\pm af'(t)\sin(f(t))=\pm ab\sin(bt+c)\implies x(t)=\pm a\cos(bt+c)+x_0$$ $$\displaystyle y'(t)=\pm af'(t)\cos(f(t))=\pm ab\cos(bt+c)\implies y(t)=\pm a\sin(bt+c)+y_0$$ Much Mahalo Very Helpful karush said: $$\tiny{243.13.01.18}$$$\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^3jv(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)jv(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)= 6tjv(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.

## 1. What is the meaning of "243.13.01.18 motion of a particle"?

The term "243.13.01.18 motion of a particle" refers to the movement or displacement of a single particle in a specific direction and at a specific time, as indicated by the given numerical values. It could represent the position, velocity, or acceleration of the particle.

## 2. What factors affect the motion of a particle?

The motion of a particle can be affected by several factors, such as the forces acting on the particle, the particle's mass, the surface or medium it is moving through, and any external constraints or barriers.

## 3. How is the motion of a particle measured?

The motion of a particle can be measured using different methods, depending on the specific parameters being observed. For example, the position of a particle can be measured using a ruler or a coordinate system, while its velocity can be measured using a stopwatch and a known distance.

## 4. What is the difference between linear and circular motion of a particle?

Linear motion of a particle refers to the movement of a particle in a straight line, while circular motion refers to the movement of a particle along a curved path, such as a circle. Linear motion can be described using one-dimensional coordinates, while circular motion requires two-dimensional coordinates.

## 5. How can the motion of a particle be predicted?

The motion of a particle can be predicted using mathematical equations and principles, such as Newton's laws of motion. By understanding the forces acting on a particle and its initial conditions, it is possible to make predictions about its future motion.

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