243.13.01.19 For which path is the particle's speed constant

[karush]

$\tiny{243.13.01.19}$
$\textsf{The following equations each describe the motion of a particle.}$
$\textsf{ For which path is the particle's speed constant?}$
\begin{align*} \displaystyle
R_1(t)&= t^7\textbf{i}+t^4\textbf{j}\\
R_2(t)&= \cos(3t)\textbf{i}+\sin(8t)\textbf{j}\\
R_3(t)&= t\textbf{i}+t\textbf{j}\\
R_4(t)&= \cos(3t^2)\textbf{i}+\sin(3t^2)\textbf{j}\\
%\textit{speed constant on}&=\color{red}{Path(3)}
\end{align*}
$\textit{By observation I would quess $ \, R_3(t)= t\textbf{i}+t\textbf{j}$ so then: }$
\begin{align*} \displaystyle
R_3(t)&= t\textbf{i}+t\textbf{j}\\
R_3^\prime (t)=V_3(t)&=\textbf{i}+\textbf{j}
\end{align*}
$\textit{so then dot product}$
\begin{align*}\displaystyle
\theta&=\cos^{-1}\left[\frac{u\cdot v}{|u||v|} \right] \\
&=\cos^{-1}\left[\frac{(t\textbf{i}+t\textbf{j})\cdot(\textbf{i}+\textbf{j})}
{|t\textbf{i}+t\textbf{j}||\textbf{i}+\textbf{j}|} \right]\\
\end{align*}

kinda maybe!

 

[MarkFL]

Here, you just want to find those choices where the magnitude of the velocity vector is constant. :)

 

[karush]

Isn't the magnitude of the velocity vector $\sqrt{2}$ which is a constant, so we don't need the the dot product?

 

[MarkFL]

Isn't the magnitude of the velocity vector $\sqrt{2}$ which is a constant, so we don't need the the dot product?

Yes, we don't need the dot product here. Your choice is correct, but I can see at least one other choice for which the speed is constant...:)

 

[karush]

I would guess $R_2$
the absense of powers > 1

 

[MarkFL]

I would guess $R_2$
the absense of powers > 1

All you're looking for is a velocity vector whose magnitude is constant, and the second choice isn't one of those. :)

 

[karush]

Not sure!

 

[MarkFL]

I was mistaken...I "eyeballed" the 4th choice and saw the argument for the sine and cosine functions was the same, and mentally declared the velocity constant when in fact it varies as the magnitude of the parameter $t$ via the chain rule. Sorry for the hasty mistake. (Bandit)

 

[karush]

no prob
happens