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  1. MHB Master
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    #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?

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    #2
    How do we know if two vectors are orthogonal?

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

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    #4
    Quote Originally Posted by karush View Post
    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?

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    #5
    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?

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    #6
    Quote Originally Posted by karush View Post
    $\textit{In the notation of the dot product,}\\$
    $\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$

  8. MHB Master
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    #7 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.

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    #8
    Quote Originally Posted by karush View Post
    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?

  10. MHB Master
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    #9 Thread Author
    \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?

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    #10
    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)$

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