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Math Help - vector valued functions

  1. #1
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    vector valued functions

    r(t) = <3t, (t-1)>

    find
    a) Dt[r(t) dot u(t)]
    b) Dt[u(t) - 2r(t)]
    c) Dt[||r(t)||], t > 0
    d) Dt[r(t) x u(t)]

    i don't understand what i am have to do for these problems
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by chris25 View Post
    r(t) = <3t, (t-1)>

    find
    a) Dt[r(t) dot u(t)]
    b) Dt[u(t) - 2r(t)]
    c) Dt[||r(t)||], t > 0
    d) Dt[r(t) x u(t)]

    i don't understand what i am have to do for these problems
    a) \frac{d}{dt}\bigg[\vec{r}(t)\cdot \hat{u}(t)\bigg]

    First of all, you're given \vec{r}(t)=\bigg<3t,t-1\bigg>

    The unit vector \hat{u}(t) is defined as \hat{u}(t)=\frac{\vec{r}(t)}{\Vert\vec{r}(t)\Vert}.

    Find \Vert\vec{r}(t)\Vert:

    \Vert\vec{r}(t)\Vert=\sqrt{(3t)^2+(t-1)^2}=\sqrt{9t^2+t^2-2t+1}=\sqrt{10t^2-2t+1}.

    Thus,
    \hat{u}(t)=\frac{\bigg<3t,(t-1)\bigg>}{\sqrt{10t^2-2t+1}}=\bigg<\frac{3t}{\sqrt{10t^2-2t+1}},\frac{t-1}{\sqrt{10t^2-2t+1}}\bigg>

    Thus,

    \vec{r}(t)\cdot\hat{u}(t)=\frac{9t^2}{\sqrt{10t^2-2t+1}}+\frac{(t-1)^2}{\sqrt{10t^2-2t+1}}=\frac{10t^2-2t+1}{\sqrt{10t^2-2t+1}}=\sqrt{10t^2-2t+1}

    Now find \frac{d}{dt}\bigg[\vec{r}(t)\cdot\hat{u}(t)\bigg]

    \frac{d}{dt}\bigg[\vec{r}(t)\cdot\hat{u}(t)\bigg]=\frac{1}{2}\bigg[10t^2-2t+1\bigg]^{-\frac{1}{2}}\cdot\bigg(20t^2-2\bigg)=\color{red}\boxed{\frac{10t^2-1}{\sqrt{10t^2-2t+1}}}

    Hope this helps. Try to do the others.

    --Chris
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by chris25 View Post
    r(t) = <3t, (t-1)>

    find
    a) Dt[r(t) dot u(t)]
    b) Dt[u(t) - 2r(t)]
    c) Dt[||r(t)||], t > 0
    d) Dt[r(t) x u(t)]

    i don't understand what i am have to do for these problems
    For the second one as Chris pointed out

    \hat{u}(t)=\frac{\left\langle{3t,(t-1)}\right\rangle}{\sqrt{10t^2-2t+1}}=\left\langle\frac{3t}{\sqrt{10t^2-2t+1}},\frac{t-1}{\sqrt{10t^2-2t+1}}\right\rangle

    So \bold{\hat{u}(t)}-2\bold{r(t)} =\left\langle\frac{3t}{\sqrt{10t^2-2t+1}}-6t,\frac{t-1}{\sqrt{10t^2-2t+1}}-2t-2\right\rangle

    So \frac{d}{dt}\bigg[\bold{\hat{u}(t)}-2\bold{r(t)}\bigg]=\left\langle\frac{-3(t-1)}{(10t^2-2t+1)^{\frac{3}{2}}}-6,\frac{9t}{(10t^2-2t+1)^{\frac{3}{2}}}-2\right\rangle

    Next we have

    ||\bold{r(t)}||=\sqrt{(3t)^2+(t-1)^2}=\sqrt{10t^2-2t+1}

    so

    \frac{d}{dt}\bigg[\bold{r(t)}\bigg]=\frac{10t-1}{\sqrt{10t^2-2t+1}}

    Lastly we have

    \frac{d}{dt}\bigg[\bold{\hat{u}(t)}\text{ x }\bold{r(t)}\bigg]=\bold{\hat{u}(t)}'\text{ x }\bold{r(t)}+\bold{\hat{u}(t)}\text{ x }\bold{r(t)}

    Which actually doesnt make sense, because you have a vector in \mathbb{R}^2 and the cross product is only defined in \mathbb{R}^3


    Off topic , how do you represent cross product, the representation I used isn't very nice looking. I tried \cross, but to no avail
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Off topic , how do you represent cross product, the representation I used isn't very nice looking. I tried \cross, but to no avail

    Use \times :

    i.e. \vec{r}(t)\times\hat{u}(t)
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