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Homework Help: Two Questions - calculus

  1. Jan 25, 2006 #1
    First question:
    This first question is kind of weird. I'm not even sure where to go with it. If anyone has a hint, that would be awesome.

    From [itex] \vec A \times \vec B = -\vec B \times \vec A [/itex] deduce [itex] \vec A \times \vec A = 0 [/itex]

    Can it be as simple as:
    let [tex] \vec B = \vec A_0 | \vec A_0 = \vec A [/tex]
    thus: [tex] \vec A \times \vec A_0 = -\vec A_0 \times \vec A = 0 [/tex]

    Second question:
    Find the minimum and maximum speed if [itex] x=t+\cos t [/itex], [itex]y=t-\sin t [/itex].

    Please allow me to take advantage of the inner space operator for sake of ease while writing the vectors :)

    [tex] \vec x = <t+\cos t,t-\sin t> [/tex]
    [tex] \vec v = <1-\sin t, 1-cos t> [/tex]

    So speed is computed as: [itex] |\vec v| [/itex]. Therefore the largest speed values that can occur are when: [itex] \vec v = <1,2> or <2,1> [/itex] and the lowest speed values that can occur are when [itex] \vec v = <1,0> or <0,1> [/itex].

    Is this reasoning even correct with this problem?
  2. jcsd
  3. Jan 25, 2006 #2


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    Homework Helper

    For the first one, I'm not sure what A0 is, but all you have to do is let B=A in the first identity. For the second, you might want to be a little more rigorous. You can find the extrema of the speed by:

    [tex]\frac{d}{dt} |\vec v|^2 = 0[/tex]

    And use:

    [tex]\frac{d}{dt} |\vec v|^2 = \frac{d}{dt} (\vec v\cdot \vec v)[/tex]

    Which will lead you to the equation:

    [tex]\vec v \cdot \vec a = 0[/tex]
    Last edited: Jan 25, 2006
  4. Jan 25, 2006 #3
    With the [itex] \vec A_0 = \vec A [/itex] I was just trying to show that I was plugging [itex] \vec A [/itex] into the expression. It really wasn't necessary and actually more confusing (I left this out of the homework).

    For the second one. Cool :smile: Thank you. That's what I wasn't doing. I needed to find the extrema of the speed not of the velocity vector. I appreciate it.
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