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Homework Help: Calculus with Parametrics problem

  1. Mar 29, 2005 #1
    Hey guys, I'm having difficulty understanding exactly what this problem is asking:

    The position vector of an object for [tex]t\geq0[/tex] is [tex]r(t) = <cos(3t),cos(2t)>[/tex].
    Find all possible values of [tex]\frac{dy}{dx}[/tex] at the point [tex](0,1/2)[/tex].

    Seriously, what does this mean?
    Last edited: Mar 29, 2005
  2. jcsd
  3. Mar 29, 2005 #2
    Well, you have [itex]y = \cos{2t}[/itex] and [itex]x = \cos{3t}[/itex]. You can use the chain rule

    [tex]\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx}[/tex]

    to find the derivative of [itex]y[/itex], then you'll need to solve [itex]\cos{2t}=\frac{1}{2}, \ \cos{3t}=0[/itex] to find the values of [itex]t[/itex] corresponding to the point [itex](0, 1/2)[/itex], substitute those into the derivative that you found (which should be in terms of [itex]t[/itex]) and find all its possible values.
  4. Mar 29, 2005 #3
    Thanks, here's what I did:

    [tex]\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2sin(2t)}{3sin(3t)}[/tex]


    [tex]cos(2t) = \frac{1}{2}[/tex] when [tex]t = \frac{\pi}{4}, \frac{3\pi}{4}[/tex], etc and

    [tex]cos(3t) = 0[/tex] when [tex]t = \frac{\pi}{6}, \frac{3\pi}{6}[/tex], etc, and

    [tex]\frac{2sin(2t)}{3sin(3t)}[/tex] has possible values: [tex]\frac{2}{3},\frac{-2}{3}[/tex].

    Is that right?
  5. Mar 29, 2005 #4


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    Valid "t" values are {t = (π/6), (5π/6), (7π/6), & (11π/6)}, for which:
    {dy/dx = (+√3)/3=(+0.577) :OR: (-√3)/3=(-0.577)}

  6. Mar 29, 2005 #5


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    The point is that you can't say "cos(2t)= 1/2)" for t= ...
    and "cos(3t)= 0 for t= ...". Each value of t must give both cos(2t)= 1/2 and sin(3t)= 0.
  7. Mar 30, 2005 #6
    Thanks guys. Got it :)
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