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Related rates

  1. Nov 12, 2004 #1
    a women is walking at a rate of 6 ft/sec across a bridge 40 ft above the water, a man in a boat is immediately below the woman on the bridge , at right angles to the bridge and at the rate of 4 ft /sec. how fast is the man and woman seprating at the end of 10 seconds

    what i have figured out is the 2 deminationally not including the 40 ft
    dx/dt = 6ft/sec dy/dt = 4 ft/sec dz/dt = ?
    z^2 = x^2 + y^2 z = sq root 52 = 7.21

    using implicate differenating dz/dt = 1/z (x(dx/dt) + y(dy/dt))

    dz/dt = 1/7.21(4(10) +6(10)
    = 100 / 7.21
    = 13.87 ft/sec

    I can drw it diagram but not sure what to do for the 40 ft

    thanks joe
     
    Last edited: Nov 13, 2004
  2. jcsd
  3. Nov 13, 2004 #2

    arildno

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    The rate of change of the distanceis found as follows:
    The position of the woman relative to the man in the boat is:
    [tex]\vec{r}(t)=6t\vec{i}+4t\vec{j}+40\vec{k}[/tex]
    The distance between them r(t), is given by the Pythagorean theorem:
    [tex]r(t)=\sqrt{36t^{2}+16t^{2}+1600}[/tex]
    The separation velocity is now given as [tex]\frac{dr}{dt}[/tex]
     
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