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Instantaneous rate of change [Trig]

  1. Jan 18, 2012 #1
    1. The problem statement, all variables and given/known data
    For the function f(x) = 3sin 2x, find the instantaneous rate of change at point A [itex](\frac{\pi}{6}, f(\frac{\pi}{6}))[/itex]


    2. Relevant equations
    iroc = instantaneous rate of change
    [tex]iroc= \lim_{h\to 0} \frac{f(a+h) - f(a)}{h}[/tex]

    3. The attempt at a solution
    iroc = instantaneous rate of change
    [tex]iroc= \lim_{h\to 0} \frac{(3sin2(\frac{\pi}{6}+0.001)) - (3sin2(\frac{\pi}{6})}{0.001}[/tex]
    [tex]iroc= \lim_{h\to 0} \frac{(3sin(\frac{\pi}{3}+0.002)) - (3sin(\frac{\pi}{3}))}{0.001}[/tex]
    [tex]iroc= \lim_{h\to 0} \frac{(3(\frac{\sqrt3}{2}+0.002)) - (3(\frac{\sqrt{3}}{2}))}{0.001}[/tex]
    [tex]iroc= \lim_{h\to 0} \frac{(3(\frac{\sqrt3 + 0.004}{2})) - (\frac{3\cdot\sqrt{3}}{2}))}{0.001}[/tex]
    [tex]iroc= \lim_{h\to 0} \frac{(\frac{3\sqrt3 + 0.012}{2})) - (\frac{3\sqrt{3}}{2}))}{0.001}[/tex]
    [tex]iroc = 6[/tex]

    Is this answer correct?
     
  2. jcsd
  3. Jan 18, 2012 #2
    You have ignored the limit, if you want to do it that way you have to vary h, so start with h=0.001 and then go down to h=0.000000001 for instance. Then once you have collected all the results you can see what it limits to, but for here just try a really small h. And you also seem to be off by a factor of 1/2, you have to take the sin of everything in the brackets.
     
    Last edited: Jan 18, 2012
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