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Posted by on Tuesday, November 23, 2010 at 8:57pm.

A certain clock has a minute hand that is 4 inches long and an hour hand is 3 inches long. How fast is the distance between the tips of the hands changing at 9:00?

  • calculus - , Tuesday, November 23, 2010 at 10:09pm

    let Ø be the angle between them

    the angular velocity of the minute hand = 2π/60 rad/min = π/30 rad/min
    the angular velicity of the hour hand = 2π/(12(60)) or π/360 rad/min

    so dØ/dt = (π/30 - π.360) rad/min
    = 11π/360 rad/min

    let the distance between the tips be x inches
    by Cosine Law
    x^2 = 16 + 9 - 2(4)(3)cosØ
    2x dx/dt = 24sinØ dØ/dt

    when the time is 9:00, Ø = 90° and x = 5
    10dx/dt = 24sin90° (11π/360)
    x = 24(1)(11π/360)/10
    = .2304 inches/min

  • calculus - , Thursday, March 22, 2012 at 1:24pm

    this is wrong

  • calculus - , Saturday, October 27, 2012 at 4:28pm

    This is right

  • calculus - , Thursday, October 17, 2013 at 7:37pm

    this is wrong, right answer is 4.4pi in/h

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