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Homework Help: Simple differentiatian of tan question

  1. Nov 30, 2008 #1
    1. The problem statement, all variables and given/known data
    What is the time rate of change of height after 5.0s?


    2. Relevant equations

    theta = 3t/(2t + 10)
    adjacent length = 1000

    Therefore: opposite length (height) h = 1000 tan (3t/(2t+10))


    3. The attempt at a solution

    I thought it would be simply finding the derivative of the second equation (dh/dt) and plugging in 5 for t.

    dh/dt = 1000 * ( sec(3t/(2t+10)) )^2 * [ (2t+10)*3 - 3t(2)/ (2t + 10)^2 ]
    dh/dt |t=5 = 1000/(cos.75)^2 * 3/40
    = 75.01...

    75 m/s

    Is this not correct?
     
  2. jcsd
  3. Nov 30, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Change your calculator from degrees to radians. Otherwise, it looks great.
     
  4. Nov 30, 2008 #3
    Thank you. :smile:
     
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