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Find k in AP Calculus BC problem

  1. Mar 20, 2016 #1
    1. The problem statement, all variables and given/known data
    Given path (x,y) described by dx/dt = 1/(t+1), dy/dt = k*exp(kt), constant k > 0. At t = 2 the tangent is parallel to y = 4x + 3. Find k. The given solution: approx. 0.495 .

    2. Relevant equations
    (dy/dt)/(dx/dt) = dy/dx = slope of tangent of path

    3. The attempt at a solution
    At t=2, (dy/dt)/(dx/dt) = (2e^2k)/(1/3) = 6e^2k = 4 (from y=4x+3), giving k=0.5*ln(2/3), or approx -0.2.
  2. jcsd
  3. Mar 20, 2016 #2


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    At t = 2, dy/dt ≠ 2e2k. Remember, dy/dt = k*exp(kt)
  4. Mar 20, 2016 #3
    Oops. Thanks very much, SteamKing.
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