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Homework Help: Determining a smooth motion when given a function

  1. Sep 4, 2010 #1
    Show from Eq. 1.1 that the below function is smooth at t = 1 and at t = 2. Is it smooth at any 1 < t < 2?

    x(t) =

    1.0 + 2.0 t 0 ≤ t ≤ 1
    3 + 4(t − 1) 1 ≤ t ≤ 2
    7 + 3(t − 2) 2 ≤ t

    for equation 1.1 my book gives me:

    dt→0 [x(t + dt) − x(t) ]/dt= 0

    This problem asks me to show that the motion is smooth, but to me it seems that it would not be smooth at points 1 and 2. the way i understand it, for a motion to be smooth all of the derivatives of x(t) must exist, but if i take the derivatives:

    2 0 ≤ t ≤ 1
    4 1 ≤ t ≤ 2
    3 2 ≤ t

    the derivatives at points 1 and 2 conflict, so i dont understand how this meets the requirements of a smooth function. i must be missing something obvious.

    Also i was confused by the formula 1.1 in the book, it tell me to use this formula to show the motion is smooth, but then next to the formula in the book it states the formula as being one used to show a function is continuous.

    Thank you in advance to anyone who responds.
  2. jcsd
  3. Sep 6, 2010 #2


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    Homework Helper

    Curious question; probably should be over on one of the math forums.
    What exactly is the definition of smooth? I graphed the function from 0 to 3 and it is continuous, but the first derivative does not exist at 1 and at 2 because the value of eqn 1.1 is different when you approach from the left or the right. We say the whole limit does not exist if the left limit is not equal to the right limit.
  4. Sep 6, 2010 #3
    thanks for the reply, the definition of a smooth motion, as given in my book, is : for a motion to be smooth all of the derivatives of x(t) must exist, but if i take the derivatives.

    my intuition was that this function is not smooth, because points one and two have trouble with derivatives, any ideas? Again, my book asks me to show the motion as smooth, but i dont understand how it could be by the definition.
  5. Sep 6, 2010 #4


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    Definitely not smooth at 1 and 2. You can see that on the graph - the slope changes suddenly at those points, so no value can be assigned to the slope or first derivative at t = 1 or t = 2. Looks like the question got mixed up.
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