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Interval with the leingth of 3

  1. Aug 28, 2004 #1
    Draw the function (I can do that) and deside an interval with the leingth of 3, there by f illustrates on an interval in the leingh of 25/4. (how??)

    (hope you can read it, I am from Denmark, so i'll try)
  2. jcsd
  3. Aug 28, 2004 #2
    I'm afraid I don't understand your question. f illustrates an interval? Maybe you could post the question in Danish instead.
    Last edited: Aug 28, 2004
  4. Aug 28, 2004 #3


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    You seem to be asking for an interval [a, a+3] (so that the interval has length 3) such that the length of the arc y= x2- 2x+ 2 is 25/4.

    Since y= x2- 2x+ 2, y'= 2x- 2 and ds= [tex]\sqrt{y'^2+1}dx[/tex] (if you don't know what that means, either you need to go back and review "arclength" or I've completely misunderstood your question!).

    This is ds= [tex]\sqrt{4x^2-8x+ 5}= \sqrt{4(x- 1)^2+1}[/tex] is 25/4

    To find the arc-length, we need find [tex]\integral_a^{a+3}\sqrt{4(x-1)^2+1}dx[/tex].
    Do that integral (I suggest the substitution u= 2(x-1) followed by a trig substitution) and then determine what a must be so that the arc-length is
  5. Aug 28, 2004 #4


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    Another, simpler, possiblity is that you just want an interval from x to x+3 such that the change in y is 25/3. Since y= x2- 2x+ 2= (x-1)2+1, the change in y between x and x+3 is ((x+2)2+ 1)- ((x-1)2+ 1)= x2+ 4x+ 5- x2- 2x+ 1= 2x+ 6= 25/4. Then 2x= 25/4- 6= 1/4 and x= 1/8.
  6. Aug 28, 2004 #5
    I guess if he can undestand post #3, it is probably this one he should solve.
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