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Homework Help: Finding the common tangent?

  1. Feb 15, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi guys, need some help finding the common tangent of two curves.
    Two curves y = 3X3+6X2+6X+3 and y = 3X3+6X+3 touch each other. Find the common tangent.


    2. Relevant equations

    y = 3X3+6X2+6X+3 and y = 3X3+6X+3

    3. The attempt at a solution
    well i made the two equations equal each other and then i found the X values. After that i subbed the values in the the original equations to get my y values. My points ended up being
    (0,3) and (-1,0)

    is the correct? what do i do after this?
    First time with calculus so not sure what to do.
     
  2. jcsd
  3. Feb 15, 2010 #2

    tiny-tim

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    Hi cruisx! :smile:
    (0,3) yes, but how did you get (-1,0) ? :confused:

    anyway, now find the tangent at (0,3). :smile:
     
  4. Feb 15, 2010 #3
    well...um i forget how i got (-1,0) so i only needed to find (0,3)? and then after i find the tangent i am done? I think i got -1,0 because my x values were -1 and 0 so i subbed -1 in to one of my equations to get (-1,0). So i should use (0,3) to find the common tangent?
     
  5. Feb 15, 2010 #4

    tiny-tim

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    Yeah, I guessed that … but how did you get x = -1?

    (I got 6X2 = 0)
    Yes. :smile:
     
  6. Feb 15, 2010 #5
    Hi, so it seems that our next unit is derivatives so i am guessing i will have to find the tangent differently? can someone tell me how to find the tangent using (0,3) because it seems that it has to lead to a derivative or what ever that is. Will find out tomorrow.
     
  7. Feb 16, 2010 #6

    tiny-tim

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    Hi cruisx! :smile:

    (just got up :zzz: …)
    hmm … without derivatives? :redface:

    ok … try this …

    you know the tangent (to either curve) is a straight line through (0,3),

    so it has to be y = Ax + 3, for some value of A.

    now try a tiny-teeny-weeny value of x (ie very close to 0) in the equation for the curve, and see what for value of A it looks most like y = Ax + 3. :smile:

    (if you prefer, change the cooridnate to y' = y - 3, so the curves meet at the origin)
     
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