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Finding a connection track to two railroads (circle and parabola)

  1. Mar 27, 2013 #1
    There are two railroads, represented by the curves y=(1/2)x2+7, and x2+y2=1. I am supposed to find a straight line connection so that a atrain can get from one curve to the other.

    I have drawn a simple diagram to help me picture things:
    http://i50.tinypic.com/2saeccx.png
    where (a,b) and (c,d) will be the points at which the line will connect with each curve.

    I've found the derivatives to the curves to be y'=x and y'=-x/y respectively. I understand that these give me the slopes of the tangent lines for each curve, so would I be correct in reasoning that x=-x/y since the slopes should be the same?
     
    Last edited: Mar 27, 2013
  2. jcsd
  3. Mar 27, 2013 #2

    Mark44

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    Probably a typo, but the 2nd one should be y' = -x/y.
    Yes.

    Note that there is another path that works, from the left side of the circle to the right side of the parabola.
     
  4. Mar 27, 2013 #3
    Yes, my mistake!

    Right, that makes sense. Thanks! How do you reckon I should proceed from here? My mind is a total blank.
     
  5. Mar 27, 2013 #4

    Mark44

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    Actually, I think this is a bit more complicated than setting the derivatives fo the two curves equal. You also want the line between (a, b) on the parabola and (c, d) on the circle to have the same slope as both derivatives you found.

    The problem is that you could have two tangents on the two curves with the same slope, but they aren't necessarily connected, if you follow what I'm saying.
     
  6. Mar 27, 2013 #5
    That's exactly why I'm having trouble working out how to make sure that the tangent line passes through (a,b) and (c,d) :/
     
  7. Mar 27, 2013 #6

    SammyS

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    What's the equation of a line tangent to y = (1/2)x2 + 7 and passing through point (a, b) which is on the parabola?

    What's the equation of a line tangent to x2 + y2 = 1 and passing through point (c, d) which is on the circle?
     
  8. Mar 27, 2013 #7
    y'=x
    m=a
    y-y1=m(x-x1)
    y-b=(a)(x-a)

    y'=-x/y
    m=-a/b
    y-y1=m(x-x1)
    y-b=(-a/b)(x-a)

    Is this what you mean?
     
  9. Mar 27, 2013 #8

    SammyS

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    That's good for the line tangent to the parabola at (a, b). But you can additionally express b in terms of a, since b = (1/2)a2 + 7 .

    For the circle, x + y2 = 1, you should use a different point, such as (c, d).

    Then y-d = (-c/d)(x-c) .

    Furthermore, as you already mentioned above, the slopes of the two lines must be equal, so now given a point, (c, d) on the unit circle, you can find the value of a, and plug that into the equation of the line tangent to the parabola.
     
  10. Mar 27, 2013 #9
    I don't really follow this last part; how do I know which point on the circle (or likewise, on the parabola) I am supposed to choose?
     
  11. Mar 27, 2013 #10

    SammyS

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    What is the slope of the line y-b = (a)(x-a) ?

    What is the slope of the line y-d = (-c/d)(x-c) ?

    Equate those slopes.
     
  12. Mar 27, 2013 #11
    a=(-c/d), but I've had that since the original post, no?
     
  13. Mar 27, 2013 #12

    SammyS

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    Beyond that, both lines must have the same y-intercept. Right?

    There is some (maybe a lot of) algebra to do.
     
  14. Mar 28, 2013 #13

    SteamKing

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    There are actually four possible connections. You can have a and c both positive or both negative, and two possibilities where a is negative and c is positive or vice versa.
     
  15. Mar 28, 2013 #14
    I've mostly lost motivation in this problem. I don't understand the meaning of what I'm doing. :(
     
  16. Mar 28, 2013 #15

    Mark44

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    You're trying to find a line that:
    1. Joins the parabola and the circle, and
    2. Is tangent to both the parabola and the circle.
     
  17. Mar 28, 2013 #16

    SammyS

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    Unfortunately, in this new format for Physics Forums, there is no button to press which can provide the necessary motivation.

    Solve the following for y, and try to simplify the right hand side.

    [itex]\displaystyle y-d= \frac{-c}{d}(x-c)[/itex]

    See if that leads you anywhere. Whether it does or not, please give your result.
     
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