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Parametrization question for my Intro. to Higher Math Class

  1. Apr 22, 2012 #1
    Two objects A and B are traveling in opposite direction on a straight line. At t=0 A and B are at positions P(A)=(-40, -20) and P(B)=(190, 980), respectively. If additionally, their paths are parameterized by directions V(A)=(3,5) and V(B)=(-24, -40), respectively. Then,

    a) find the point where these two object intersect.

    b) how long does it take these two objects to intersect?
  2. jcsd
  3. Apr 22, 2012 #2


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    Have you tried unwrapping the definitions?

    I'm not sure I understand your layout; what do P(A)=(-40,-20), and V(A)=(3,5) mean, given that movement is along a line?
  4. Apr 22, 2012 #3


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    hi vanitymdl! :wink:

    show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
  5. Apr 22, 2012 #4


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    I, for one, am not certain what you mean by "parameterized by directions V(A)=(3,5) and V(B)=(-24, -40)".

    I imagine you mean that A contains the point (-40, -20) and extends parallel to the vector 3i+ 5j but there are still and infinite number of parameterizations. The most "natural" would be to take t= 0 at point (-40, -20) and t=1 at (-40+ 3, -20+ 5)= (-37, -15). That would be given by x= -40+ 3t, y= -20+ 5t. But, as I said, there are an infinite number of parameterizations giving that same line.
  6. Apr 22, 2012 #5
    I attached a picture of the problem, it's number six.
    How should I even attempt to start this problem?

    Attached Files:

  7. Apr 22, 2012 #6
    I know that my parameters for A are:
    x = -40 + 3t
    y = -20 +5t

    and for B are:
    x = 190 -24t
    y = 980 -40t

    now how can I find where the two intersect?
  8. Apr 22, 2012 #7


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    O.K, so, at time t=0:

    A is at (-40,-20)

    B is at (190,980)

    At time t=1:

    A is at (-40+30,-20+5)=(-10,-15) , B is at ( 166,948),etc.

    Now, by intersecting we mean that A,B must have the same x- and y- coords.

    How do we figure out when the coords. are trhe same?
  9. Apr 22, 2012 #8
    I was plugging in different values of t from 1-12 but I noticed that the points weren't getting any nearer to eachother. Is there another way then trying to plug in different values for t
  10. Apr 22, 2012 #9


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    Well, how can you tell if/when A,B have the same x- and the same y- values? You know how their respective x-, y- values are defined. When do the curves y=t^2 and y=t meet? How can you tell?

    Sorry, I need to leave know.
  11. Apr 22, 2012 #10
    well if y=t^2 and y=t then they meet at the origin (0,0) but I don't see how that's going to help me?
  12. Apr 22, 2012 #11


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    Can you see how to formally figure out?

    What if we had y=t-1 and y=t^2-4 ?

    I don't mean to be obtuse about this; I am just trying to lead you to the answer and not just give it to you.

    You want :

    190-24t= ?

  13. Apr 22, 2012 #12
    No no I appreciate what you are trying to do, thank you.
    Well no wonder why I wasn't getting anything near it...

    but I got (45.375, -995) for that intersection
  14. Apr 22, 2012 #13


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    First, recognize that while x and y are coordinates in the plane, the "t" in each set of equations simply mark points on the line and are NOT necessarily the same in both sets. So that you don't confuse them, call the parameter in the second set "s" rather than "t".

    So we have x= -40+ 3t and y= -20+ 5t for one line and x= 190- 24s, y= 980- 40s for the other. They will intersect where the x and y values on one line are the same as on the other line: x= -40+ 3t= 190- 24s and y= -20+ 5t= 980- 40t, two equations to solve for t and s. Once you have found them, put either into the appropriate x, y equations to find the coordinates of the point of intersection.
  15. Apr 22, 2012 #14


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    It ultimately comes down to the definition of 'A,B intersecting'. Intersecting means they are in the same location in the plane. This means that , they will be in the exact same coordinates.
  16. Apr 23, 2012 #15
    okay okay I'm getting confused even more confused to how the find the intersection...From what i was doing the A and B never intersect?
  17. Apr 23, 2012 #16
    I've gone through t=100 and I'm still not seeing anything near an intersection. Its getting further and further apart
  18. Apr 23, 2012 #17
    Ah I finally got an intersection its approximately it's (-14.4444, 91.1111) can someone check that for me?
  19. Apr 24, 2012 #18


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    No, that isn't correct. The two equations (using t for one parameter and s for the other as I suggested before) are
    -40+ 3t= 190- 24s and -20+ 5t= 980- 40s which we can rewrite as
    24s+ 3t= 230 and 40s+ 5t= 20. If we multiply both sides of the first equation by 5 and the second by 3 we get
    120s+ 15t= 1150 and 120s+ 15t= 60.

    The left hand sides are the same but the right hand sides are not. The are NO values of s and t that make both equations true. As you were told before, these paths do NOT intersect.
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