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Parametrization question for my Intro. to Higher Math Class.. |
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| Apr22-12, 02:13 PM | #1 |
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Parametrization question for my Intro. to Higher Math Class..
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? |
| Apr22-12, 02:41 PM | #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? |
| Apr22-12, 02:42 PM | #3 |
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hi vanitymdl!
![]() show us what you've tried, and where you're stuck, and then we'll know how to help!
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| Apr22-12, 02:51 PM | #4 |
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Parametrization question for my Intro. to Higher Math Class..
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. |
| Apr22-12, 03:04 PM | #5 |
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I attached a picture of the problem, it's number six.
How should I even attempt to start this problem? |
| Apr22-12, 05:02 PM | #6 |
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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? |
| Apr22-12, 05:23 PM | #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? |
| Apr22-12, 05:25 PM | #8 |
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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
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| Apr22-12, 05:45 PM | #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. |
| Apr22-12, 05:49 PM | #10 |
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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?
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| Apr22-12, 05:53 PM | #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= ? 980-40t=? |
| Apr22-12, 06:05 PM | #12 |
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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 |
| Apr22-12, 09:00 PM | #13 |
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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. |
| Apr22-12, 09:34 PM | #14 |
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| Apr23-12, 08:38 PM | #15 |
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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?
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| Apr23-12, 09:42 PM | #16 |
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I've gone through t=100 and I'm still not seeing anything near an intersection. Its getting further and further apart
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| Apr23-12, 09:57 PM | #17 |
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Ah I finally got an intersection its approximately it's (-14.4444, 91.1111) can someone check that for me?
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