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A slingshot rotates counterclockwise on the circle x^2+y^2=9

  1. Dec 7, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose a slingshot rotates clockwise along the circle x^2 + y^2 = 9 and the rock is released at the point (2.99,0.77). If the rock travels 200 feet, where does it land?


    2. Relevant equations



    3. The attempt at a solution
    I think you might have to find the tangent at the point and use the distance formula. Or something alone those lines. If that thought is wrong, I am completely lost :| lol
     
  2. jcsd
  3. Dec 7, 2011 #2

    Char. Limit

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    Well, that sounds about right, but remember that even though you need the tangent for the circle at that point, the rock's starting position is NOT on the circle. As long as you keep that in mind, though, you should be fine. First take the tangent to figure out a direction, then figure out what the vector is for 200 feet in that direction.
     
  4. Dec 7, 2011 #3
    First, I mean't 300 feet. Sorry. But anyways, I haven't learned anything about vectors yet so I don't know how to use your method :\
     
  5. Dec 7, 2011 #4

    Char. Limit

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    First thing's first, figure out the tangent direction. That's just finding the slope at that point, and you should be able to do that easily. Then you just need to travel 300 feet along that line. You'll probably get a couple of equations like this:

    [tex]x^2 + y^2 = 300[/tex]

    [tex]\frac{y}{x} = s[/tex]

    Where s is whatever the slope is. The answer to those two equations will tell you where your new point is, from the origin at least. Then you just need to factor in that you didn't start at (0,0), but at (2.99, 0.77). Sorry about the vectors... they aren't strictly necessary. Is there any specific part that you don't understand?
     
  6. Dec 7, 2011 #5
    Ok. I messed up again. The point is (2.9,0.77) sorry, It's good now though. I think I understand everything except how to factor in that we started at (0,0) and not( 2.9,0.77). I found the tangent line at (2.9,0.77): y=-3.7755x + 11.02595. Now how do I figure where the new point is?
     
  7. Dec 7, 2011 #6

    Char. Limit

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    All right, you've got the slope (-3.7755). Now, you know that the point has to satisfy two conditions. One, it has to be 300 feet from the point (2.9, 0.77). You can represent that as the equation of a circle with a radius of 300 feet, centered at that point:

    [tex](x - 2.9)^2 + (y - 0.77)^2 = 90,000[/tex]

    The other condition is that the slope between the two points has to be -3.7755. You can represent this by the following equation:

    [tex]\frac{y - 0.77}{x - 2.9} = -3.7755[/tex]

    Now, the second equation you can multiply by (x-2.9) and it'll give you an expression for y-0.77, which you can then substitute into the first equation to give you an equation for the x-coordinate. (Note that there will be TWO solutions, so you'll have to use common sense to figure out which one you want) Once you solve that, you can find the y-coordinate easily. Then you'll have the point you need!
     
  8. Dec 7, 2011 #7
    My man: you must have a way with words! Thank you so much. You prolly went on to another problem but if you actually solved this can check my answer? I got (-76.794, 301.654). If you haven't done it, it's no big deal. This is graded so I'm trying to get it right haha. Thanks again!
     
  9. Dec 7, 2011 #8

    Char. Limit

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    I got (-73.9114, 290.77). With that sort of difference, there's probably some difference on when we rounded. If you could show your work, I could probably identify the error, if there is one. But I'm willing to wager that there's just rounding differences.
     
  10. Dec 7, 2011 #9
    As long as I have the right concept, I'm sure my prof. won't be too unhappy, but I'll still put it up. How do you insert equations like you did instead of typing it out like y/x=s?
     
  11. Dec 7, 2011 #10

    Char. Limit

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    LaTeX tags, man. [tex*] (get rid of the asterisk) does the trick, as long as you know LaTeX format. (Like, to write a fraction, you do \frac{numerator}{denominator}). Most of it's pretty intuitive, though. Like to write my first equation, I used:

    [tex*](x-2.9)^2 + (y-0.77)^2 = 90000[/tex]

    [tex](x-2.9)^2 + (y-0.77)^2 = 90000[/tex]

    EDIT: Note that tex tags are used when you devote an entire line to an equation. When you want to put a short simple equation inline with the text, like if I were to write [itex]y=x^2[/itex], you'd use [itex*] tags.
     
  12. Dec 7, 2011 #11
    Well I'll tell you now I was goin' pretty quick trying to finish it and rounded in a bizzare manner, lol.

    [tex]y-0.77 = -3.7755(x-2.9)[/tex]
    [tex](x-2.9)^2 +(-3.7755(x-2.9)^2) = 90000[/tex]
    [tex]x^2 - 5.8x + 8.41 + (-3.7755x + 10.94895)^2[/tex]
    [tex]x^2 - 5.8x + 8.41 +14.2544 - 82.67575 + 119.87951 = 90000[/tex]
    [tex]15.2544x^2 - 88.47575x - 89871 = 0[/tex]
    Then I used my calculator to find x= +- 76.79375304.
    Then
    [tex]y-0.77 = -3.755(-76.79375304)[/tex]
    [tex]y=301.6537646[/tex]
     
  13. Dec 7, 2011 #12
    in the fourth step, I meant 14.2544x^2 − 82.67575x. Soo many mistakes..
     
  14. Dec 7, 2011 #13

    Char. Limit

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    Ah, there's the problem. You've got the quadratic equation right, but the roots are not. I'd check those again.
     
  15. Dec 7, 2011 #14
    Ok I got x=-73.911. But then when I put it back in I get

    [tex]y-.0.77 = -3.7755(-73.911)[/tex]
    [tex]y=279.82[/tex]
     
  16. Dec 7, 2011 #15
    Wait, nvm forgot to subtract 2.9, I got it. Thanks so much for explaining everything and not just giving me the answer, now I actually understand the problem!
     
  17. Dec 7, 2011 #16

    Char. Limit

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    That actually looks correct. I think I might have been wrong myself!
     
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