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Did I assume too much in this problem? Stuck on some trig, simple mirror.

  1. Mar 15, 2006 #1
    Hello everyone. I was hinted by a member that I should use pythagoris and some trig to figure the length out. It all made sense but now i'm not getting the right answer and i'm wondering if its becuase I assumed the first triangle had a length of 3.10, because it said the bird is 3.10m from the mirror, is that info just useless and they wanted me to use the information of the 2 other sides, 4.10 and 4.90 and find the ohter side? 4.90^2 = 4.10^2 + b^2?
    Here is the problem and my drawing:
    You look through a camera toward an image of a hummingbird in a plane mirror. The camera is 4.30 m in front of the mirror. The bird is at camera level, 4.90 m to your right and 3.10 m from the mirror. What is the distance between the camera and the apparent position of the bird's image in the mirror?

    [​IMG]


    I also asked my professor and he said yeah this is how u do it, which isn't right.
     
  2. jcsd
  3. Mar 16, 2006 #2

    Doc Al

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    I believe you are interpreting the given information incorrectly (and thus drawing the diagram incorrectly). I'll describe how I would draw the diagram in terms of coordinates.

    Let's have the plane of the mirror be at y = 0. You are standing 4.3m in front of the mirror, thus your coordinates are x = 0; y = +4.3m. The bird is 4.9m to your right, thus it's at x = -4.9m; it's 3.1m in front of the mirror, so y = +3.1m. Using this interpretation, find the coordinates of the bird's image and then the distance between image and camera.
     
  4. Mar 16, 2006 #3
    Hey thanks alot Doc Al, your method is alot cleaner and makes alot more senes than mine but I still got it wrong. I don't see what I messed up...Can you check my work please? Thanks!
    [​IMG]

    If i use the distance formula for those 2 points i get:
    sqrt(4.9^2+(-3.1-4.3)^2) = 8.88;
    Do i use that method instead of c^2 = a^2+b^2?
     
    Last edited: Mar 16, 2006
  5. Mar 16, 2006 #4

    Doc Al

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    Your bird image is way off! Plane mirrors only reverse front to back: only the y-coordinate will change, not the x-coordinate.
     
  6. Mar 16, 2006 #5
    THanks a ton doc, I think i got it this time, but can you make sure I didn't screw it up again? I only have 2 submissions left so I'm just checking!
    [​IMG]
     
  7. Mar 16, 2006 #6

    Doc Al

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    You have the correct coordinates of the camera and the bird's image. Find the distance between them. (Don't really know what all those calculations are that you are doing. Why are you finding the camera's image? You need to find the distance between the real camera and the bird's image: the image of the camera has nothing to do with that.)
     
  8. Mar 16, 2006 #7
    Well you see where I drew that triangle wit h sides 8.6, 5.04 and C? well C would be the distance from the birds image to the cammera. I found the distance between the cammera's image because for me to draw a triangle of 8.6, 5.04, and C, i need at least 2 of the sides to find the 3rd, so by finding the Cammera's image, that gave me the big side 8.6. The 5.04 I found was from above the axis if you can see where I wrote 5.04, i assumed I can bring it down. I submitted 9.97 but that was wrong. Hm...
     
  9. Mar 16, 2006 #8

    Doc Al

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    I see what you did, but it doesn't make sense. (I hope you realize that the camera-camera image-bird image do not form a right triangle.) Do you know how to find the distance between two points, given their coordinates?
     
  10. Mar 17, 2006 #9
    Sorry about not responding sooner! Thanks for the help!

    Ohh so I can't apply c^2=a^2+b^2 to find the sides? Yes I know how to find the distance between 2 points I believe. I used it a few times in my work. To find the triangle with sides, 8.6, and 5.04 I used d = sqrt((y2-y1)^2+(x2-x1)^2);
    So would I just find the distance between points:
    (-4.9,-3.1) and (0,4.3) and that will give me the distance between the birds image and the cammera?
     
  11. Mar 17, 2006 #10

    Doc Al

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    The Pythagorean theorem only applies to right triangles. That's not a right triangle.

    Sure, since those are the coordinates of the bird's image and the camera.
     
  12. Mar 18, 2006 #11
    THanks for the help Doc Al, it worked great!!!
    The professor showed us another way of doing it which required 1 right triangle that had sides of, 7.4, 4.90 and u just use pytogris to find the hyptonus and thats ur answer! Actually the professor had no idea how to do it, he looked at how the book did it while he gave us a quiz :P
     
    Last edited: Mar 19, 2006
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