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Math Help - Componets of vectors

  1. #1
    Junior Member cinder's Avatar
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    Componets of vectors

    Let the angle \theta be the angle that the vector \vec{A} makes with the +x-axis, measured counterclockwise from that axis. Find the angle \theta for a vector that has the following components.

    First one is A_{x}=1.30 m, A_{y}= -3.40.

    I've read the section, but it's not helping. It lists a couple of equations such as \frac{A_{x}}{A}=\cos\theta, but I don't see how that helps me.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cinder View Post
    Let the angle \theta be the angle that the vector \vec{A} makes with the +x-axis, measured counterclockwise from that axis. Find the angle \theta for a vector that has the following components.

    First one is A_{x}=1.30 m, A_{y}= -3.40.

    I've read the section, but it's not helping. It lists a couple of equations such as \frac{A_{x}}{A}=\cos\theta, but I don't see how that helps me.
    I assume by A_x you mean the horizontal displacement, and by A_y you mean the vertical displacement. If so, draw a diagram of it. You will end up with a right-triangle (with legs 1.3 and 3.40), use the appropriate trig ratio to find the desired angle, which would end up being the one you were given i believe: \cos \theta = \frac {A_x}{A}, where A is the length of the hypotenuse. remember, you want to describe the angle as the angle obtained by going anti-clockwise from the positive x-axis
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  3. #3
    Junior Member cinder's Avatar
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    Ugh. Nevermind. I keep asking stupid questions and answering them myself.
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  4. #4
    Junior Member cinder's Avatar
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    Okay, I'm not sure where \theta is.

    \frac{A_{x}}{A}=0.36=\cos\theta, but what's next?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cinder View Post
    Okay, I'm not sure where \theta is.
    the hypotenuse is in the fourth quadrant, correct? \theta is the angle that goes from the positive x-axis counterclockwise (that is it goes from the 1st quad to the second to the third to the fourth and stops on the hypotenuse of the right-triangle.
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  6. #6
    Junior Member cinder's Avatar
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    I keep getting 1.20, but I know that's not right.
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  7. #7
    Junior Member cinder's Avatar
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    The answer is 291, but I have no idea how they got it.
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  8. #8
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    You are having bioth Domain/Range and Calculator Setting problems.

    1) Once you have determined the Quadrant, do NOT be dissuaded from that quadrant. The Range of the arccosine function is [0,\pi]. There is no Fourth Quadrant in there. It is the end user's responsibility to translate the result to the proper quadrant if [0,\pi] is unsatisfactory. The inverse cosine function on your calculator WILL NOT give you a result in the Fourth Quadrant.

    2) Your calculator is NOT telling you 1.20. It is TRYING to tell you 1.20 Radians. Cconvert it to degrees. Better yet, if you want degrees, feel free to put your calculator in degrees mode. It should give you 69.075, which still is not correct because that's in the First Quadrant. You have to fix it by point 1).

    3) Why cosine when you've a perfectly good inverse tangent to solve the problem? Are you doing vectors and leaning on dot products defined as cosines? I guess that would motivate such behavior. It's still a little irritating to force a more difficult solution.
    Last edited by TKHunny; August 27th 2007 at 04:37 AM. Reason: Add point 3)
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  9. #9
    Junior Member cinder's Avatar
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    Well, I can see that 360 - 69 = 291, but I really can't see why that gives me the correct result.

    I had to do three more like these, and I'm given four tries each. I missed two completely and got the other two right after a few tries. The remaining three problems put a triangle into one of the remaining three quadrants (I, II, and III).
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  10. #10
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    Inverse Functions give only hints. The end user muat get the answer in the correct quadrant.

    I would do it this way (being forced to use the cosine).

    y-component is negative
    x-component is positive
    The result is in Quadrant IV

    acos(1.3/13.25) = 1.206 rad = 69.075

    This is a reference angle in Quadrant I.

    The angle with identical reference in Quadrant IV is 360 - 69.075 = 290.925

    If you would like, you can define your own Inverse Cosine function. Give it whatever Range you need. You are not stuck with the standard definition.
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