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Math Help - Angle between 2 curves

  1. #1
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    Angle between 2 curves

    Here's another one that he didn't go over & I am totally lost on.
    Determine the angle between the two curves to the nearest degree for the curves at each point of intersection. Note they are not orthogonal trajectories.

    a. y=x^2 and y=(x-2)^2
    b. x^2 - y^2=3 and x^2 - 4x+y^2 +3=0

    Thanks again!
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  2. #2
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    a)

     y = x^2

    y = (x-2)^2

    Point of intersection:

    x^2 = (x-2)^2
     -4x + 4 = 0
     x = 1 \Rightarrow y = 1

    Derivative:
    1. y' = 2x
    2. y' = 2x - 4

    1. y'(1) = 2 = k_1
    2. y'(1) = -2 = k_2

     tan \phi = |\frac{k_1 - k_2}{1+ k_1k_2}| = |\frac{2+2}{1-4}| = \frac {4}{3}

     \phi = 53^0 8'
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  3. #3
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    Yeah, he didn't go over that at all. Thanks for the help with a. I am trying to figure out b the same way but I am pretty sure I'm not going to be getting that one on my own.
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  4. #4
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    b)

    1. x^2 - y^2 = 3 \Rightarrow y=\sqrt{x^2 -3}
    2. x^2 - 4x + y^2 + 3 = 0 \Rightarrow y = \sqrt{-x^2+4x-3}

    point of intersection:

    x^2 - 3 = -x^2 +4x -3
    2x^2 = 4x
    x_1 = 0 -> not suitable
    x_2 = 2 \Rightarrow y_1 = 1, y_2 = -1

    Derivatives:

    1st curve:  2x - 2y \cdot y' = 0
     y' = \frac {x}{y} = \frac{x}{\sqrt{x^2-3}}

    2nd curve:  2x -4 + 2y \cdot y' = 0
     y' = \frac{-x+2}{y} = \frac{-x+2}{\sqrt{-x^2+4x3}}

    from the first curve we get 2 k's.
    k_1 = f'_1(2) = \frac {x}{y} = \frac{x}{\sqrt{x^2-3}} = \frac {2}{1} = 2
    k_2 = f'_2(2) = \frac {x}{y} = \frac{x}{\sqrt{x^2-3}} = \frac {2}{-1} = -2

    from the second curve:
    k_3 = f'_3(2) = {-x+2}{\sqrt{-x^2+4x3}} = \frac {0}{\pm 1} = 0

     tan \Phi = |\frac {k_{1,2} - k_3}{1+k_{1,2}k_3}| = 2
     \Phi = 63^0 26'

     k_{1,2} means you can take any of the 2 k's and the result will be the same..
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  5. #5
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    hahahaha yeah. Wasn't even close to getting "b" right by myself. Thank you so very, very much for the help, even though I don't understand it!
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  6. #6
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    what is it that you don't understand?

    i can explain
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  7. #7
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    Quote Originally Posted by metlx View Post
    what is it that you don't understand?

    i can explain
    The POI for B. How do you get x1 and x2 from the problem 2x^2=4x?
    If I am not mistaken, you are assigning x1 to 2x^2 and x2 to 4x.
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  8. #8
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     <br />
x^2 - 3 = -x^2 + 4x - 3<br />
    <br />
2x^2 - 4x = 0<br />

    <br />
x(2x-4) = 0<br />
    <br />
x_1 = 0 ; x_2 = 2

    But x_1 = 0 does not fit into the first equation x^2 - y^2 = 3

    0^2 - y^2 = 3
     3 + y^2 = 0

    and since anything squared within R cannot be negative then you can assume that the above equation is not right..
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