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Thread: 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)

    $\displaystyle y = x^2$

    $\displaystyle y = (x-2)^2$

    Point of intersection:

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

    Derivative:
    $\displaystyle 1. y' = 2x $
    $\displaystyle 2. y' = 2x - 4 $

    $\displaystyle 1. y'(1) = 2 = k_1$
    $\displaystyle 2. y'(1) = -2 = k_2$

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

    $\displaystyle \phi = 53^0 8'$
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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)

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

    point of intersection:

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

    Derivatives:

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

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

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

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

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

    $\displaystyle k_{1,2}$ 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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    $\displaystyle
    x^2 - 3 = -x^2 + 4x - 3
    $
    $\displaystyle
    2x^2 - 4x = 0
    $

    $\displaystyle
    x(2x-4) = 0
    $
    $\displaystyle
    x_1 = 0 ; x_2 = 2$

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

    $\displaystyle 0^2 - y^2 = 3$
    $\displaystyle 3 + y^2 = 0$

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