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Math Help - finding the angle

  1. #1
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    finding the angle

    I need help finding the angle of the curves r_1(t) = (t, 1-t, 3+t^2), \ \  r_2(s) = (3-s, s-2, s^2) I found that they intersect at x=1, \ \ y = 0, \ \ z= 4 , and would the angle between them be cos^{-1} \frac{(3)(0)+(1)(-2)+(3)(1)}{(\sqrt{1+9})(\sqrt{9+4+1})} = cos^{-1} \frac{1}{\sqrt{140}}=1.48618 ?
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  2. #2
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    Quote Originally Posted by lllll View Post
    I need help finding the angle of the curves r_1(t) = (t, 1-t, 3+t^2), \ \  r_2(s) = (3-s, s-2, s^2) I found that they intersect at x=1, \ \ y = 0, \ \ z= 4 , and would the angle between them be cos^{-1} \frac{(3)(0)+(1)(-2)+(3)(1)}{(\sqrt{1+9})(\sqrt{9+4+1})} = cos^{-1} \frac{1}{\sqrt{140}}=1.48618 ?
    The intersection point corresponds to t = 1 and s = 2.

    So I think you want the angle between the vectors \left[ \frac{dr_1}{dt}\right]_{t = 1} and \left[ \frac{dr_2}{ds}\right]_{s = 2} .....
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    So I think you want the angle between the vectors \left[ \frac{dr_1}{dt}\right]_{t = 1} and \left[ \frac{dr_2}{ds}\right]_{s = 2} .....
    why would we have to differentiate to find the angle? once you get the derivative you only get:

    r'(t) = (1,-1,2t), \ \ r'(s) = (-1,1,2s) and if I put t=1 and s=2 that gives r'(t)= (1,-1,2), \ \ r'(s) = (-1,1,4). I don't see how I can use those coordinates to find the angle. Would you have to use cos(\theta) \frac{a \cdot b}{|a||b|} ?
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    Quote Originally Posted by lllll View Post
    why would we have to differentiate to find the angle? once you get the derivative you only get: r'(t) = (1,-1,2t), \ \ r'(s) = (-1,1,2s) and if I put t=1 and s=2 that gives r'(t)= (1,-1,2), \ \ r'(s) = (-1,1,4). I don't see how I can use those coordinates to find the angle. Would you have to use cos(\theta) \frac{a \cdot b}{|a||b|} ?
    The derivative gives the direction of the curve at any time.
    The angle between the curves is the angle between the directions.
    If r'(t) = (1,-1,2t), \ \ r'(s) = (-1,1,2s) then r'(1) = (1,-1,2), \ \ r'(2) = (-1,1,4) are the two directions.
    The angle is \phi  = \arccos \left( {\frac{{ < 1, -1, 2 >  \cdot  <  - 1,1,4 > }}{{\left\| { < 1,-1, 2 > } \right\|\left\| { < -1,1, 4 > } \right\|}}} \right).
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