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Angle of Intersection of Space Curves

  1. Apr 25, 2008 #1
    1. The problem statement, all variables and given/known data

    The curves `bar r_1(t) = < 2t,t^(4),5t^(6) >` and `bar r_2(t) = < sin(-2t),sin(4t),t - pi >` intersect at the origin.

    Find the angle of intersection, in radians on the domain `0<=t<=pi`, to two decimal places.

    3. The attempt at a solution

    Well, I tried to do it in the same way I would with two vectors. I needed to find the dot product and the magnitudes for starters, so...

    Dot Product:
    (2t)(sin(-2t) + (t^4)(sin(4t)) + (5t^6)(t - pi)

    Magnitudes:
    sqrt((2t)^2 + (t^4)^2 + (5t^6)^2) * sqrt((sin(-2t))^2 + (sin(4t))^2 + (t - pi)^2)

    But...now I'm not sure what to do. Am I even going in the right direction?
     
  2. jcsd
  3. Apr 25, 2008 #2
    you need the tangent vectors at the point of intersection -- do you know how to find these?
     
  4. Apr 25, 2008 #3
    I think so...

    T(0) = r'(0)/the magnitude of r'(0), so...

    r'(0) = <2, 0, 0>

    magnitude of r'(0) = sqrt(4) = 2

    so

    T(0) = <1, 0, 0>

    And for the other one we have...

    r'(0) = <cos(0), -cos(0), 1> = <1, 1, 1>

    Magnitude of r'(0) = 1.732 = sqrt(3)

    T(0) = <1/sqrt(3), 1/sqrt(3), 1/sqrt(3)>

    Now what?
     
  5. Apr 25, 2008 #4
    now you can use the vector identity that relates the dot product to the angle between two vectors to get your answer. I didnt check your work but the tangent vector is:

    [tex] \vec{T}=\frac{\frac{dr}{dt}}{|\frac{dr}{dt}|}[/tex]
    but it looks like you are on the right track.
     
  6. Apr 25, 2008 #5
    I tried that, didn't work...

    I tried dot product over magnitudes = cos(theta) and tan(theta), but neither was the right answer.

    FTR, the right answer was 2.022.
     
  7. Apr 25, 2008 #6
    you're not differentiating the trig terms correctly.
     
  8. Apr 25, 2008 #7
    I did it and got the right answer. just be careful
     
  9. Apr 25, 2008 #8
    Well, t=0, right?

    Derivative of sin is cosine...

    So r'1(t) = <2, 4t^3, 30t^5>

    r'1(0) = 2, 0, 0

    and

    r'2(t) = <cos(-2t)*-2, cos(4t)*4, 1>

    r'2(0) = <-2, 4. 1>

    Now dot product...

    (2)(-2) + (4)(0) + (1)(0) = -4

    magnitude

    2 * 4.5825 = 9.16515139

    cos(theta) = -4/9.16515139

    theta = 2.022.

    Aha! Got it.

    Thanks!
     
  10. Apr 25, 2008 #9
    r'(t) = <-2, 5t^4, -2t>

    r'(0) = <-2, 0, 0>

    r'(t) = <cos(3t)*3, cos(-3t)*-3, 1>

    r'(0) = <3, -3, 1>

    dot product

    -6

    magnitude

    2*4.3588989 = 8.717797887

    cos(theta) = -6/8.717797889

    theta = 2.33

    Why isn't this right?

    This is driving me crazy...
     
  11. Apr 26, 2008 #10
    It's probably just a stupid minor error, but I swear I did the exact same thing above that I did before when I got it right...
     
  12. Apr 26, 2008 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since you did not state the original problem, it is impossible to tell whether that is right or wrong or where you might have made a mistake.
     
  13. Apr 26, 2008 #12
    The curves `bar r_1(t) = < -2t,t^(5),-1t^(2) >` and `bar r_2(t) = < sin(3t),sin(-3t),t - pi >` intersect at the origin.

    Find the angle of intersection, in radians on the domain `0<=t<=pi`, to two decimal places.
     
  14. Apr 27, 2008 #13

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, they do intersect at the origin. In particular, r_1(0)= < 0, 0, 0>. But r_2(0) = < 0, 0, -pi>, not <0, 0, 0>. For what value of t is r_2(t)= < 0, 0, 0>?

    No, you have the wrong t.

     
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