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Homework Help: 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
    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
    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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