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Angle of Intersection of Space Curves

  • Thread starter the7joker7
  • Start date
113
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1. Homework Statement

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?
 

Answers and Replies

208
0
you need the tangent vectors at the point of intersection -- do you know how to find these?
 
113
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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?
 
208
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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.
 
113
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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.
 
208
0
you're not differentiating the trig terms correctly.
 
208
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I did it and got the right answer. just be careful
 
113
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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!
 
113
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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...
 
113
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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...
 
HallsofIvy
Science Advisor
Homework Helper
41,732
893
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...
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.
 
113
0
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.
 
HallsofIvy
Science Advisor
Homework Helper
41,732
893
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.
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>?

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>
No, you have the wrong t.

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...
 

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