Angle of Intersection of Space Curves

In summary: 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...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>
  • #1
the7joker7
113
0

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.

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?
 
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  • #2
you need the tangent vectors at the point of intersection -- do you know how to find these?
 
  • #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?
 
  • #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.
 
  • #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.
 
  • #6
you're not differentiating the trig terms correctly.
 
  • #7
I did it and got the right answer. just be careful
 
  • #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!
 
  • #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...
 
  • #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...
 
  • #11
the7joker7 said:
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.
 
  • #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.
 
  • #13
the7joker7 said:
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>?

the7joker7 said:
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...
 

What is the angle of intersection of space curves?

The angle of intersection of space curves is the angle formed between two curves in three-dimensional space. It is the angle between the tangent lines of the two curves at the point of intersection.

How is the angle of intersection of space curves calculated?

The angle of intersection of space curves can be calculated using the dot product of the tangent vectors of the two curves at the point of intersection. The formula is given by cosθ = (u · v)/(|u||v|), where θ is the angle of intersection, u and v are the tangent vectors, and |u| and |v| are the magnitudes of the vectors.

Can the angle of intersection of space curves be negative?

No, the angle of intersection of space curves is always positive. It represents the smallest angle between the two curves, and is measured in the counterclockwise direction.

Are there any special cases when calculating the angle of intersection of space curves?

Yes, there are two special cases to consider when calculating the angle of intersection of space curves. The first is when the two curves are parallel, in which case the angle of intersection is 0. The second is when the two curves are perpendicular, in which case the angle of intersection is 90 degrees.

What is the significance of the angle of intersection of space curves?

The angle of intersection of space curves is important in understanding the relationship between different curves in three-dimensional space. It can be used in various applications, such as in physics and engineering, to determine the direction of motion or the forces acting on an object. It also has applications in computer graphics and animation, where it is used to create realistic 3D images and animations.

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