# Angle of Intersection of Space Curves

1. Apr 25, 2008

### the7joker7

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. Apr 25, 2008

### EngageEngage

you need the tangent vectors at the point of intersection -- do you know how to find these?

3. Apr 25, 2008

### the7joker7

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. Apr 25, 2008

### EngageEngage

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:

$$\vec{T}=\frac{\frac{dr}{dt}}{|\frac{dr}{dt}|}$$
but it looks like you are on the right track.

5. Apr 25, 2008

### the7joker7

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. Apr 25, 2008

### EngageEngage

you're not differentiating the trig terms correctly.

7. Apr 25, 2008

### EngageEngage

I did it and got the right answer. just be careful

8. Apr 25, 2008

### the7joker7

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. Apr 25, 2008

### the7joker7

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. Apr 26, 2008

### the7joker7

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. Apr 26, 2008

### HallsofIvy

Staff Emeritus
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. Apr 26, 2008

### the7joker7

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. Apr 27, 2008

### HallsofIvy

Staff Emeritus
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.