Finding the point of intersection between two curves

1. Oct 26, 2008

jheld

1. The problem statement, all variables and given/known data
At what point do the curves r1(t) = <t, 1 - t, 3 + t^2> and r2(s) = <3 - s, s - 2, s^2> intersect? Find their angle of intersection correct to the nearest degree.

2. Relevant equations

3. The attempt at a solution
I set t = 3 -s
1 - t = s - 2
3 + t^2 = s^2

I got s = s and t =t, and I should of course assume so, but I wasn't able to find their exact numerical values.

Then I thought to differentiate both functions and see about that.
1 = -1, they don't equal
-1 = 1, they don't equal
2t = 2s, take out the 2, and t = s, but they don't from the previous two equations.

Also I am not sure how to calculate the angle. I realize I would use some inverse trigonometric function, but I am not sure how to get to that step.

2. Oct 26, 2008

Dick

The first two equations are basically the same, they just tell you t=3-s. Put that into the third equation. So 3+(3-s)^2=s^2. You should be able to solve that for s.

3. Oct 26, 2008

jheld

Okay. I found t = 3 - sqrt(6) and s = sqrt(6)
So, what do I need to do to calculate the angle?

4. Oct 26, 2008

Dick

You'd better show how you found those values, because it's not correct. Once you've found the correct point you find the tangent vectors at that point and use the dot product.

5. Oct 26, 2008

jheld

Well, I understand why it's wrong, because I ended up making an illegal mathematical move changing (3-s)^2 to (9 - s^2), which doesn't work. I can't find the right value for s. all i get is that 12 = 0. (s^2 - s^2).
3 + (3 - s)^2 = s^2.

6. Oct 26, 2008

Dick

(3-s)^2=(3-s)*(3-s)=9-6s+s^2. Multiply it out.

7. Oct 26, 2008

jheld

Okay. I did that, got t = -1, and s = 2.
then I took the derivatives of each function, and then took the dot product and solved for theta.
I got theta = 164.21 degrees.
cos^-1(-10/sqrt(108))

8. Oct 26, 2008

Dick

I get s=2, t=+1. Can you show more steps in your solution if you want us to check it?

9. Nov 18, 2008

donald1403

I also got s = 2 and t = 1. <<< so what is the point of intersection (1,2)??

Where do i go from here to get the angle of intersection?

10. Nov 18, 2008

donald1403

r1'(t) = < 1, -1, 2t>

r2'(s) = <-1, 1, 2s>

solving for t and s, I got t = 1 and s =2.

I assume I have to put t and s into r1'(t) and r2'(t)?

so, r1'(t) = < 1, -1, 1> and r2'(s) = < -1, 1, 4>.

I used the dot product to find the angle between two vectors.

cos∅=(<1,-1,1> ∙ <-1,1,4>)/(√(3 ) ∙ √18)

and ∅ =74.2068 degree.

can anyone confirm if it is correct?

11. Nov 18, 2008

Dick

That seems fine to me.

12. Nov 18, 2008

donald1403

thanks

13. Nov 18, 2008

donald1403

sorry. one more question.. i did this problem based on what you said above..

when i use the dot product, i use the derivative r1 and r2. why can't i use just r1 and r2? what would be the different? if i just plug t=1 and s=2 into r1(t) and r2(t), the angle of intersection would be different?

can u explain a lil bit?

14. Nov 18, 2008

Dick

r1(t) and r2(s) are points on the curves. In fact, you found the intersection so r1(1)=r2(2), right? The angle between the two curves at that point is the angle between their tangent vectors, isn't it? "Tangent vector" = "derivative".

15. Nov 18, 2008

donald1403

so how do i write the point of intersection. can i write that in (x,y) form like (1,2) or (2,1) or I just write down that the two curves intersect when t =1 and s =2?

16. Nov 18, 2008

Dick

Now you are confusing me. r1(1)=(1,0,4), r2(2)=(1,0,4). THAT'S the point of intersection. They are three dimensional vectors, aren't they? Just put t=1 and s=2 into the original formulas. 1 and 2 are just the values of the t and s parameters. They aren't components of points on the curves. What are you thinking??

17. Nov 18, 2008

donald1403

arrr.. sorry to confuse you.. i got it... i guess i m jz going crazy with the test tomorrow.. anyway thanks again for your help...

18. Feb 8, 2010

geno921

I believe your angle is wrong. Check the derivate of r1'(1)= <1,-1,2t>, thus giving a vector of <1,-1,2> not <1,-1,1>. The angle comes out to pi/6.