Register to reply

Finding the point of intersection between two curves

by jheld
Tags: curves, intersection, point
Share this thread:
jheld
#1
Oct26-08, 05:32 PM
P: 81
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.
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
Dick
#2
Oct26-08, 06:06 PM
Sci Advisor
HW Helper
Thanks
P: 25,228
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.
jheld
#3
Oct26-08, 06:26 PM
P: 81
Okay. I found t = 3 - sqrt(6) and s = sqrt(6)
So, what do I need to do to calculate the angle?

Dick
#4
Oct26-08, 06:28 PM
Sci Advisor
HW Helper
Thanks
P: 25,228
Finding the point of intersection between two curves

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.
jheld
#5
Oct26-08, 06:35 PM
P: 81
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).
I had:
3 + (3 - s)^2 = s^2.
Dick
#6
Oct26-08, 06:38 PM
Sci Advisor
HW Helper
Thanks
P: 25,228
(3-s)^2=(3-s)*(3-s)=9-6s+s^2. Multiply it out.
jheld
#7
Oct26-08, 06:55 PM
P: 81
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))
Dick
#8
Oct26-08, 09:37 PM
Sci Advisor
HW Helper
Thanks
P: 25,228
Quote Quote by jheld View Post
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))
I get s=2, t=+1. Can you show more steps in your solution if you want us to check it?
donald1403
#9
Nov18-08, 08:29 PM
P: 16
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?
donald1403
#10
Nov18-08, 08:48 PM
P: 16
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?
Dick
#11
Nov18-08, 10:19 PM
Sci Advisor
HW Helper
Thanks
P: 25,228
Quote Quote by donald1403 View Post
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?
That seems fine to me.
donald1403
#12
Nov18-08, 10:21 PM
P: 16
thanks
donald1403
#13
Nov18-08, 10:26 PM
P: 16
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?
Dick
#14
Nov18-08, 10:33 PM
Sci Advisor
HW Helper
Thanks
P: 25,228
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".
donald1403
#15
Nov18-08, 10:37 PM
P: 16
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?
Dick
#16
Nov18-08, 10:43 PM
Sci Advisor
HW Helper
Thanks
P: 25,228
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??
donald1403
#17
Nov18-08, 10:45 PM
P: 16
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...
geno921
#18
Feb8-10, 05:26 AM
P: 4
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.


Register to reply

Related Discussions
Alg Geom: Rational curves with self-intersection -2 General Math 1
Angle of Intersection of Space Curves Calculus & Beyond Homework 12
Finding the point of intersection of two lines Precalculus Mathematics Homework 4
Intersection of 2 curves Precalculus Mathematics Homework 4
Intersection of two 3D parametric curves General Math 4