Finding the point of intersection between two curves

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Homework Statement


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.


Homework Equations





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.
 

Answers and Replies

  • #2
Dick
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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
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Okay. I found t = 3 - sqrt(6) and s = sqrt(6)
So, what do I need to do to calculate the angle?
 
  • #4
Dick
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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
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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.
 
  • #6
Dick
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(3-s)^2=(3-s)*(3-s)=9-6s+s^2. Multiply it out.
 
  • #7
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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
Dick
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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?
 
  • #9
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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
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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
Dick
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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.
 
  • #13
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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
Dick
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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
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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
Dick
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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
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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
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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.
 

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