Find intersection of parametric curve and line

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Homework Statement



I'm trying to find when that parametric curve intersects with the line x=20

Homework Equations



x(t)=(2t^3)/(t^2-1) ; y(t)=(2t^3)/((t^2+1)^2)

The Attempt at a Solution


I tried representing the line as y=t ; x=20

35=2t^3/(t^2-1) ; t=2t^3/((t^2+1)^2)

I also ended up with this equation:
20(t^2-1)=t(t^2+1)^2
but solving for it did not end up with the answers I expect to have. The y-values should be between 0 and 1.

Solving for t in terms of x, then plugging into the y(t) equation resulted in imaginary numbers. So how do you find the 2 intersections?
note: I need the answers by tonight. Thank you!
 
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Answers and Replies

  • #2
HallsofIvy
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Homework Statement



I'm trying to find when that parametric curve intersects with the line x=20

Homework Equations



x(t)=(2t^3)/(t^2-1) ; y(t)=(2t^3)/((t^2+1)^2)

The Attempt at a Solution


I tried representing the line as y=t ; x=20
Better to use (20, y) or y= s, x= 20. There is no reason to assume the parameter for the line is the same as the parameter for the curve.

35=2t^3/(t^2-1) ;
I assume you mean [itex]20= 2t^3/(t^2- 1)[/itex]. If not, I have no idea where you got "35" from!

t=2t^3/((t^2+1)^2)
No. Again the two parameters are not the same. Better, solve the first equation for t, then use that value of t in [itex]y= 2t^3/(t+1)^2[/itex]

I also ended up with this equation:
20(t^2-1)=t(t^2+1)^2
but solving for it did not end up with the answers I expect to have. The y-values should be between 0 and 1.

Solving for t in terms of x, then plugging into the y(t) equation resulted in imaginary numbers. So how do you find the 2 intersections?
note: I need the answers by tonight. Thank you!
 
  • #3
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My bad, it is indeed 20 = ...

But, I tried solving for the first equation and it resulted in imaginary answers. Plugging THOSE into y=2t^3/(t+1)^2 just resulted in imaginary answers, as well. Looking at the parametric graph and the line together, though, I know there are two intersections with y-values between 0 and 1. I just can't figure out the exact answers.
 
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  • #4
HallsofIvy
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That's impossible. That's a cubic equation with real coefficients- since complex roots of polynomial equations with real coefficents come in conjugate pairs, any such cubic equation has at least one real root.

[itex]20= 20t^3/(t^2-1)[/itex] is the same as [itex]20t^2- 20= 2t^3[/itex] or [itex]2t^3- 20t^2+ 20= 0[/itex]. I notice that if t= 1, [itex]2t^3- 20t^2+ 20= 2- 20+20= 2> 0[/itex] while if t= 2, [itex]2t^3- 20t^2+ 20= 16- 80+ 20= -44< 0[/itex] so there is a real root between 1 and 2.
 
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