Parametric equations for a loop

[ILoveBaseball]

The following parametric equations trace out a loop
[tex] x = 8 - 3/2t^2[/tex]
[tex] y = -3/6t^3+3t+1[/tex]

1.) Find the [tex]t[/tex] values at which the curve intersects itself.
wouldn't i just have to solve for t for one of the equaltion to find t? also, can you find the intersects using a TI-83 plus to check your answer?

[tex] 8 - 3/2t^2=0[/tex]
[tex]3/2t^2=8[/tex]
[tex]t^2=16/3[/tex]
[tex]t = \pm\sqrt(16/3)[/tex] which is wrong

 

[whozum]

You can use the ti83 to find them.
Put it in polar mode, enter your equations, and from there find intersects like you normally would.

 

[ILoveBaseball]

there is no "intersect" function when i set it in parametic mode. How would i find the T values manually?

 

[whozum]

Well what you did was find the y intercepts of the graph. I looked in my calc book and can't find anything on intersections of parametric curves.

 

[ILoveBaseball]

so how would i find the [tex]t[/tex] values?

 

[xanthym]

The following parametric equations trace out a loop
[tex] x = 8 - 3/2t^2[/tex]
[tex] y = -3/6t^3+3t+1[/tex]

1.) Find the [tex]t[/tex] values at which the curve intersects itself.
wouldn't i just have to solve for t for one of the equaltion to find t? also, can you find the intersects using a TI-83 plus to check your answer?

[tex] 8 - 3/2t^2=0[/tex]
[tex]3/2t^2=8[/tex]
[tex]t^2=16/3[/tex]
[tex]t = \pm\sqrt(16/3)[/tex] which is wrong

The curve will intersect itself when the same "x" and "y" values are produced by 2 different "t" values, say "t1" and "t2". Thus:

[tex] x \ = \ 8 \ - \ (3/2) t_{1}^{2} \ = \ 8 \ - \ (3/2)t_{2}^{2} [/tex]
[tex] y \ = \ (-3/6) t_{1}^{3} \ + \ 3 t_{1} + 1 \ = \ (-3/6) t_{2}^{3} \ + \ 3 t_{2} + 1 [/tex]

Solve for DIFFERENT values of "t1" and "t2" to find the crossing point.
(Solve the equations simultaneously. Both equations must be satisfied.)


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[xanthym]

The curve will intersect itself when the same "x" and "y" values are produced by 2 different "t" values, say "t₁" and "t₂". Thus:

[tex] x \ = \ 8 \ - \ (3/2) t_{1}^{2} \ = \ 8 \ - \ (3/2)t_{2}^{2} [/tex]

[tex] y \ = \ (-3/6) t_{1}^{3} \ + \ 3 t_{1} + 1 \ = \ (-3/6) t_{2}^{3} \ + \ 3 t_{2} + 1 [/tex]

Solve for DIFFERENT values of "t₁" and "t₂" to find the crossing point.
(Solve the equations simultaneously. Both equations must be satisfied.)

Here are some HINTS:

[tex] \color{blue} a): \ \ \ \ \mathsf{ Eq \ for \ "x" \ indicates \ t_{1} = -t_{2} \ \ \ \ \ \ \ Eq \ for \ "y" \ uses \ next \ hint } [/tex]

[tex] \color{blue} b): \ \ \ \ \frac { t_{1}^{3} \ - \ t_{2}^{3} } { t_{1} \ - \ t_{2} } \ = \ t_{1}^{2} \ + \ t_{1}t_{2} \ + \ t_{2}^{2} \ \ \ \ \ \ \ ( t_{1} \ \ne \ t_{2} )[/tex]

[tex] \color{blue} c): \ \ \ \ (Answers) \ \longrightarrow \ \ (t_{1} \ = \ \sqrt{6}) \ \ and \ \ (t_{2} \ =\ -\sqrt{6}) [/tex]


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