self intersecting parametric loop

[renegade05]

:?: How do you find the intersection points on a self intersecting parametric loop analytically?

Say:

\(\displaystyle x(t)=3t-t^3\) and \(\displaystyle y(t)=3t^2\)

 

[tkhunny]

Pick two and solve.

\(\displaystyle x(t_{1}) = x(t_{2})\)
\(\displaystyle y(t_{1}) = y(t_{2})\)

 

[renegade05]

elaborate

 

[soroban]

Hello, renegade05!

How do you find the intersection points on a self intersecting parametric loop analytically?

. . \(\displaystyle \begin{array}{ccc} x(t) \:=\:3t-t^3 \\ y(t) \:=\:3t^2 \end{array}\)

\(\displaystyle \text{Choose two }di\!f\!f\!erent\text{ values of }t\text{, say }a\text{ and }b.\)

\(\displaystyle \text{Then we have: }\:\begin{Bmatrix}x(a) \:=\:3a-a^3 \\ y(a) \:=\:3a^2\end{Bmatrix} \;\;\text{ and }\;\; \begin{Bmatrix}x(b) \:=\:3b-b^3 \\ y(b) \:=\:3b^2 \end{Bmatrix}\)

\(\displaystyle \text{Equate }x\text{'s and }y\tet{'s: }\;\begin{Bmatrix}3a-a^3 \:=\:3b-b^3 & [1] \\ 3a^2 \:=\:3b^2 & [2] \end{Bmatrix}\)


\(\displaystyle \text{From [1]: }\:3a - 3b - a^3 + b^3 \:=\:0 \quad\Rightarrow\quad 3(a-b) - (a^3-b^3) \:=\:0\)

. . \(\displaystyle 3(a-b) - (a-b)(a^2+ab + b^2) \:=\:0 \quad\Rightarrow\quad (a-b)(3 - a^2 - ab - b^2) \:=\:0\)

\(\displaystyle \text{Since }a \ne b\text{, we have: }\:3-a^2-ab-b^2 \:=\:0 \quad\Rightarrow\quad a^2+ab + b^2 \:=\:3\;\;[3]\:\)


\(\displaystyle \text{From [2]: }\:3a^2 - 3b^2 \:=\:0 \quad\Rightarrow\quad 3(a - b)(a + b) \:=\:0\)

\(\displaystyle \text{Since }a \ne b\text{, we have: }\:a + b \:=\:0 \quad\Rightarrow\quad b \:=\:-a\;\;[4]\)


\(\displaystyle \text{Substitute [4] into [3]: }\:a^2 + a(\text{-}a) + (\text{-}a)^2 \:=\:3 \quad\Rightarrow\quad a^2 \:=\:3 \quad\Rightarrow\quad a \:=\:\pm\sqrt{3}\)

\(\displaystyle \text{Substitute into [4]: }\:b \:=\:-(\pm\sqrt{3}) \quad\Rightarrow\quad b \:=\:\mp\sqrt{3}\)


\(\displaystyle \text{If }t = \sqrt{3}\!:\;\begin{Bmatrix} x(\sqrt{3}) \;=\;3(\sqrt{3}) - (\sqrt{3})^3 \;=\;3\sqrt{3} - 3\sqrt{3} \;=\;0 \\ \\[-3mm] y(\sqrt{3}) \;=\;3(\sqrt{3})^2 \;=\;3\cdot3 \;=\;9 \end{Bmatrix}\)

\(\displaystyle \text{If }t = \text{-}\sqrt{3}\!:\;\begin{Bmatrix}x(\text{-}\sqrt{3}) \;=\;3(\text{-}\sqrt{3}) - (\text{-}\sqrt{3})^3 \;=\;\text{-}3\sqrt{3} + 3\sqrt{3} \;=\;0 \\ \\[-3mm] y(\text{-}\sqrt{3}) \;=\;3(\text{-}\sqrt{3})^2 \;=\;3\cdot 3 \;=\;9 \end{Bmatrix}\)


\(\displaystyle \text{The point of intersection is: }\,(0,9)\;\)

 

[tkhunny]

elaborate

Did you manage to think about it AT ALL before soroban dumped it on you?

 

[mmm4444bot]

elaborate

Please refrain from posting commands. We are not dogs.

Your learning experience will improve, if you exert sufficient effort to post specific questions about things that you do not understand.

Cheers

 

[renegade05]

elaborate

Please refrain from posting commands. We are not dogs.

Your learning experience will improve, if you exert sufficient effort to post specific questions about things that you do not understand.

Cheers

How is my question not a specific question? I didn't know how to find intersection points on parametric curves, so I asked specifically that.

I only ask questions on here after I have searched on Google, youtube and have tried numerous problems for many hours. So your assumption about me not "exerting sufficient effort" has no validity whatsoever.

In other words, yes, I did manage to think about it tkhunny. I tried for many hours to do it myself with your very brief post.

I posted "elaborate" because firstly, i was on my iPhone and typing with that thing sucks. Secondly, I felt tkhunny also didn't put much effort into his post.

Don't read into "elaborate" to much. I didn't mean it in a condescending way - it meant just that (to elaborate PLEASE). It wasn't a command.

 

[tkhunny]

No matter how you spin it (and I certainly believe the part about the iPhone), the single word does not show anyone that you did any work at all. In order to help you, we need to see what you can do and have done. We need to see you respond to a hint.

Time and effort is not the standard or the goal. What will cause your thinking to change so that you can figure it out in your own mind - This should be the goal. Obviously, not everyone agrees.

Let's see what else you have. In particular, do another one of these for us so we can help YOU do it.