Eliminate Parameter with Sin and Cos

[Mikeylikesit182]

x = 2 − π cos t
y = 2t − π sin t
−π ≤ t < π
I understand how to eliminate parameter using sin^2 + cos^2 = 1.
I can't figure out how to deal with the "2t" in the y equation, if you solve for sin(t) and square, you get
((2t-y)/π )^2 which leaves the parameter. Is there a way to get it into the argument of sin?
I need this to then figure out intersection points and tangents later. Thanks in advance for any insight!

 

[fzero]

x = 2 − π cos t
y = 2t − π sin t
−π ≤ t < π
I understand how to eliminate parameter using sin^2 + cos^2 = 1.
I can't figure out how to deal with the "2t" in the y equation, if you solve for sin(t) and square, you get
((2t-y)/π )^2 which leaves the parameter. Is there a way to get it into the argument of sin?
I need this to then figure out intersection points and tangents later. Thanks in advance for any insight!

You will not be able to completely eliminate $t$, but you can use these 2 equations to write a single equation that depends on $x,y$ and $t$ (but not $\sin t,\cos t$). There is an interesting geometric interpretation of the equation that we can discuss after you try to find it.

 

[Mikeylikesit182]

You will not be able to completely eliminate $t$, but you can use these 2 equations to write a single equation that depends on $x,y$ and $t$ (but not $\sin t,\cos t$). There is an interesting geometric interpretation of the equation that we can discuss after you try to find it.

Thank you for the help! I can't figure what that equation would be ( I can see one involving arccos and substitution but I think that would be a dead end.)
I thought maybe finding the derivative dy/dx for the parametric eqn and realizing that at the point of intersection the 2 slopes would be dy/dx and -dy/dx might lead me closer?

 

[fzero]

Thank you for the help! I can't figure what that equation would be ( I can see one involving arccos and substitution but I think that would be a dead end.)
I thought maybe finding the derivative dy/dx for the parametric eqn and realizing that at the point of intersection the 2 slopes would be dy/dx and -dy/dx might lead me closer?

Using the derivatives might be of help later, but for now do something simpler. Rewrite the equations in the form

$$\begin{split}
x -2 & = -\pi \cos t, \\
y -2t & = -\pi \sin t\end{split}$$

and then square both and add them (you were close to this in your original attempt). Can you interpret the resulting quadratic equation?

 

[Mikeylikesit182]

Using the derivatives might be of help later, but for now do something simpler. Rewrite the equations in the form

$$\begin{split}
x -2 & = -\pi \cos t, \\
y -2t & = -\pi \sin t\end{split}$$

and then square both and add them (you were close to this in your original attempt). Can you interpret the resulting quadratic equation?

(x-2)^2 = (pi^2)(cos^2(t))
(y-2t)^2 = (pi^2)(sin^2(t))
added =
(x-2)^2 + (y-2t)^2 = (pi^2)(cos^2(t)) + (pi^2)(sin^2(t))
(x-2)^2 + (y-2t)^2 = (pi^2)[(cos^2(t)) + (sin^2(t))]
(x-2)^2 + (y-2t)^2 = (pi^2)
[(x-2)^2]/(pi^2) + [(y-2t)^2]/(pi^2) = 1

It seems to look like an ellipse adjusted upwards by the parameter? I really appreciate the help!

 

[fzero]

Correct. To be precise, it's actually a circle with radius $\pi$.

 

[Mikeylikesit182]

Correct. To be precise, it's actually a circle with radius $\pi$.

So it seems at this point to find where the graph would intersect itself, we would ask what value for t would give us 2 sets of identical x,y coordinates?

 

[fzero]

So it seems at this point to find where the graph would intersect itself, we would ask what value for t would give us 2 sets of identical x,y coordinates?

Actually you want to find values $t_1$ and $t_2$ for which $( x(t_1),y(t_1))=( x(t_2),y(t_2))$. In this case there is only one pair of such $t_1,t_2$, but in general there can be more.

Also, I should warn that the graph itself is not a circle, despite the appearance of that equation above.

 

[Svein]