Nickolas Taylor

Answered

2022-07-04

Show that $x(t)=2r{\mathrm{cos}}^{2}(t)$ , $y(t)=2r\mathrm{sin}t\mathrm{cos}t$ is a regular parametrization of the real circle of radius r, centre (r, 0).

Answer & Explanation

zlepljalz2

Expert

2022-07-05Added 22 answers

Step 1

Because of the identities ${\mathrm{cos}}^{2}u=(1+\mathrm{cos}2u)/2$ and $\mathrm{sin}2u=2\mathrm{sin}u\mathrm{cos}u$ sinucosu, your parametric equations are equivalent to

$\begin{array}{rl}x(t)& =r+r\mathrm{cos}2t\\ y(t)& =r\mathrm{sin}2t\end{array}$x(t)y(t)=r+rcos2t=rsin2t

And now you can see that this describes the circle centred (r, 0) with radius r.

Yes, the parameterisation is twice as fast as you'd expect (with 2t in the expressions), but this is okay as parametrisations are not unique.

Because of the identities ${\mathrm{cos}}^{2}u=(1+\mathrm{cos}2u)/2$ and $\mathrm{sin}2u=2\mathrm{sin}u\mathrm{cos}u$ sinucosu, your parametric equations are equivalent to

$\begin{array}{rl}x(t)& =r+r\mathrm{cos}2t\\ y(t)& =r\mathrm{sin}2t\end{array}$x(t)y(t)=r+rcos2t=rsin2t

And now you can see that this describes the circle centred (r, 0) with radius r.

Yes, the parameterisation is twice as fast as you'd expect (with 2t in the expressions), but this is okay as parametrisations are not unique.

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