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Could anyone help me with this

$x = 1 + \cos t$, $y = −2 + \sin t$, $π ≤ t ≤ 2π$;

$x = t$, $y = −2 −\sqrt{2t − t^2}$, $0 ≤ t ≤ 2$

For the following parametric equations, how do I determine whether they both represent the same curve? And how to represent the curve in a single equation? What I did is to sub in the value of x and y from both equations to find the value of t, but i don't know how to proceed. Or should i draw the curve out and check whether it is the same curve?

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2 Answers 2

up vote 5 down vote accepted

$x = 1 + \cos t, \ \ \ y = −2 + \sin t, \ \ \ \pi \leq t \leq 2 \pi$:

$$(x-1)^2+(y+2)^2=(1 + \cos t-1)^2+(−2 + \sin t+2)^2=\cos^2 t+\sin^2 t=1 \\ \Rightarrow (x-1)^2+(y+2)^2=1$$

$x = t, \ \ \ y = −2 −\sqrt{2t − t^2}, \ \ \ 0 \leq t \leq 2$:

$$(x-1)^2+(y+2)^2=(t-1)^2+(−2 −\sqrt{2t − t^2}+2)^2=t^2-2t+1+2t-t^2=1 \\ \Rightarrow (x-1)^2+(y+2)^2=1$$

Therefore, the parametric equations represent the same curve.

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3  
thanks for the answer –  ys wong Aug 16 at 10:59

Let $$\begin{align} 1+\cos t&=T\\ \Rightarrow \cos t&=T-1\\ \Rightarrow \sin t&=\sqrt{1-(T-1)^2} =\sqrt{2T-T^2}\\ \Rightarrow -2+\sin t&=-2+\sqrt{2T-T^2} \end{align}$$

Therefore the parametric equations $$x=1+\cos t,\quad y=-2+\sin t$$ and $$x=T,\quad y=-2+\sqrt{2T-T^2}\\ \text{or}\\x=t,\quad y=-2+\sqrt{2t-t^2}\ $$ represent the same curve.

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