Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer
thanks for the answer – ys wong Aug 16 '14 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.