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If there is a parametric equation $x=2\cos{2t}$ and $y=6\sin{t}$, $0\le t \le \frac{\pi}{2}$ the Cartesian equation is $y=3\sqrt{2-x}$.

How do I find the domain of the Cartesian equation? I tried:

$$0\le t \le \frac{\pi}{2}$$ $$2\cos{2(0)}\le 2\cos{2t} \le 2\cos{2(\frac{\pi}{2})}$$ $$2\cos{0}\le x \le 2\cos\pi$$ $$2(1)\le x \le 2(-1)$$ $$2\le x \le-2$$

Which can't be true, as both inequalities can't be satisfied at the same time?

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  • $\begingroup$ Rectify $\le$ with $\ge$ $\endgroup$ Jan 13, 2013 at 13:47
  • $\begingroup$ $x\leq y $ implies $f(x)\leq f(y)$ only if $f$ is an increasing function. Is $cos$ increasing in $[0,\frac{\pi}{2}]$? $\endgroup$
    – Git Gud
    Jan 13, 2013 at 13:48

1 Answer 1

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When $0\leq t\leq\pi/2$ then $0\leq2t\leq\pi$. Between $0$ and $\pi$ the function of cosine is decreasing so $$\cos(\pi)\leq\cos(2t)\leq\cos(0)$$ or $$-1\leq\cos(2t)\leq 1$$ then $$-2\leq2\cos(2t)\leq 2$$ or $$-2\leq x\leq 2$$.enter image description here

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  • $\begingroup$ Why does the function decreasing mean you must swap the inequalities? Also what would happen if in that range the function was both increasing and decreasing. $\endgroup$
    – Jonathan.
    Jan 13, 2013 at 14:04
  • $\begingroup$ You'd have to check case by case. Why does the function decreasing mean you must swap the inequalities? It's just the way it is: If $f$ is decreasing, then $x\leq y$ implies $f(y)\leq f(x)$. It's a consequence of the definition of decreasing function. $\endgroup$
    – Git Gud
    Jan 13, 2013 at 14:16
  • $\begingroup$ @Jonathan.: Whenever you have $I:a<x<b$ and a decreasing well-defined function on $I$, then $f(b)<f(x)<f(a)$. The cosine function $\cos(2t)$ is decreasing on $[0,\pi]$ and that's why the inequalities were swapped. $\endgroup$
    – Mikasa
    Jan 13, 2013 at 14:19
  • $\begingroup$ @GitGud: If $0\leq 2t$ then $\cos(2t)\leq\cos(0)$ and if $2t\leq\pi$ then $\cos(\pi)\leq\cos(2t)$. Now consider the two resulted inequalities simultaneously on $[0,\pi]$. $\endgroup$
    – Mikasa
    Jan 13, 2013 at 14:27
  • $\begingroup$ @Jonathan.: Did you read my comment above? $\endgroup$
    – Mikasa
    Jan 13, 2013 at 16:08

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