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A question in my text book (I am self-teaching) begins: The parametric equations of a curve are $x=\cos 2t, y=4 \sin t$. Sketch the curve for $0 \leq t\leq \frac{1}{2}\pi $

I have proceeded as follows:

enter image description here

But when I plot in Desmos I get:

enter image description here

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  • $\begingroup$ Well as you can see from the last column, the point $(-1,4)$ must be on the curve, which is the case in the Desmos plot but doesn't seem to be the case in your plot. $\endgroup$ Jul 24, 2021 at 10:13
  • $\begingroup$ Your table is correct; you just drew the graph wrong. Which $(x,y)$ points were you trying to graph? $\endgroup$
    – TonyK
    Jul 24, 2021 at 10:18
  • $\begingroup$ you just plotted the points incorrectly. I just plot the same points on a sheet of paper and I get the correct curve. Also note that your desmos graph is not correct either. For the given interval of $t$, $y$ must be positive. In Desmos, you should put $t \in (0, \pi/2)$ $\endgroup$
    – Math Lover
    Jul 24, 2021 at 10:23

3 Answers 3

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It will be better if we can convert the equations from parametric from to cartesian form

Now

$x=\cos 2t$ and y = $4\sin t$

$\displaystyle x=1-2\sin^2t=1-2(\frac{y}{4})^2=1-\frac{y^2}{8}$

Therefore $\frac{y^2}{8}=1-x$ which is a parabola opening towards negative $x-$ axis and can be graphed easily.

Also since $t\in [0,\frac{\pi}{2}]$, therefore domain of function is $[-1,1]$

The correct graph is the following one :

enter image description here

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While you can covert it into equation of parabola in terms of $(x, y)$, there is nothing wrong with the approach you took to sketch. In fact you found points correctly but you plotted them incorrectly. Please see below Desmos graph that shows the curve and I have also marked the points from your table on it.

enter image description here

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  • $\begingroup$ It would be better to write $2\sqrt2$ instead of $\dfrac4{\sqrt2}.$ $\endgroup$
    – Angelo
    Jul 24, 2021 at 11:46
  • $\begingroup$ Thanks. I can't believe I didn't see it! It's so obvious I almost feel ashamed to have put the question up. But I stared at it for over an hour and didn't see that the true connection was between x and y, not t and anything else. I should have gone for a walk and come back. $\endgroup$
    – Steblo
    Jul 24, 2021 at 11:55
  • $\begingroup$ @Steblo happens:) also my suggestion - when you sketch, first draw both x and y axes and mark $0, 1, 2, ..$ etc on both axes. $\endgroup$
    – Math Lover
    Jul 24, 2021 at 11:58
  • $\begingroup$ I did. The one I posted was just to show the shape $\endgroup$
    – Steblo
    Jul 24, 2021 at 12:00
  • $\begingroup$ @Angelo of course but I wanted to exactly write the way OP did in their table. $\endgroup$
    – Math Lover
    Jul 24, 2021 at 12:00
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Hint: Since $$\begin{cases}x = 1-2\sin^2t \\ y = 4\sin t\end{cases}$$ we have $$x = 1 - \frac{y^2}{8}$$

Cut the graph with the following restrictions: $$\begin{cases} -1 \le (\cos2t)_x \le 1, \\ 0 \le (4\sin t)_y \le 4\end{cases}$$ since $0 \le t \le \pi/2$.

The correct graph is the following one :

enter image description here

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  • $\begingroup$ Your graph is too big: $y=4\sin t$, so $|y|\le 4$. $\endgroup$
    – TonyK
    Jul 24, 2021 at 10:17
  • $\begingroup$ Since $0\leqslant t\leqslant\dfrac{\pi}2$, actually $0\leqslant y\leqslant4$. $\endgroup$
    – Angelo
    Jul 24, 2021 at 10:19
  • $\begingroup$ @TonyK Sure, that was why I said it was a hint. $\endgroup$
    – VIVID
    Jul 24, 2021 at 10:21
  • $\begingroup$ The correct graph is the arc of the parabola $x=1-\dfrac{y^2}8$ from the point $(1,0)$ to the point $(-1,4)$. The correct graph is the following one : i.stack.imgur.com/jlzNX.jpg $\endgroup$
    – Angelo
    Jul 24, 2021 at 10:31
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    $\begingroup$ VIVID, I pointed out your mistake, and you threw it back in my face. ("What, me wrong? Impossible!") So of course I downvoted. $\endgroup$
    – TonyK
    Jul 24, 2021 at 13:35

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