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    A curve is given by the parametric equations x=sint, y=tant.

    For : - pi/2 < t < pi/2

    Draw a graph of the curve.

    I did:

    t = -pi/2

    Put it into calc and got x = 1 but for y, I get math error.

    Slightly confused how to go about this, any help?
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    Well that means you have an asymptote at t=-\frac{\pi}{2} (incidentally there's also another one), which is why the < signs are there, rather than \le.

    Your best bet is to substitute in t=-\frac{\pi}{3}, -\frac{\pi}{4}, -\frac{\pi}{6}, 0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3} and look at the behaviour as t \to \pm\frac{\pi}{2}. When you've got a few points plotted, the shape of the graph should become clear.
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    (Original post by nuodai)
    Well that means you have an asymptote at t=-\frac{\pi}{2} (incidentally there's also another one), which is why the < signs are there, rather than \le.

    Your best bet is to substitute in t=-\frac{\pi}{3}, -\frac{\pi}{4}, -\frac{\pi}{6}, 0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3} and look at the behaviour as t \to \pm\frac{\pi}{2}. When you've got a few points plotted, the shape of the graph should become clear.
    The answers have (-1, -4) and (1, 4) through which the curve passes through. I don't know how they get this though?
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    (Original post by Deep456)
    The answers have (-1, -4) and (1, 4) through which the curve passes through. I don't know how they get this though?
    Bizarre... well it doesn't :p:
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    (Original post by nuodai)
    Bizarre... well it doesn't :p:
    Thanks. Back to C4.
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    Alternatively note that:
    \sin t = x \implies \cos t = \sqrt{1-x^2} for -\frac{\pi}{2}&lt;t&lt;\frac{\pi}{2}.

    Also note that this condition on t implies that -1&lt;x&lt;1, y \in \mathbb{R}.

    Therefore y=\tan t = \frac{\sin t}{\cos t} \implies y = \frac{x}{\sqrt{1-x^2}}.

    Which clearly passes through the origin and has asymptotes at x=\pm 1.

    Additionally \frac{dy}{dx}=\cos ^3t, so by considering the value of this as t varies, we can find out how the gradient behaves over the required interval.
 
 
 
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Updated: December 23, 2010
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