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I have a task given to me in my homework I can not figure out what asks of me. The task is worded like this:

A curve in a plane is given by

$$ x(t) = 3(t - \sin(t)) $$

$$ y(t) = 3(1 - \cos(t)) $$

Find the parametric normal-curve $x(s, t), y(s, t)$ through $(x(t), y(t))$, where $s$ is the normal's parameter.

I am not interested in a solution to this task, just an interpretation of what it means so that I will be able to solve it myself. Thanks in advance =)

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This is how I interpreted the problem:

Consider an arbitrary point $(x_t,y_t)$ on the curve. There is a unique line that passes through that point and is normal (perpendicular) to the curve. Find a parametrization $(u_t(s),v_t(s))$ of that normal line. Now your parametric normal curve will be

$$ x(s,t) = u_t(s),\\ y(s,t) = v_t(s). $$

However, this interpretation fails at the cusps of the curve. For example, at $(6\pi,0)=(x(2\pi),y(2\pi))$ there is no normal line.

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  • $\begingroup$ So what you say is that it asks me to find a general parametric equation that for every point on the curve $t$ gives me a normal-line through that point by varying the parameter $s$? $\endgroup$ – Andreas Hagen Mar 24 '13 at 17:18
  • $\begingroup$ Yes, that's how I interpreted it. However, since this interpretation doesn't make sense at every point on the curve, I recommend asking for clarification from the person who gave you the question. $\endgroup$ – Snowball Mar 24 '13 at 17:20
  • $\begingroup$ Yes, I would have asked my teacher in the first place if it was not for the fact that it is easter-vacation here. Thanks for your two cents on this =) $\endgroup$ – Andreas Hagen Mar 24 '13 at 17:32

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