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 =)


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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