Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a prolate cycloid: $$\begin{align*} x &= 2 - \pi\cos(t)\\ y &= 2t - \pi\sin(t) \end{align*}$$ over the interval $-\pi \leq t \leq \pi$, crossed itself at point $P$ on the $x$-axis

a) Find the equations of the 2 tangent lines at $P$

b) find the points on curve where tangent line is horizontal.

c) find the point on curve where tangent line is vertical.

So I know for part $b$ and $c$, you just need to use the derivatives $dx$ and $dy$ and then solve for when $dx=0$ and $dy=0$... however for part a), do I just simply take the derivative of both $x$ and $y$ and that is my solution?

share|cite|improve this question

Note that $dx$ is not a derivative, it's a differential. Same with $dy$.

What you mean, presumably, is that you will take $\frac{dx}{dt}$ and $\frac{dy}{dt}$ for parts (b) and (c). Note that $$\frac{dy}{dx} = \frac{\quad\frac{dy}{dt}\quad}{\frac{dx}{dt}}$$ so you can use this for (a), (b), and (c). For (a), this can be used to get the slope of the tangent, but to find the equation of the tangent you'll have to do a bit more work. For (b), you want $\frac{dy}{dx}$ to be $0$, so you want $\frac{dy}{dt}=0$ and $\frac{dx}{dt}\neq 0$. For (c), you want $\frac{dy}{dt}\neq 0$ and $\frac{dx}{dt}=0$.

share|cite|improve this answer
Alright, so I think I got part b and c down alright.. but for part a, i solved for dy/dx and I get (2-picos(t))/(pisin(t))... but not sure what the next step is to find the EQN of the tangent? – Nick Jun 10 '12 at 19:59
@Nick: If you know the point and you know the slope(s), you should be able to get the equation. – Arturo Magidin Jun 10 '12 at 20:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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