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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?

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1 Answer 1

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 $\frac{dx}{dt}=0$.

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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

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