Find the equation of the tangent line at parameter values $\theta=\pi/6$ and $\theta =5\pi/4$ to the cycloid given by
$$x(t)=r\theta-r\sin \theta$$
and
$$y(t)= r-r\cos \theta$$
with $\theta\in [0,2pi]$
At which parameter values is the tangent line to the cycloid given above horizontal? At
which values is it vertical? Write down the equations of all horizontal and vertical tangent lines.Is the cycloid concave up or concave down? Prove your answers.
Can anyone give me the basic information on how to solve these kind of problem. or suggest a website i can have a look at?
Best Answer
If you have a parametric equation $x=x(t)$ and $y=y(t)$, then
$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.$$
In your case, you have $x(\theta)=r\theta-r\sin\theta$ and $y(\theta)=r-r\cos\theta$. Therefore, $\dfrac{dx}{d\theta} = \ldots$ and $\dfrac{dy}{d\theta} = \ldots$, and hence
$$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \ldots$$
Then you'll want to compute $\left.\dfrac{dy}{dx}\right|_{\theta=\pi/6}$ and $\left.\dfrac{dy}{dx}\right|_{\theta=5\pi/4}$ to get the slopes of the two tangent lines. Once you have those, you'll be able to compute the tangent line equations like usual using point-slope form; in particular, your two tangent line equations will be
$$y=\left(\left.\dfrac{dy}{dx}\right|_{\theta=\pi/6}\right)(x-x(\pi/6)) + y(\pi/6)$$
and
$$y=\left(\left.\dfrac{dy}{dx}\right|_{\theta=5\pi/4}\right)(x-x(5\pi/4)) + y(5\pi/4)$$
To see where the tangent line is horizontal, you'll want to solve $\dfrac{dy}{d\theta}=0$.
To see where the tangent line is vertical, you'll want to solve $\dfrac{dx}{d\theta}=0$.
To see if the cycloid is concave up or down, you'll need to compute the second derivative:
$$\frac{d^2y}{dx^2} = \dfrac{d}{dx}\left(\frac{dy}{dx}\right) = \dfrac{\dfrac{d}{d\theta}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{d\theta}}$$
and see where it's positive and/or negative.
I hope this provides you with enough information to at least start the problem.