91影视

The cycloid is$x=r胃-r\sin 胃,y=r-r\cos 胃,胃鈭�\mathrm{鈩�}$

Consider the cycloid, $x=r胃-r\sin 胃,y=r-r\cos 胃胃鈭�\mathrm{鈩�}$.

The objective is to prove that the cycloid has a vertical tangent at each even multiple of $\mathrm{蟺}$.

First find the derivatives of the parametric equations and equate them to zero to get the points where it is vertical or horizontal.

A vertical tangent line occurs when $\frac{dx}{dt}=0$w.

To find the vertical tangent line of the cycloid the denominator is zero and the numerator is not zero.

Take the equation, $x=r胃-r\sin 胃$.

Differentiate with respect to$胃$

$$\begin{gathered} \frac{dx}{d胃}=\frac{d}{d胃}(r胃-r\sin 胃) \\ \frac{dx}{d胃}=\frac{d}{d胃}r胃-r\frac{d}{d胃}\sin 胃 \\ \frac{dx}{d胃}=r-r\cos 胃 \end{gathered}$$

Now take the equation, $y=r-r\cos 胃$.

Differentiate with respect to$胃$.

$$\begin{gathered} \frac{dy}{d胃}=\frac{d}{d胃}(r-r\cos 胃) \\ \frac{dy}{d胃}=\frac{d}{d胃}r-r\frac{d}{d胃}\cos 胃 \\ \frac{dy}{d胃}=r\sin 胃 \end{gathered}$$

Now the derivative is, $\frac{dy}{dx}=\frac{\frac{dy}{d胃}}{\frac{dx}{d胃}}$

$$\begin{gathered} \frac{dy}{dx}=\frac{r\sin 胃}{r-r\cos 胃}\left[since\frac{dx}{d胃}=r-r\cos 胃,\frac{dy}{d胃}=r\sin 胃\right] \\ \frac{dy}{dx}=\frac{r\sin 胃}{r(1-\cos 胃)} \end{gathered}$$

Then,

$\frac{dy}{dx}=\frac{\sin 胃}{(1-\cos 胃)}$

Now the limit of the derivative at $2k蟺$.

$\begin{array}{rcl}\underset{胃鈫�2k{蟺}^{*}}{\lim }\frac{dy}{dx}& =& \underset{胃鈫�2k蟺}{\lim }\frac{\sin 胃}{(1-\cos 胃)}\\ & =& \underset{胃鈫�2kz}{\lim }\frac{\cos 胃}{\sin 胃}\text{By using L'Hospitals}\\ & =& 鈭�\end{array}$

Now the limit of the derivative at $2k{蟺}^{-}$.

For the vertical tangent line the denominator is zero.

$\begin{array}{rcl}\underset{胃鈫�2k{蟺}^{-}}{\lim }\frac{dy}{dx}& =& \underset{胃鈫�2k{蟺}^{-}}{\lim }\frac{\sin 胃}{(1-\cos 胃)}\\ & =& \underset{胃鈫�2k{蟺}^{-}}{\lim }\frac{\cos 胃}{\sin 胃}\text{By using L'Hospitals}\\ & =& -鈭�\end{array}$

Thus,

$\underset{胃鈫�2kx}{\lim }\frac{dy}{dx}$does not exists.

The cycloid has a vertical tangent line at even multiple of $胃$.

Hence Proved.