## Problem Statement

Derive the equations of the tangent and normal to the curve at the point corresponding to the parameter value $t=t_{0}$.

\[
\left\{
\begin{array}{l}
x=a(t-\sin t) \\
y=a(1-\cos t)
\end{array}
\right.
\]

$t_{0}=\frac{\pi}{3}$
## Solution

Since $t_{0}=\frac{\pi}{3}$, then

$x_{0}=a\left(\frac{\pi}{3}-\sin \frac{\pi}{3}\right)=a \cdot\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right)$

$y_{0}=a\left(1-\cos \frac{\pi}{3}\right)=a \cdot\left(1-\frac{1}{2}\right)=\frac{a}{2}$

Find the derivatives:

$x_{t}^{\prime}=(a(t-\sin t))^{\prime}=a(1-\cos t)$

$y_{t}^{\prime}=(a(1-\cos t))^{\prime}=a \cdot \sin t$

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{a \cdot \sin t}{a(1-\cos t)}=\frac{\sin t}{1-\cos t}$

Then:

$y_{0}^{\prime}=\frac{\sin \frac{\pi}{3}}{1-\cos \frac{\pi}{3}}=\frac{\frac{\sqrt{3}}{2}}{1-\frac{1}{2}}=\sqrt{3}$

Equation of the tangent line:

$y-y_{0}=y_{0}^{\prime}\left(x-x_{0}\right)$

$y-\frac{a}{2}=\sqrt{3} \cdot\left(x-a \cdot\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right)\right)$

$y=\sqrt{3} \cdot x-\frac{\sqrt{3} \cdot \pi \cdot a}{3}+\frac{3 a}{2}+\frac{a}{2}$

$y=\sqrt{3} \cdot x-\frac{\sqrt{3} \cdot \pi \cdot a}{3}+2 a$

$y=\sqrt{3} \cdot x-\left(\frac{\pi}{\sqrt{3}}+2\right) \cdot a$

## Equation of the normal line:

$$
\begin{aligned}
& y-y_{0}=-\frac{1}{y_{0}^{\prime}}\left(x-x_{0}\right) \\
& y-\frac{a}{2}=-\frac{1}{\sqrt{3}}\left(x-a \cdot\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right)\right) \\
& y=-\frac{x}{\sqrt{3}}+\frac{a \cdot \pi}{3 \sqrt{3}}-\frac{a}{2}+\frac{a}{2}
\end{aligned}
$$

$y=-\frac{x}{\sqrt{3}}+\frac{a \cdot \pi}{3 \sqrt{3}}$

## Kuznetsov Differentiation Problem 17-3