## Problem Statement

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=\sqrt{3} \cdot \cos t \\
y=\sin t
\end{array}\right. \\
& t_{0}=\frac{\pi}{3}
\end{aligned}
$$
## Solution

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

$x_{0}=\sqrt{3} \cdot \cos \frac{\pi}{3}=\frac{\sqrt{3}}{2}$

$y_{0}=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$

Let's find the derivatives:

$x_{t}^{\prime}=(\sqrt{3} \cdot \cos t)^{\prime}=-\sqrt{3} \cdot \sin t$

$y_{t}^{\prime}=(\sin t)^{\prime}=\cos t$

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{\cos t}{-\sqrt{3} \cdot \sin t}=-\frac{\operatorname{ctg} t}{\sqrt{3}}$

Then:

$y_{0}^{\prime}=-\frac{\operatorname{ctg} \frac{\pi}{3}}{\sqrt{3}}=-\frac{1}{3}$

Equation of the tangent line:

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

$y-\frac{\sqrt{3}}{2}=-\frac{1}{3} \cdot\left(x-\frac{\sqrt{3}}{2}\right)$

$y=-\frac{x}{3}+\frac{\sqrt{3}}{6}+\frac{\sqrt{3}}{2}$

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

Equation of the normal line:

$y-y_{0}=-\frac{1}{y_{0}^{\prime}}\left(x-x_{0}\right)$

$y-\frac{\sqrt{3}}{2}=-\frac{1}{\left(-\frac{1}{3}\right)}\left(x-\frac{\sqrt{3}}{2}\right)$
$y=3 x-\frac{3 \sqrt{3}}{2}+\frac{\sqrt{3}}{2}$

$y=3 x-\sqrt{3}$

## Problem Kuznetsov Differentiation 17-2