## Problem Statement

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

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

$t_{0}=0$
## Solution

Since $t_{0}=0$, then

$x_{0}=0 \cdot(1-\sin 0)=0$

$y_{0}=0 \cdot \cos 0=0$

Let's find the derivatives:

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

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

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

Then:

$y_{0}^{\prime}=\frac{0 \cdot \cos 0-\sin 0}{1-\sin 0-0 \cdot \cos 0}=\frac{0}{1}=0$

The equation of the tangent line:

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

$y-0=0 \cdot(x-0)$

$y=0$

Since $y_{0}^{\prime}=0$, the equation of the normal line:

$x=x_{0}$

$x=0$

## Problem Kuznetsov Differentiation 17-21