## Problem Statement

Derive the equations of the tangent and normal to the curve at the point corresponding to the parameter value $t=t_{0}$
\[
\left\{
\begin{array}{l}
x=a \cdot t \cdot \cos t \\
y=a \cdot t \cdot \sin t
\end{array}
\right.
\]

$t_{0}=\frac{\pi}{2}$
## Solution

Since $t_{0}=\frac{\pi}{2}$, then

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

$y_{0}=a \cdot \frac{\frac{\pi}{2}}{2} \cdot \sin \frac{\frac{\pi}{2}}{2}=\frac{a \cdot \pi}{2}$

Let's find the derivatives:

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

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

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

Then:

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

Equation of the tangent line:

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

$y-\frac{a \cdot \pi}{2}=-\frac{2}{\pi} \cdot(x-0)$

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

Equation of the normal line:

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

$y-\frac{a \cdot \pi}{2}=-\frac{1}{\left(-\frac{2}{\pi}\right)} \cdot(x-0)$

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