## Problem Statement

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=t(t \cdot \cos t-2 \sin t) \\
y=t(t \cdot \sin t+2 \cos t)
\end{array}\right. \\
& t_{0}=\frac{\pi}{4}
\end{aligned}
$$
## Solution

Since $t_{0}=\frac{\pi}{4}$, then

$x_{0}=\frac{\pi}{4} \cdot\left(\frac{\pi}{4} \cdot \cos \frac{\pi}{4}-2 \sin \frac{\pi}{4}\right)=\frac{\pi}{4} \cdot\left(\frac{\pi}{4} \cdot \frac{\sqrt{2}}{2}-2 \cdot \frac{\sqrt{2}}{2}\right)=$

$=\frac{\pi}{4} \cdot\left(\frac{\pi \cdot \sqrt{2}}{8}-\sqrt{2}\right)$

$y_{0}=\frac{\pi}{4} \cdot\left(\frac{\pi}{4} \cdot \sin \frac{\pi}{4}+2 \cos \frac{\pi}{4}\right)=\frac{\pi}{4} \cdot\left(\frac{\pi}{4} \cdot \frac{\sqrt{2}}{2}+2 \cdot \frac{\sqrt{2}}{2}\right)=$

$=\frac{\pi}{4} \cdot\left(\frac{\pi \cdot \sqrt{2}}{8}+\sqrt{2}\right)$

Let's find the derivatives:

$x_{t}^{\prime}=(t(t \cdot \cos t-2 \sin t))^{\prime}=(t \cdot \cos t-2 \sin t)+t(\cos t-t \cdot \sin t-2 \cos t)=$ $=t \cdot \cos t-2 \sin t+t \cdot \cos t-t^{2} \cdot \sin t-2 t \cdot \cos t=-2 \sin t-t^{2} \cdot \sin t=-\sin t \cdot\left(2+t^{2}\right)$ $y_{t}^{\prime}=(t(t \cdot \sin t+2 \cos t))^{\prime}=(t \cdot \sin t+2 \cos t)+t(\sin t+t \cdot \cos t-2 \sin t)=$ $=t \cdot \sin t+2 \cos t+t \cdot \sin t+t^{2} \cdot \cos t-2 t \cdot \sin t=2 \cos t+t^{2} \cdot \cos t=\cos t \cdot\left(2+t^{2}\right)$ $y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{\cos t \cdot\left(2+t^{2}\right)}{-\sin t \cdot\left(2+t^{2}\right)}=-\operatorname{ctg} t$

Then

$y_{0}^{\prime}=-\operatorname{ctg} \frac{\pi}{4}=-1$

Equation of the tangent line:

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

$y-\frac{\pi}{4} \cdot\left(\frac{\pi \cdot \sqrt{2}}{8}+\sqrt{2}\right)=-1 \cdot\left(x-\frac{\pi}{4} \cdot\left(\frac{\pi \cdot \sqrt{2}}{8}-\sqrt{2}\right)\right)$

$y=-x+\frac{\pi^{2} \cdot \sqrt{2}}{32}-\frac{\pi \cdot \sqrt{2}}{4}+\frac{\pi^{2} \cdot \sqrt{2}}{32}+\frac{\pi \cdot \sqrt{2}}{4}$

$y=-x+\frac{\pi^{2} \cdot \sqrt{2}}{16}$

Equation of the normal line:

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

$y-\frac{\pi}{4} \cdot\left(\frac{\pi \cdot \sqrt{2}}{8}+\sqrt{2}\right)=-\frac{1}{-1} \cdot\left(x-\frac{\pi}{4} \cdot\left(\frac{\pi \cdot \sqrt{2}}{8}-\sqrt{2}\right)\right)$

$y=x-\frac{\pi^{2} \cdot \sqrt{2}}{32}+\frac{\pi \cdot \sqrt{2}}{4}+\frac{\pi^{2} \cdot \sqrt{2}}{32}+\frac{\pi \cdot \sqrt{2}}{4}$

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

## Problem Kuznetsov Differentiation 17-7