## Problem Statement

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

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

$t_{0}=\frac{\pi}{4}$
## Solution

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

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

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

Let's find the derivatives:

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

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

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

Then:

$y_{0}^{\prime}=\tan \frac{\pi}{4}=1$

Equation of the tangent line:

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

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

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

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

$y=x+\frac{\sqrt{2} \cdot a \cdot \pi}{4}$

Equation of the normal line:

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

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

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

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

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

## Problem Kuznetsov Differentiation 17-17