## Problem Statement

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

\[
\left\{
\begin{array}{l}
x=\ln \left(1+t^{2}\right) \\
y=t-\operatorname{arctg} t
\end{array}
\right.
\]

$t_{0}=1$
## Solution

Since $t_{0}=1$, then

$x_{0}=\ln \left(1+1^{2}\right)=\ln 2$

$y_{0}=1-\operatorname{arctg} 1=1-\frac{\pi}{4}=\frac{4-\pi}{4}$

Let's find the derivatives:

$x_{t}^{\prime}=\left(\ln \left(1+t^{2}\right)\right)^{\prime}=\frac{1}{1+t^{2}} \cdot 2 t=\frac{2 t}{1+t^{2}}$

$y_{t}^{\prime}=(t-\operatorname{arctg} t)^{\prime}=1-\frac{1}{1+t^{2}}=\frac{1+t^{2}-1}{1+t^{2}}=\frac{t^{2}}{1+t^{2}}$

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{t^{2}}{1+t^{2}}\right) /\left(\frac{2 t}{1+t^{2}}\right)=\frac{t^{2}}{2 t}=\frac{t}{2}$

Then:

$y_{0}^{\prime}=\frac{1}{2}$

Equation of the tangent line:

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

$y-\frac{4-\pi}{4}=\frac{1}{2} \cdot(x-\ln 2)$

$y=\frac{x}{2}-\frac{\ln 2}{2}+\frac{4-\pi}{4}$

$y=\frac{x}{2}+\frac{4-\pi-2 \ln 2}{4}$

$y=\frac{x}{2}+\frac{4-\pi-\ln 4}{4}$

Equation of the normal line:

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

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

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

$y=-2 x+2 \ln 2+\frac{4-\pi}{2}$

$y=-2 x+\frac{4-\pi+4 \ln 2}{2}$

$y=-2 x+\frac{4-\pi+\ln 16}{2}$

## Problem Kuznetsov Differentiation 17-20