## Problem Statement

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

\[
\left\{\begin{array}{l}
x=\arcsin \frac{t}{\sqrt{1+t^{2}}} \\
y=\arccos \frac{1}{\sqrt{1+t^{2}}}
\end{array}\right.
\]

$t_{0}=-1$
## Solution

Since $t_{0}=-1$, then

$x_{0}=\arcsin \frac{-1}{\sqrt{1+(-1)^{2}}}=\arcsin \frac{-1}{\sqrt{2}}=-\frac{\pi}{4}$
$y_{0}=\arccos \frac{1}{\sqrt{1+(-1)^{2}}}=\arccos \frac{1}{\sqrt{2}}=\frac{\pi}{4}$

Find the derivatives:

$$
\begin{aligned}
& x_{t}^{\prime}=\left(\arcsin \frac{t}{\sqrt{1+t^{2}}}\right)^{\prime}=\frac{1}{\sqrt{1-\left(\frac{t}{\sqrt{1+t^{2}}}\right)^{2}}} \cdot \frac{1 \cdot \sqrt{1+t^{2}}-t \cdot \frac{1}{2 \sqrt{1+t^{2}}} \cdot 2 t}{1+t^{2}}= \\
& =\frac{\sqrt{1+t^{2}}}{1+t^{2}-t^{2}} \cdot \frac{1+t^{2}-t^{2}}{\left(1+t^{2}\right) \cdot \sqrt{1+t^{2}}}=\frac{1}{1+t^{2}}
\end{aligned}
$$

$$
\begin{aligned}
& y_{t}^{\prime}=\left(\arccos \frac{1}{\sqrt{1+t^{2}}}\right)^{\prime}=\frac{1}{\sqrt{1-\left(\frac{1}{\sqrt{1+t^{2}}}\right)^{2}}} \cdot \frac{1}{2 \sqrt{1+t^{2}}} \cdot 2 t= \\
& =\frac{\sqrt{1+t^{2}}}{\sqrt{1+t^{2}-1}} \cdot \frac{t}{\sqrt{1+t^{2}}}=\frac{t}{\sqrt{t^{2}}}=1 \\
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{1}{\left(\frac{1}{1+t^{2}}\right)}=1+t^{2}
\end{aligned}
$$

Then:

$y_{0}^{\prime}=1+(-1)^{2}=2$

Equation of the tangent line:

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

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

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

$y=2 x+\frac{3 \pi}{4}$

Equation of the normal line:

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

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

$y=-\frac{x}{2}-\frac{\pi}{8}+\frac{\pi}{4}$

$y=-\frac{x}{2}+\frac{\pi}{8}$

Kuznetsov Differentiation Problem 17-6