## Problem Statement

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

\[
\left\{
\begin{array}{l}
x=\frac{1+\ln t}{t^{2}} \\
y=\frac{3+2 \ln t}{t}
\end{array}
\right.
\]

$t_{0}=1$
## Solution

Since $t_{0}=1$, then

$x_{0}=\frac{1+\ln 1}{1^{2}}=1$

$y_{0}=\frac{3+2 \ln 1}{1}=3$

Let's find the derivatives:

$$
\begin{aligned}
& x_{t}^{\prime}=\left(\frac{1+\ln t}{t^{2}}\right)^{\prime}=\frac{\frac{1}{t} \cdot t^{2}-(1+\ln t) \cdot 2 t}{t^{4}}= \\
& =\frac{1-2-2 \ln t}{t^{3}}=\frac{-1-2 \ln t}{t^{3}} \\
& y_{t}^{\prime}=\left(\frac{3+2 \ln t}{t}\right)^{\prime}=\frac{\frac{2}{t} \cdot t-(3+2 \ln t) \cdot 1}{t^{2}}=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{2-3-2 \ln t}{t^{2}}=\frac{-1-2 \ln t}{t^{2}} \\
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{-1-2 \ln t}{t^{2}}\right) /\left(\frac{-1-2 \ln t}{t^{3}}\right)=\frac{t^{3}}{t^{2}}=t
\end{aligned}
$$

Then:

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

Equation of the tangent line:

$$
\begin{aligned}
& y-y_{0}=y_{0}^{\prime}\left(x-x_{0}\right) \\
& y-3=1 \cdot(x-1) \\
& y=x-1+3 \\
& y=x+2
\end{aligned}
$$

Equation of the normal line:

$$
\begin{aligned}
& y-y_{0}=-\frac{1}{y_{0}^{\prime}}\left(x-x_{0}\right) \\
& y-3=-\frac{1}{1} \cdot(x-1) \\
& y=-x+1+3 \\
& y=-x+4
\end{aligned}
$$

## Problem Kuznetsov Differentiation 17-14