## Problem Statement

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=\frac{1+t}{t^{2}} \\
y=\frac{3}{2 t^{2}}+\frac{2}{t}
\end{array}\right. \\
& t_{0}=2
\end{aligned}
$$
## Solution

Since $t_{0}=2$, then

$x_{0}=\frac{1+2}{2^{2}}=\frac{3}{4}$
$y_{0}=\frac{3}{2 \cdot 2^{2}}+\frac{2}{2}=\frac{3}{8}+1=\frac{11}{3}$

Let's find the derivatives:

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

Then:

$y_{0}^{\prime}=\frac{3+2 \cdot 2}{2+2}=\frac{7}{4}$

Equation of the tangent line:

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

$y-\frac{11}{3}=\frac{7}{4} \cdot\left(x-\frac{3}{4}\right)$

$y=\frac{7 x}{4}-\frac{21}{16}+\frac{11}{3}$

$y=\frac{7 x}{4}-\frac{63}{48}+\frac{176}{48}$

$y=\frac{7 x}{4}+\frac{113}{48}$

$y=\frac{7 x}{4}+2 \frac{17}{48}$

Equation of the normal line:

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

$y-\frac{11}{3}=-\frac{1}{\left(\frac{7}{4}\right)} \cdot\left(x-\frac{3}{4}\right)$

$y=-\frac{4 x}{7}+\frac{3}{7}+\frac{11}{3}$

$y=-\frac{4 x}{7}+\frac{9}{21}+\frac{77}{21}$

$y=-\frac{4 x}{7}+\frac{86}{21}$

$y=-\frac{4 x}{7}+4 \frac{2}{21}$

## Problem Kuznetsov Differentiation $17-15$