## Problem Statement

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

\[
\left\{\begin{array}{l}
x=2 e^{t} \\
y=e^{-t}
\end{array}\right.
\]

$t_{0}=0$
## Solution

Since $t_{0}=0$, then

$x_{0}=2 e^{0}=2$

$y_{0}=e^{-0}=1$

Let's find the derivatives:

$x_{t}^{\prime}=\left(2 e^{t}\right)^{\prime}=2 e^{t}$

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

Then:

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

Equation of the tangent line:

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

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

$y=-\frac{1}{2} \cdot x+2$

Equation of the normal line:

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

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

$y=2 x-3$

## Problem Kuznetsov Differentiation 17-31