## Problem Statement

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=\sin t \\
y=a^{t}
\end{array}\right. \\
& t_{0}=0
\end{aligned}
$$
## Solution

Since $t_{0}=0$, then

$x_{0}=\sin 0=0$

$y_{0}=a^{0}=1$

Let's find the derivatives:

$x_{t}^{\prime}=(\sin t)^{\prime}=\cos t$

$y_{t}^{\prime}=\left(a^{t}\right)^{\prime}=a^{t} \cdot \ln a$

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{a^{t} \cdot \ln a}{\cos t}$

Then:

$y_{0}^{\prime}=\frac{a^{0} \cdot \ln a}{\cos 0}=\frac{1 \cdot \ln a}{1}=\ln a$

Equation of the tangent line:

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

Equation of the normal line:

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

## Problem Kuznetsov Differentiation 17-29