Tangent line of parametric curve I have not seen a problem like this so I have no idea what to do.
Find an equation of the tangent to the curve at the given point by two methods, without elimiating parameter and with.
$$x = 1 + \ln t,\;\; y = t^2 + 2;\;\; (1, 3)$$
I know that $$\dfrac{dy}{dx} = \dfrac{\; 2t\; }{\dfrac{1}{t}}$$
But this give a very wrong answer. I am not sure what a parameter is or how to eliminate it.
 A: *

*Eliminating the parameter $t$. The given system of two parametric equations $$\begin{eqnarray*}\left\{ 
\begin{array}{c}
x=1+\ln t \\ 
y=t^{2}+2
\end{array}
\right. \end{eqnarray*} \tag{A}$$ is equivalent successively to
$$\begin{eqnarray*}
\left\{ 
\begin{array}{c}
x-1=\ln t \\ 
y=t^{2}+2
\end{array}
\right. \Leftrightarrow \left\{ 
\begin{array}{c}
t=e^{x-1} \\ 
y=t^{2}+2
\end{array}
\right.  
\Leftrightarrow \left\{ 
\begin{array}{c}
t=e^{x-1} \\ 
y=\left( e^{x-1}\right) ^{2}+2
\end{array}
\right. \end{eqnarray*}$$ and finally to $$\left\{ 
\begin{array}{c}
t=e^{x-1} \\ 
y=e^{2\left( x-1\right) }+2.
\end{array}
\right. \tag{B}
$$
We have thus eliminated the parameter $t$. 
Here is a plot of $y=e^{2\left( x-1\right) }+2$

Differentiating the equation of the curve $$y=e^{2\left( x-1\right) }+2\tag{C}$$ gives by the chain rule applied to $e^u$, with $u=2(x-1)$
$$\begin{eqnarray*}
\frac{dy}{dx} &=&\frac{d}{dx}\left( e^{2\left( x-1\right) }+2\right) =\frac{d
}{dx}e^{2\left( x-1\right) }=\left( \frac{
d}{du}e^{u}\right) \frac{du}{dx} 
=e^{u}\times 2=2e^{2\left( x-1\right) }.
\end{eqnarray*}$$
At $x=1$ this derivative has the value 
$$\begin{equation*}
\left. \frac{dy}{dx}\right\vert _{x=1}=\left. 2e^{2\left( x-1\right)
}\right\vert _{x=1}=2e^{2(1-1)}=2.\tag{1}
\end{equation*}$$
Since at $x=1$, $$y=e^{2(1-1)}+2=3,$$ we have confirmed that the point $(x,y)=(1,3)$ belongs to the graph of the curve given by equation $(C)$. Hence the equation of the tangent line to the graph of the curve at $(1,3)$ is
$$\begin{equation*}
y-3=2(x-1)\Leftrightarrow y=2x+1\tag{2}
\end{equation*}$$

*Without eliminating the parameter $t$. (Reformulated in view of OP's comment.) To compute the derivative we use now the parametric equations $(A)$ and the formula $$\frac{dy}{dx} =\frac{dy}{dt}\frac{dt}{dx}=\frac{dy}{dt}/\frac{dx}{dt}.\tag{D}$$  We have
$$\begin{eqnarray*}
\frac{dy}{dx} &=&\frac{dy}{dt}/\frac{dx}{dt}=
\left( \frac{d}{dt}\left( t^{2}+2\right) \right) /\frac{d}{dt}\left( 1+\ln
t\right)  \\
&=&2t/\frac{1}{t}=2t^2,\tag{3}
\end{eqnarray*}$$
which confirms your result. Since for $x=1$ the equation $$x=1+\ln t $$ gives $$1=1+\ln t\Leftrightarrow 0=\ln t \Leftrightarrow t=1,$$   we get the same value as in $(1)$ for the derivative $$\begin{equation*}
\left. \frac{dy}{dx}\right\vert _{x=1}=\left. 2t^2\right\vert
_{t=1}=2\cdot 1^2=2.\tag{3a}
\end{equation*}$$
The equation of the tangent line is as above
$$\begin{equation*}
y=2x+1.\tag{4}
\end{equation*}$$
In terms of the parameter $t$ the tangent line at $t=1$, i.e. at $(x,y)=(1,3)$ is given by the parametric equations
$$\begin{equation*}
\left\{ 
\begin{array}{c}
x=t \\ 
y=2t+1,
\end{array}
\right.\tag{4a} 
\end{equation*}$$
because
$$\begin{equation*}
\left. \frac{dx}{dt}\right\vert _{t=1}=\left. \frac{1}{t}\right\vert _{t=1}=1
\end{equation*}$$
and
$$\begin{equation*}
\left. \frac{dy}{dt}\right\vert _{t=1}=\left. 2t\right\vert _{t=1}=2.
\end{equation*}$$
A: One way to do this is by considering the parametric form of the curve:
$(x,y)(t) = (1 + \log t, t^2 + 2)$, so $(x,y)'(t) = (\frac{1}{t}, 2t)$
We need to find the value of $t$ when $(x,y)(t) = (1 + \log t, t^2 + 2) = (1,3)$, from where we deduce $t=1$. The tangent line at $(1,3)$ has direction vector $(x,y)'(1) = (1,2)$, and since it passes by the point $(1,3)$ its parametric equation is given by: $s \mapsto (1,2)t + (1,3)$.
Another way (I suppose this is eliminating the parameter) would be to express $y$ in terms of $x$ (this can't be done for any curve, but in this case it is possible). We solve for $t$: $x = 1 + \log x \Rightarrow x = e^{x-1}$, so $y = t^2 + 2 = (e^{x-1})^2 + 2 = e^{2x-2} + 2$. The tangent line has slope $\frac{dy}{dx}=y_x$ evaluated at $1$: we have $y_x=2e^{2x-2}$ and $y_x(1)=2e^0 = 2$, so it the line has equation $y=2x +b$. Also, it passes by the point $(1,3)$, so we can solve for $b$: $3 = 2 \cdot 1 + b \Rightarrow b = 1$. Then, the equation of the tangent line is $y = 2x + 1$.
Note that $s \mapsto (1,2)t + (1,3)$ and $y = 2x + 1$ are the same line, expressed in different forms.
A: Method 1 Eliminating. I think they want to write everything in terms of x first. 
$x = 1 + ln(t) \iff e^{x - 1} = t$
$y = t^2 + 2 = e^{2x -2} + 2 \implies y' = 2e^{2x -2}$
At (1,3) $y' = 2$. So the tangent line is $y = 2(x- 1) + 3$ or parametrically let $x - 1 = t 
\iff x = 1 + t$ and $y = 2t + 3$
Method 2 No eliminating. 
Let $r = (1 + ln(t),2 + t^2) \implies r' = (1/t, 2t)$. At (1,3), $t = 1$, therefore $r'(1) = (1,2)$
$r = (1,3) + s(1,2)$ which gives you $x = 1 + s$ and $y = 3 + 2s$
A: 2nd method: eliminating parameter:
$x=1+ \ln t , y=t^2+2 \Leftrightarrow  t=\exp(x-1), y=2+\exp(2x-2)$
Consider the function:  $f(x)=2+ \exp(2x-2)$, then: $f'(x)=2 \exp(2x-2)$
The  tangent at $x=1$ has equation: $Y=f'(1)(X-1)+ f(1)$, thus: $Y=2(X-1)+3$, thus:$$Y=2X+1$$
