What does it mean for a function to increase along a curve? I think that if we were to say that, for instance, $y$ increases along the curve, (with no specific rate) then this means for the derivative to simply be positive. Or does it mean to choose the positive value of a square root?
I'm given the functions $x = y^2 + 2$, and $z=x+4$. I simply parametrized $t=y$ and had $\langle t^2 + 2, t, t^2 + 6\rangle$. However, I could have put $ y = \sqrt{x - 2}$ parametrizing $t=x$ to have $\langle t,\sqrt{x - 2}, t + 4 \rangle$. 
I don't get the same derivative obviously, and despite using the same point, which gives me different $t$ values, I don't get the same tangent vector. So does which parametrization I choose depend on the statement for $y$ to increase along a curve? I don't think it should though.
 A: The derivative of the first parametrization can be represented as $\langle 2y,1,2y\rangle$, the one of the second one as $\langle 1, \frac{1}{2y},1\rangle$, so the derivatives are paralell, albeit not the same. This is typical for parametrizations. 
A simple example: suppose you want to parametrize the line
$$\{(x,y)\in \mathbb{R}^2: x=y,0\leq x \leq 1\}.$$
Two possible parametrizations are
$$\phi_1(t)=\left(\begin{array}{lll}t \\ t\end{array}\right),\ t\in [0,1]$$
and
$$\phi_2(t)=\left(\begin{array}{lll}2t \\ 2t\end{array}\right),\ t\in [0,0{,}5].$$
Those curves have a different derivatives since $\phi_2$ moves "faster" inside the set.
A: The mention of square roots in your first paragraph is incomprehensible.
That $y$ is increasing along the curve means that if you go from point to point in a certain direction along the curve, the value of $y$ always gets bigger.
I would be an error to think that "increasing" is defined in a way that relies on knowledge of derivatives. If one says that $f(x)$ is an increasing function of $x$, that means that for any two points $x_1,x_2$ in the domain of $f$, if $x_1<x_2$ then $f(x_1)<f(x_2)$.  Derivatives are not mentioned in that definition.  It is true that if $a<b$ and $f$ continuous in the whole closed interval from $a$ to $b$ and $f'(x)>0$ in the whole open interval, then $f$ is increasing in that interval.  But that is a theorem, not a definition.
A: Think of what happens when you parametrise a curve in $\mathbb{R}^3$ with some variable $t$ - you imagine a point sliding along the curve with motion described by a function of $t$. When you compute the derivative along the curve in terms of $t$, you are finding its velocity. 
If you choose a different parametrisation, your point moves along the curve with potentially different velocities at different places, so your derivative (in terms of $t$) looks different.
Is this the source of your confusion?
