# What does the letter $t$ stands for in “$t$-value of a curve”?

The "$t$-value of a curve" is the parameter of a point on a curve. For example if a curve's domain is form $0$ to $1$, then a t-value of $0.5$ would be a mid point. What does the letter $t$ stand for?

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$t$ is the underlying parameter. Any "nice" one-dimensional curve can be parameterized by a single parameter. For instance, if you want a line connecting pointing a point $\vec{r}_1$ and $\vec{r}_2$, then the equation of the line is given as $$\vec{r}(t) = (1-t) \vec{r}_1 + t \vec{r_2}$$ What this means is for any value of $t$, $\vec{r}(t)$ will lie on the straight line connecting the points $\vec{r}_1$ and $\vec{r}_2$.

Taking $t=0$, we get that $\vec{r}(0) = \vec{r}_1$, which essentially says that the point $\vec{r}_1$ lies on the line connecting $\vec{r}_1$ and $\vec{r}_2$, which is not surprising. Similarly, taking $t=1$, we get that $\vec{r}(1) = \vec{r}_2$, which essentially says that the point $\vec{r}_2$ lies on the line connecting $\vec{r}_1$ and $\vec{r}_2$, which again is not surprising. The midpoint is given by taking $t = 1/2$ i.e. we get that $\vec{r}(1/2) = \dfrac{\vec{r}_1 + \vec{r}_2}2$ which is nothing but the mid-point of the line-segment connecting $\vec{r}_1$ and $\vec{r}_2$.

Similarly, if we want to parameterize a unit circle in $2$D, we can do as follows. We know that the equation of the unit circle in $2$D is $x^2 + y^2 = 1$. This suggests that we can take $x = \cos(t)$ and $y = \sin(t)$. Hence, the vector $$\vec{r}(t) = (\cos(t),\sin(t))$$ is a point on the circle and parameterizes the circle since every point on the circle can be written as $(\cos(t), \sin(t))$ and conversely all the points of the form $(\cos(t), \sin(t))$ lie on a unit circle.

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$t=0.5$ doesn't have to be a midpoint. What we're doing here is parameterizing a curve. Like so: instead of saying $y$ is a function of $x$, say $y$ and $x$ are functions of a third variable, $t$. Physically you can think of a particle's $(x,y)$ position as a function of time. take

$x(t)=\cos(t)$

$y(t)=\sin(t)$

This generates a circle of radius 1. Here, $t$ is the angle counter-clockwise from the x-axis, but it doesn't have to be:

$x(t)=\cos(t^2)$

$y(t)=\sin(t^2)$

will generate the same circle, but now the $t$s aren't spread 'evenly' around the circle.

So the $t$ is just a variable that your curve is a function of. The $t$-value is a point on the curve, in the particular representation you've chosen. You can parameterize with respect to arclength, where $t$ represents how far (in terms of length) you are from some starting point, but you don't have to. This does seem to be the case in your example. The first circle is parameterized with respect to arc length: the length of a circle from $t=0$ to $t=T$ is $T$ itself, for any $T$. The second case however, is not.

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I think it is the parametrization of the curve. Take this curve

$$\begin{cases}\tag{1} x(t)=\cos(t)\\ y(t)=\sin(t) \end{cases}$$

where $t\in[0,2\pi]$. This curve describes a circle centered at the origin. The variable $t$ is a parameter that describes where the point is on the curve. Given the value of $t$ you can then evaluate the $x$ and $y$ coordinates of the points through the equations given in (1).

In your case, if $t\in[0,1]$, it means that there is a set of equations that express every component of a point on the curve as a function of $t$, with the starting point of the curve corresponding to the point whose coordinates are given by the values of this set of equations at $t=0$, and analogously for the end point, corresponding to $t=1$. If you specify a value for $t$, you can then compute the components of the point and establish where the point is on the curve. Therefore, the middle point of the curve corresponds to the point whose coordinates are given by these functions at the point $t=1/2$.

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