A parametric curve $C$ can be defined as follows

$$ C(p) = \{x(p), y(p) \}, \; p \in [0, 1] $$

where $p$ is the parameter.

We can define the unnormalised tangent to the point of the curve corresponding to $p$ as follows

$$ \frac{d C}{d p} = \left[\frac{d x}{d p}, \frac{d y}{d p}\right] $$

Why is that? What's the meaning of the derivative of a curve? It is apparently the tangent, but mathematically why is that? Is $\frac{d C}{d p}$ considered a gradient?

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    $\begingroup$ It you would write the independent variable as $t$ (instead of $p$), then you could think of position as a function of $t$ (e.g., time) and the derivative would be the velocity vector. $\endgroup$ – Ted Shifrin May 21 at 1:43

Let's write the derivative of a curve at point $p_0$ as follows: \begin{equation} \begin{split} \frac{dC}{dp}(p_0) &= \lim_{h \to 0} \left[ \frac{x(p_0+h) - x(p_0)}{h}, \frac{y(p_0+h) - y(p_0)}{h}\right] = \\ &= \lim_{h \to 0} \frac{1}{h} \left[ x(p_0+h) - x(p_0), y(p_0+h) - y(p_0)\right] = \\ &= \lim_{h \to 0} \frac{1}{h} \left( C(p_0 + h) - C(p_0) \right) \end{split} \end{equation} From the expression above, you can see that the derivative of the curve $C(p)$ can be considered as a difference between two points on a curve: $C(p_0 + h)$ and $C(p_0)$. This difference defines a vector. As $h \to 0$ this vector becomes shorter, however multiplying by the factor $\frac{1}{h}$ "rescales" it back. Therefore, tangent vector is defined as the vector between infinitely close points on a curve.

Hope that helps a bit.

  • $\begingroup$ Why we can even define the derivative of a curve (with respect to the parameter $p$), if a curve is not a function? $\endgroup$ – nbro May 20 at 23:06
  • $\begingroup$ btw, the term gradient is rather relevant to multivariate functions: (e.g., $\nabla F(x,y) = \left[ \frac{ \partial F}{ \partial x}, \frac{ \partial F}{\partial y} \right]$). However, parametric curve $C(p)$ is rather considered as a 1-dimensional vector-valued function. $\endgroup$ – dm_k May 20 at 23:06
  • $\begingroup$ you can treat the curve $C(p)$ as two independent functions $x(p)$ and $y(p)$, the values of which are stacked together in a vector. That's why the curve is a vector-valued function. $\endgroup$ – dm_k May 20 at 23:12
  • $\begingroup$ So we are taking the derivative of a 1-dimensional vector-valued function (even though I had heard that vector-valued functions are not really functions). $\endgroup$ – nbro May 20 at 23:20
  • $\begingroup$ how do you define a function? $\endgroup$ – dm_k May 20 at 23:36

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