# Derivative of a vector field along a curve

Given is the curve

$x(p)=sin(p)$

$y(p)=cos(p)$

$z(p)=p$

and a vector field $\mathbf{A}(p)$ defined on it as $\mathbf{A}(p)=\left(\begin{array}{c} cos(p)\\ -sin(p)\\ 1\\ \end{array}\right)$

Then $\mathbf{A}(p)=\left(\begin{array}{c} y(p)\\ -\sqrt{1-y^2(p)}\\ 1\\ \end{array}\right)$, or perhaps $\mathbf{A}(p)=\left(\begin{array}{c} \sqrt{1-x^2(p)}\\ -x(p)\\ 1\\ \end{array}\right)$

What is the derivative $\frac{\partial \mathbf{A}}{\partial x}$ - is it

$\frac{\partial \mathbf{A}}{\partial x}=\frac{\partial}{\partial x}\left(\begin{array}{c} y(p)\\ -\sqrt{1-y^2(p)}\\ 1\\ \end{array}\right)=\left(\begin{array}{c} 0\\ 0\\ 0\\ \end{array}\right)$, or is it

$\frac{\partial \mathbf{A}}{\partial x}=\frac{\partial}{\partial x}\left(\begin{array}{c} \sqrt{1-x^2(p)}\\ -x(p)\\ 1\\ \end{array}\right)=\left(\begin{array}{c} \frac{-x}{\sqrt{1-x^2}}\\ -1\\ 0\\ \end{array}\right)$

This is confusing...

$\frac{\partial \mathbf{A}}{\partial x}=\frac{\partial}{\partial x}\left(\begin{array}{c} y(p)\\ -\sqrt{1-y^2(p)}\\ 1\\ \end{array}\right)$
$y(p)$ is not independent of $x(p)$, as we are moving in a one-dimensional space. If $x(p)$ changes, $p$ changes. If $p$ changes, so does $y(p)$. Therefore, using chain rule, you should write it as
$\frac{\partial \mathbf{A}}{\partial x}=\frac{\partial}{\partial x}\left(\begin{array}{c} y(p)\\ -\sqrt{1-y^2(p)}\\ 1\\ \end{array}\right)=\frac{\partial}{\partial p}\left(\begin{array}{c} y(p)\\ -\sqrt{1-y^2(p)}\\ 1\\ \end{array}\right)\times \frac{\partial p}{\partial x}=\left(\begin{array}{c} -sin(p)\\ -cos(p)\\ 0\\ \end{array}\right)\times \frac{1}{cos(p)}=\left(\begin{array}{c} -tan(p)\\ -1\\ 0\\ \end{array}\right)$
• Where does the $1/\cos(p)$ come from? Jun 4, 2021 at 15:38