What is an arc length derivative?

Question

I have come across this problem which gives you the following vector function: $x(t)= <t, \frac{2}{3}t^{3/2}, -\frac{2}{3}t^{3/2}>; t\geq 0$

and then provides a function: $f(x,y,z) = xy^2-x^2 $

Now the question asks to give the Arc Length Derivative, $\frac{df}{ds}$, along the curve x at t=1. I'm confused what $\frac{df}{ds}$ is. I first thought it was curvature but it can't be because they gave a function to use somehow. Does anyone know what this is asking for?

Answer 1

Presumably they are asking for the directional derivative of $f$ along the direction of the curve $x(t)$. You need to work out the unit vector tangent to $x(t)$ at the point $t=1$, then form the dot product of this with the gradient of $f$ at that point.

(But surely they should have told you what these terms mean and given you some idea how to do this before asking you to do this question?)

(Thanks to Ted Shifrin for corrections.)

Answer 2

It is possible (but by no means certain) that this is a typo, and what is intended is for you to find the derivative $\frac{df}{dt}$. If so, here is what that means:

$f$ is explicitly given as a function of three variables $x, y$ and $z$; each of these is, in turn, a function of a single parameter $t$. Thus you can interpret $f$ (implicitly) as a function of $t$, and calculate the derivative $\frac{df}{dt}$. There are two distinct approaches that can be used here:

1. You could explicitly write out $f(x(t),y(t), z(t)$ (i.e., substitute the formulas for $x(t), y(t), z(t)$ into the formula for $f(x,y,z)$ and then compute its derivative using normal single-variable Calculus methods.
2. Alternatively, you can use the multivariable Chain Rule to build $\frac{df}{dt}$ up from the partial derivatives $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}$ and the derivatives $\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}$.

Now it's possible that they really do mean for you to find $\frac{df}{ds}$, where $s$ denotes the arc length along the curve. In this case, you can (again) use the (single-variable) Chain Rule:

$$\frac{df}{dt} = \frac{df}{ds} \frac{ds}{dt}$$

and complete the problem using the fact that $\frac{ds}{dt} = \| \vec v(t) \|$, the speed of the moving object.