# Conceptual trouble with derivatives with respect to Arc Length

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1. Oct 17, 2014

Hi,

So I'm working through a bunch of problems involving gradient vectors and derivatives to try to better understand it all, and one specific thing is giving me trouble.

I have a general function that defines a change in Temperature with respect to position (x,y). So for example, dT/dt would be the change in temperature with respect to time, which is implied to mean (dT/dx*dx/dt + dT/dy*dy/dt) by the Chain Rule. I understand this, and understand that dx/dt or dy/dt refers to the velocity (change in x & y position with respect to time).

However, it then asks for dT/ds, the change in temperature with respect to arc length.

I know that dT/ds would equate to dT/dt*dt/ds, where dt is velocity, and so dT/dt = (grad)T * v, but the answer shows that dt/ds is v / |v|, which implies that the derivative of time with respect to arc length is the unit tangent vector. I can't wrap my head around that, and I'm having a hard time seeing this either graphically or conceptually. Why is the derivative of time with respect to arc length the unit tangent vector? Thanks~

2. Oct 17, 2014

### SteamKing

Staff Emeritus
Your post appears to be confusing two different things.

ds/dt can be used to represent a velocity.

dT/ds does not have to be expanded using the chain rule into dT/dt* dt/ds (note: dt/ds is not the same as ds/dt) unless there is some reason for this. A temperature which is varying with time would be one reason.

dT/ds could simply represent a temperature gradient along some arbitrary path, whose differential length element is ds. Knowing something about the geometry of this arbitrary path could lead to expressions for dT/dx and dT/dy by applying the chain rule.