Calculus Question that should be really easy I know that the derivative of a distance function is velocity. That is, if a car has distance function $f(t)=t^2$ then at time $t$ it's velocity is $2t \text{ m/s}$. So let's say I observe some particle that appears to moving along the graph $f(t)=e^t$ towards positive infinity. I can easily calculate it's distance function from the fixed point $(0,1)$ on the graph. This is just $$\|(t,e^t)-(0,1)\|=\sqrt{t^2+(e^t-1)^2}$$The derivative of this function can be easily computed. But what is it representing? It can't be the velocity of the particle, because I have no idea how the particle is moving and I haven't given any initial conditions. So what is the derivative?
 A: Hello user and welcome to StackExchange!
To answer your question the derivative literally measures "the rate of change of" the function being derived.
In this case you are literally measuring "the rate of change of distance from the point (0,1) at a given time t). 
This quantity is a scalar (meaning number without reference to direction) so velocity in the formal sense is ruled out, but it still encodes the idea of how 'fast' the particle is moving away from a given point in simpler terms
"the speed with which the particle moves AWAY from the point (0,1)" 
Notice that if hypothetically our particle starts to run circles around the point (0,1) the velocity indeed may not be zero but the speed-moving-away is (since the distance is fixed).
Hopefully that clarifies what is being measured.
A: As frogeyedpeas explained, the derivative is the rate of change of distance from the point (0,1). A policeman's radar gun measures exactly that: it measures how quickly you are headed toward or away from the policeman, but it does not measure any movement sideways. If you drove in a circle around the policeman, the radar gun would measure zero speed. So your physical meaning is the speed measured by a radar gun at the point (0,1).
