# I Instant distance visualization

1. Feb 23, 2017

### LLT71

how can one visualize an idea of "instant distance". seems a bit abstract that at some time "t" we can evaluate "instant distance" by *speed(t)*dt". imagine a speed versus time graph.
now, there are two scenarios paradoxing in my head: if I assume that "dt" is a point on a time axis it seems that "no time has passed" (time freezed) and I can't visualize that something "moved" at speed(t) if there is no time interval. but if "dt" is some point (dt=0) surely we are gonna get instant speed(t) (like plugging "x" in some function f(x) and evaluating value of f(x) at that point "x")

if I let some time "dt" to "pass" after "t" => t+dt than, surely, something "moved" by small amount and it seems reasonable to obtain distance traveled (but than it is not really "instant" distance at time t). looking at the function "speed(t)" if we have some time interval [t,t+dt] than there is some difference between speed(t) and speed(t+dt) and even if we calculate distance(t)=speed(t)*dt it doesn't look "instant" to me because of interval thing...

help!

2. Feb 23, 2017

### Drakkith

Staff Emeritus
I'm sorry, I've never heard of 'instant distance'. How is this different from just 'distance'?

3. Feb 23, 2017

### phinds

Sounds like maybe he just means "the distance at a specific instant in time" but I'm not sure because the whole post makes no sense to me.

4. Feb 23, 2017

### LLT71

yeah "the distance at a specific instant in time".
distance=integral of speed(t)*dt from "a" to "b". because the speed is changing over time we should sum all instant's speed(t)*dt to get area under curve speed(t) and get how much distance we traveled on the interval [a,b]. for simplicity let it be [0,a].

5. Feb 23, 2017

### phinds

Yes, that's what we should do. What's the problem?

6. Feb 23, 2017

### LLT71

here is the picture for the first post

7. Feb 23, 2017

### gmax137

Find the part of the book where they tell you about "limits" and what they mean when they say "the limit as dt goes to zero."

8. Feb 23, 2017

### Drakkith

Staff Emeritus
Okay, now your post makes a bit more sense.

$dt$ represents an interval, not a point. It is commonly said that it represents a 'small change in $t$', which is true. The full answer has to do with how calculus defines differentials. The basic idea is that when you differentiate position vs time to get speed, you are finding how the position changes over some small interval $dt$. If you let $dt$ get infinitely small (but remain non-zero), you end up finding how position changes as $t$ changes for any value of $t$.

9. Feb 24, 2017

### LLT71

awesome, thanks!

10. Feb 24, 2017

### jbriggs444

$dx(t) = v(t) dt$

If you are taking a path integral to determine work done over a path, the notion is common.