I Instant distance visualization

  • I
  • Thread starter Thread starter LLT71
  • Start date Start date
  • Tags Tags
    Visualization
AI Thread Summary
The discussion revolves around the concept of "instant distance" and its visualization through calculus, particularly using speed versus time graphs. Participants clarify that "instant distance" refers to the distance at a specific moment, which can be calculated using the formula distance = speed(t) * dt, where dt represents a small time interval. The challenge lies in understanding that while dt can approach zero, it must remain non-zero to reflect actual movement. The conversation emphasizes the importance of limits in calculus to reconcile the idea of instantaneous measurements with continuous motion. Overall, the thread highlights the complexities of visualizing instantaneous distance within the framework of calculus.
LLT71
Messages
73
Reaction score
5
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 going to 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!
 
Physics news on Phys.org
I'm sorry, I've never heard of 'instant distance'. How is this different from just 'distance'?
 
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.
 
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].
 
LLT71 said:
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].
Yes, that's what we should do. What's the problem?
 
phinds said:
Yes, that's what we should do. What's the problem?
here is the picture for the first post
speed.jpg
 
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."
 
  • Like
Likes LLT71
Okay, now your post makes a bit more sense.

LLT71 said:
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 going to get instant speed(t) (like plugging "x" in some function f(x) and evaluating value of f(x) at that point "x")

##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##.
 
  • Like
Likes LLT71
Drakkith said:
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##.

awesome, thanks!
 
  • #10
LLT71 said:
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.
##dx(t) = v(t) dt##

If you are taking a path integral to determine work done over a path, the notion is common.
 
  • Like
Likes LLT71
Back
Top