Understanding how time derivative = acceleration

  • #1
SlowLearner1218
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I'm having a hard time understanding some concepts and would really appreciate some help(not super smart so I need some things basically dumbed down). In my physics lab we're going over Newton's Second Law. There's a statement in the lab papers I don't understand. It states "As you should know by now, the time derivative (or change in velocity over a time interval) is equivalent to acceleration, which gives the familiar F=ma".

Ok so as I have learned this past summer semester that a derivative is the slope of a tangent line or a single point in a function or basically the instantaneous rate of change. I looked up some YouTube videos and came to understand that it's not exactly a single point but rather the difference between 2 points that are so so close to each other that they're basically taken as a single point. So I'd like to know if I have this right. The derivative is the rate of change between two points that are close to each other in proximity that the distance between them inches them closer and closer to zero but never ACTUALLY the same position because at that point there wouldn't be 2 points to compare.

So if I have the previous statement above correct my next question is this. Am I interpreting this right in that time derivative in an acceleration graph means the very very small distance, almost zero but never quiet zero, between two points in the x axis? If interpretation is correct then the time derivative would not equal the acceleration right? Since time derivative is only talking about the change in the x-axis and acceleration is the change of y-axis over the change in the x axis.
 

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  • #2
PeroK
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Ok so as I have learned this past summer semester that a derivative is the slope of a tangent line or a single point in a function or basically the instantaneous rate of change.
Yes, exactly.
I looked up some YouTube videos and came to understand that it's not exactly a single point but rather the difference between 2 points that are so so close to each other that they're basically taken as a single point.
No. This is not correct. Mathematically, the derivative is a limit.

So if I have the previous statement above correct my next question is this. Am I interpreting this right in that time derivative in an acceleration graph means the very very small distance, almost zero but never quiet zero, between two points in the x axis?
No. The videos have confused and misled you

If interpretation is correct then the time derivative would not equal the acceleration right? Since time derivative is only talking about the change in the x-axis and acceleration is the change of y-axis over the change in the x axis.
This is completely wrong now.

Basically, in this post you started with a correct description of the derivative and ended up with nonsense!
 
  • #3
SlowLearner1218
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Here is a video where I understood that a derivative of x (or y for that matter) is a small change between x's (or could be y's). it's right at 5:24. Hopefully this helps aid in how I came to the conclusions above.
 
  • #4
Borek
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Key point: derivative is a limit. So your understanding starts with a correct idea of a sequence of smaller and smaller changes, but you didn't take the last step, which is taking the limit of that sequence.
 
  • #5
anuttarasammyak
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It states "As you should know by now, the time derivative (or change in velocity over a time interval) is equivalent to acceleration, which gives the familiar F=ma".

Just to clarify the statement as I think it shoud be
time derivative of coordinate x(t) is velocity v(t).
time derivative of velocity v(t) is acceleration a(t).
 
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  • #6
PeroK
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Here is a video where I understood that a derivative of x (or y for that matter) is a small change between x's (or could be y's). it's right at 5:24. Hopefully this helps aid in how I came to the conclusions above.

The video is a typical example of the tangle you can get into trying to define the limit intuitively by making ##\Delta x## "super-super-small" and calling it ##dx##. What we have, formally, is:
$$\frac{dy}{dx} \equiv \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$$And the informal way to describe this is: the smaller we make ##\Delta x##, then the better that ##\frac{\Delta y}{\Delta x}## approximates ##\frac{dy}{dx}##.
 
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  • #7
PeroK
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The critical flaw in that video is that he never uses the word "approximation". Instead, he gives the impression that ##\frac{dy}{dx} = \frac{\Delta y}{\Delta x}## for some sufficiently small ##\Delta x##. Which is wrong. In general, the slope of a secant line is only ever an approximation to the derivative (slope of tangent line).
 
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  • #8
jbriggs444
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The critical flaw in that video is that he never uses the word "approximation". Instead, he gives the impression that ##\frac{dy}{dx} = \frac{\Delta y}{\Delta x}## for some sufficiently small ##\Delta x##. Which is wrong. In general, the slope of a secant line is only ever an approximation to the derivative (slope of tangent line).
While it is technically incorrect to think of a derivative as the slope between a pair of very very nearby points on the graph, the idea will not lead you very far astray.

I got through two years of calculus using precisely this intuition. It is, I believe, the same intuition that Newton used. The idea of infinitesimals.

I knew enough not to treat the idea as being rigorously correct or to promote it to my instructors. But it was quite handy and always delivered the right answer.

As I understand it, Robinson's non-standard Analysis and the transfer principle allows one to put the notion of infinitesimals on a footing that is as rigorous as the notion of limits. It is just not the standard formalism.
 
  • #9
PeroK
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As I understand it, Robinson's non-standard Analysis and the transfer principle allows one to put the notion of infinitesimals on a footing that is as rigorous as the notion of limits. It is just not the standard formalism.
Yes, but handling infinitesimals correctly is potentially tricky. And you can't mix and match the two. We are either using the Real numbers or the Surreal or Hyperreal numbers.

Most advanced physics textbooks I've seen fall back on standard real analysis when they need to. Ultimately, despite its critics, standard analysis is the simplest way to introduce rigour.
 
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  • #10
jack action
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Since time derivative is only talking about the change in the x axis
No, the time derivative is the instantaneous change in the y-axis with respect to a given point on the x-axis (which must represent time). You said it yourself:
I have learned this past summer semester that a derivative is the slope of a tangent line or a single point in a function or basically the instantaneous rate of change.

Am I interpreting this right in that time derivative in an acceleration graph means the very very small distance, almost zero but never quiet zero, between two points in the x axis? If interpretation is correct then the time derivative would not equal the acceleration right?
This is a very confusing statement. An "acceleration graph" would be, to me, a plot of acceleration (y axis) vs time (x axis). the time derivative of that graph doesn't, of course, represent acceleration, but something called jerk.

The acceleration is found with the time derivative applied to the "velocity graph", which is a plot of velocity (y axis) vs time (x axis).

If on the "acceleration graph" at time ##t## the acceleration is, say, 10 m/s², then on the "velocity graph", the velocity at time ##t## will increase by 10 m/s per second of a time interval that is basically of length zero. That is what people refer to by saying ##dt = \lim_{\Delta t \to 0} \Delta t##. We're so close to zero that we can assume it is zero.

In real life, we take discrete measurements. A clock tells the time, then an instant later it tells another time. At each time, we measure the velocity ##v##. The closer both measurements are, the better the approximation of the time derivative can be (instantaneous acceleration). But the average acceleration ##\bar{a}## between those two points will always be:
$$\bar{a} = \frac{\Delta v}{\Delta t}$$
 
  • #12
PeroK
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@SlowLearner1218 it may be worth giving an explicit example of a derivative, in this case of the function ##y = x^2##. We take two points: ##x## and ##x + \Delta x##, with function values ##x^2## and ##(x + \Delta x)^2## respectively. We have:
$$\frac{\Delta y}{\Delta x} = \frac{(x + \Delta x)^2-x^2}{\Delta x} = \frac{x^2 + 2x\Delta x + (\Delta x)^2 -x^2}{\Delta x} = 2x + \Delta x$$Now, we find the derivative by taking the limit as ##\Delta x \to 0##:
$$\frac{dy}{dx} =\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} (2x + \Delta x) = 2x$$And that is how a derivative is calculated from first principles. Note that:
$$\frac{dy}{dx} = 2x$$is exact. It is not an approximation. It is not a "never-ending process". It is not an equation where ##dx## is "super-super small". It is an exact statement that the function ##2x## is the derivative of the function ##x^2##.

We can do this more generally for any power of ##x## and then we have the rule for derivatives:
$$\frac{d}{dx}x^n = nx^{n-1}$$This can be derived from first principles, but generally it's something you remember, as it is one of the most useful results in all of mathematics!
 
  • #13
PeroK
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PS this is just my opinion, but I think the Khan Academy is doing you a disservice by avoiding this. If you cannot cope with that level of mathematics, then you are not in a position to study calculus, IMHO.
 

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