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How did the instantaneous speed pop out of this?

  1. Jul 23, 2011 #1
    If the function of a falling ball is s = 4t^2 where s is distance and t is time then to find the instantaneous speed you'd use:

    s + k = 4(t + h)^2

    k = additional distance, h = additional time

    So if t=3 then you get:

    s + k = 4(3+h)^2
    s + k = 4(9+6h+h^2)
    s + k = 36 + 24h + 4h^2

    Subtract s from both sides to get additional distance

    k = 24h + 4h^2

    Then when you divide by h to obtain the velocity and you let h approach 0, you get 24 the instantaneous speed at t=3, my question is essentially why 24 is the instantaneous speed, how did it we get that from the equation (not how as in we plugged numbers in and that's what we got), but its relationship to the other variables and the constant and what it means physically...know what I'm saying?

    By the way, this example is from Morris Kline's Calculus text, I just changed the constant to 4 to make easier to work out in my head
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  3. Jul 23, 2011 #2


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    It's just what we call velocity. The average velocity of the object from t = 3 to t = 3 + h approaches 24 as h approaches 0, so it makes sense to decide to call this the velocity of the object.
  4. Jul 23, 2011 #3


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    One way of thinking of instantaneous speed is the speed the object would have if suddenly its acceleration became zero. I like to think of of a ball being swung in a circle at the end of a rope. Suddenly the rope breaks. The balls heads off at its instantaneous velocity - in a direction tangent to its circular path.

    Try to convince yourself that the limit of 24 is the speed that the object would have if its acceleration suddenly stopped.

    Newtons's Laws of motions express the idea of instantaneous velocity. A body will remain in a state of uniform motion unless acted upon by a force. So if the forces suddenly stop acting - i,e, the accelerations suddenly stop - then it will continue afterwards in uniform motion. Its uniform speed is the instantaneous velocity it has at the moment when the forces stop.
    Last edited: Jul 23, 2011
  5. Jul 23, 2011 #4
    If you are asking, what does instantaneous velocity mean ... it's what's measured by the speedometer in your car. As you drive, your position is given by some function p(t). At any instant of time your speedometer tells you the first derivative of your position function. So for example if you took one hour to drive 60 miles, then your average speed was 60/mph. But at any moment you may have been going 55mph or 75mph.

    Is that what you meant? The physical meaning of instantaneous speed? It's just how fast you're going right at a particular instant.
  6. Jul 23, 2011 #5
    No, I probably shouldn't have said that. I'd just like to know how 24h came out and why it's the instantaneous velocity. Obviously you can just say 24h is the bx value that results from the expansion of 4(3+h)^2 but I don't get the physical relationship between the numbers that give this and why it is the instantaneous velocity (after dividing by h)

    For example, speed = distance/time, I can think of it intuitively, if I move 10m north in 2 seconds then it makes sense that I've moved an average of one second. But for the above, I can't get my head around it, to me, it seems to have popped out of nowhere really.
  7. Jul 23, 2011 #6


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    SteveL27, physical measurement as an example are not very good in so far all they ever will measure is average velocity. Speedometers measure (I believe, I may be wrong) the distance in rapid succesion and divide by the respective time intervals. It doesn't really make sense to measure instantaneous speed, as this is the result of a limit where the time interval approaches zero.

    I liked lavina's example, by comparing to the situation where acceleration instantaneously drops to 0, and it is easy to imagine.
  8. Jul 23, 2011 #7
    Conceptually, the speedometer is the derivative of your position function. Of course it's a physical measurement so it's approximate. Your objection is technically correct but applies equally to anything at all that physicists can measure.

    But actually as I understand it, there is not a little differentiator inside your car. There is not a little computer program continually computing the difference quotient of your position and taking a limit. Nothing of the sort, in fact.

    Rather, there is a direct analog measurement ... the rotation of one of your wheels drives an electromagnet, whose current increases in proportion to your velocity. I find that interesting -- we can measure a derivative directly, not by computation.

    Again, all measurement in the physical world is approximate, so I don't think it's fair to say that the speedometer doesn't measure your instantaneous speed, unless you're willing to apply the same criticism to every physical measurement anyone can make.
  9. Jul 25, 2011 #8
    on the left k represents length so 24h + 4h^2 must also represent length. The constant 4 must have dimensions length divided by time squared. 24 has dimensions length divided by time, the exponent 2 is dimensionless. h represents time, the 'magic' happens when you divide, length divided by time is speed. However you still need to cancel the h otherwise the limit gives 0 over 0. Thats the real magic, you cancel the problem and are left with the right answer.:smile:
    Last edited: Jul 25, 2011
  10. Jul 25, 2011 #9
    ^ Thanks for your input, but I don't think that's what I'm after...not sure why I can't edit my the initial post...but...:

    No, I probably shouldn't have said that. I'd just like to know how 24h came out and why it's the instantaneous velocity. Obviously you can just say 24h is the bx value that results from the expansion of 4(3+h)^2 but I don't get the physical relationship between the numbers that give this and why it is the instantaneous velocity (after dividing by h)

    The value that ultimately gives the instantaneous speed is the t=3, plus itself, times the constant 4, I don't understand what this means...mathematically, I get it (on a very basic level perhaps) but not physically...time + time x constant...

    For example, speed = distance/time, I can think of it intuitively, if I move 10m north in 2 seconds then it makes sense that I've moved an average of one second. But for the above, I can't get my head around it, to me, it seems to have popped out of nowhere really.

    I hope I'm making sense
  11. Jul 25, 2011 #10


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    the instantaneous speed is the average speed over an infinitesimally small time interval. This is computed as the limit of the Newton quotient.
  12. Jul 25, 2011 #11
    You didn't read my post #8 on dimensional analysis or i wasn't clear. In formulas everybody leaves out the units so in s = at^2, s has dimension of length, t has dimension time so by dimensional analysis the equation is...

    length = a(time )^2 how can a length be equal to time sqared? It is impossible so a must have dimensions (length)/(time)^2 in your case a = 4(length)/(time)^2. Lets use m for length and s for time to avoid long words.

    s(m) + k(m) = 4(m)/(s^2)[t(s) + h(s)]^2

    when t = 3

    s(m) + k(m) = 4(m)/(s^2)[3(s) + h(s)]^2

    s(m) + k(m) = 4(m)/(s^2)[9(s)^2 + 6(s)h(s) +h^2(s)^2]

    s(m) + k(m) = 36(m) + 24h(m) + 4h^2(m)

    k(m) = 24h(m) + 4h^2(m)

    notice that the dimensions work out in all equations above, but the dimensions is length, how do you get speed? Divide by the dimension of time!

    divide both sides by (s)

    k(m)/(s) = 24h(m)/(s) + 4h^2(m)/(s)

    so now the left side of the equation is speed and the right side of the equation is speed. You are not done yet, you still need to divide the right hand side by h, cancel h and then take the limit as h goes to zero.

    k(m)/(s) = [24h(m)/(s) + 4h^2(m)/(s)]/h

    k(m)/(s) = 24(m)/(s) + 4h(m)/(s)


    There is a subtle point here... as h goes to zero, h^2 goes to zero much faster so it is plausible you can divide out an h that cancels the h in 24h and then let 4h = 0 because 4h^2 will get to zero much faster than 24h.:smile:
    Last edited: Jul 25, 2011
  13. Jul 25, 2011 #12
    here is the dimensional analysis with just the units so less confusing.

    L = (L/T^2)(T^2)

    (L/T) = (L/T)

    Thats what is going on in s = 4t^2, the rest is mathematical gymnastics.:smile:

    So speed 'pops out' when you divide the dimension of length by the dimension of time.

    Of course if Kline and many others explained every subtle point, texts would be 10000 pages long and most people would be too bored to read them. I still wish there was someone out there who would explain every subtle point, I would not be bored.:smile:
    Last edited: Jul 25, 2011
  14. Jul 25, 2011 #13
    Nice insight!

    What is the 6 seconds before you multiply it out?

  15. Jul 26, 2011 #14
    The units are seconds squared because you got that by 3(s)h(s) + 3(s)h(s) = 6h(s^2)

    In my analysis the numbers are pure and the dimension multiplies the number right after. Since multiplication is a COMMUTATIVE operation 6(s)h(s) can be re-arranged without fear to give 6h(s^2)

    In the brackets all terms have dimension (s^2) BUT THEY ARE NOT ALIKE TERMS because of h and h^2 and absent h.

    I am very happy you find my insight useful, perhaps i have reduced the total entropy of the universe in a small way.:smile:
    Last edited: Jul 26, 2011
  16. Jul 27, 2011 #15
    ^ Thanks, I'm gonna think about that (my brain takes time to understand)
  17. Jul 31, 2011 #16
    s=4t² ⇒ ds/dt=8t. Hence, ds/dt|(t=3) = 8(3)= 24.
  18. Aug 4, 2011 #17
    For this:

    s(m) + k(m) = 36(m) + 24h(m) + 4h^2(m)

    Doesn't h have a a time dimension? If I put in 1s, it's ms which doesn't make sense to me. Or do you only input numbers with no dimensions, so 1s just put in 1 (and h is dimensionless, I forgot the cancelling).

    So then, am I correct in saying the ball falling, and at 3s, it's falling at 24m/s and accelerating at 4m/s^2, distance given by 24h + 4h^2?

    And I've got a somewhat similar question, why 24? Where does 24 come from? Why does this method work? - I get the limit as h->0 but the number 24 comes out and is sitting there, I can't get my head around how that works

    I can see that 24 was from 4m/s^2(6hs^2) = 24h(m), but why do these relations give the derivative?

    I guess I'm not asking in just the context of this physical problem, like y=x^2, when you differentiate it, you have ((x+h)^2 - x^2)/h = x^2 + 2hx + h^2 - x^2, and right there, after simplifying, you get 2x + h, and right there's the formula to find the slope at any point, why - HOW!? What is the relationship??

    Thank you

    @matphysik: I haven't been introduced to that notation yet, so I don't fully understand you
  19. Aug 4, 2011 #18
    1. Given, s(t)=4t². You seek the `instantaneous speed` when t=3 by way of first principles:

    Let s(t+δt) and δt be increments of the variables `s` and `t`, respectively. Consider, s(t+δt)-s(t)=4(t+δt)²-4t²=4[2tδt+δt²]. Let ds/dt≡ lim [s(t+δt)-s(t)]/δt as δt→0 denote the `instantaneous speed`. Then ds/dt=4[2t]=8t, and when t=3 ds/dt=24.

    2. The quantity `s(t+δt)-s(t)` is the chord through the curve s: I⊂ℝ⁺→ℝ⁺ defined by s(t)=4t² from `s(t)` to `s(t+δt)`. As δt→0, [s(t+δt)-s(t)]/δt gives the tangent to the curve at `s(t)`.
  20. Aug 5, 2011 #19
    Yes, we just put in the numbers without the dimensions. I did the dimensional analysis to convince you that it makes sense on a deeper philosophical level.

    I wish it was as simple as saying 4m/s^2 is the acceleration but it isn't. Acceleration is the 2nd derivative of distance. The acceleration is actually 8m/s^2

    You still have to take the limit as h-->0 in the expression 2x + h even though it seems to work if you don't.

    Perhaps a geometric analysis will help you understand why the derivative gives the slope of the tangent line.




    Hope this helps. The concept is fairly difficult so if you don't get it right away do not get discouraged.

    By the way, understanding the derivative using dimensional analysis is harder than understanding the derivative without dimensional analysis but your original post appeared to me it was asking a deeper question. One more thing, 24h(m) still represents distance. You have to divide by the dimension of time, then cancel h using the limit, thats when you get speed. :smile:
    Last edited by a moderator: Sep 25, 2014
  21. Aug 7, 2011 #20
    Second derivative?

    So as h->0, you get 2x...I can't seem to get my head around how 2x popped out...I don't mean the computational part is simple enough....hopefully I'll be able to understand all this soon cause it's driving me nuts! I hate using something but not fully understanding how it works.

    No doubt! I had no problems understanding differentiation when I watched the Khan Academy videos but learning it from Kline's book whereby he illustrates it using a physical problem raised all sorts of questions. If I hadn't read his book I probably wouldn't be wondering about the relationship between the derivative and the function and all this haha

    Thanks for the links, I'll have a look at them. Hopefully they'll bring me closer to the answer I'm seeking
  22. Aug 8, 2011 #21
    I've only skimmed the posts so sorry if this isn't quite what you're after but...
    I believe you said you understand the idea of average velocity, taking a distance traveled and dividing it by the time you took, but you weren't quite sure how this related to a derivative. Take a car travelling down a road, you measure its position at the start and its position at the end and divide it by the time it took to get there and you have the average velocity, but say this road has a bunch of traffic lights and Volvo drivers so the car's speed is constantly changing, the average velocity won't tell you a whole lot about what it was doing right at the start.
    So to get a better idea about the velocity of this car during the start of the trip you shorten the length and time you measure, this gives you a bit of a better idea about what is happening just in that short time frame.
    Now you say you want to know it more accurately so you take smaller and smaller chunks and each time you get a velocity closer and closer to some value. You keep taking smaller and smaller chunks of distance until it's basically zero but not quite, then it's only taken you basically zero time so the velocity you get is the time it takes to travel almost no distance in almost no time at all, you get what we call instantaneous velocity.
    This process of taking a chunk of almost zero change in distance and dividing it by an almost zero change in time is exactly what the derivative does. We say h is some infinitely small time period and X(t) is the position of the car at time t.
    Then if we take our formula for velocity (distance over time) we see we get...
    X(t+h) - X(t)

    Where X(t+h) is our end position and X(t) is our start position, and we can see that this is also the formula for a derivative when h is basically zero
  23. Aug 8, 2011 #22
    Yes, I hate using things without fully understanding how they work also. At some point i realized i was going to use them anyway, because there are so many things in life i don't fully understand and if i want to get something done, i have to become good at using the tools. Keep on questioning, that's good, but don't let unanswered questions stop you from using the tools. The derivative is an excellent tool that provides accurate answers to many questions.

    Good luck on your voyage of discovery.:smile:

    Something I hope you find interesting...

    Before we learn limits and calculus we need 2 points to get the slope of a straight line. That's a total of 4 numbers needed. After we learn how to take derivatives we only need HALF A POINT! We only need the x value of the point on the function whose slope we wish to determine. The slope of a function is 'hidden' within the function itself. The derivative is the tool (process) used to uncover this 'hidden' phenomenon.:biggrin:

    I guess this answers your question hopefully. The instantaneous speed 'pops out' because it was in there to begin with, hidden of course, but the derivative uncovers it.
    Last edited: Aug 8, 2011
  24. Aug 9, 2011 #23
    Not quite, basically I'm curious about how the terms are related to the derivative and WHY it gives the derivative (I can see why it's not obvious from my first few posts!)...but I appreciate you taking the time out to type up the post

    Hmmm...well it's the most satisfying answer for me so far :p

    Just out of curiosity, does real analysis address anything similar to this?
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