Problem with differentiation

  • #26
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Of course it will since ##4.0 = 4.0 × 10^{-10} = 4.0 × 10^{-100} = 4.0 × 10^{-1000} ~\dots~\rm {etc}## . But if you want the negative exponent of ##10^{~n}## to stop, nobody can ever do it. Exactly the value of 4.0 means ##4.0 × 10^{-\infty}##. If you count from 1 to ##\infty##, you will never stop counting.
 
  • #27
rudransh verma
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Of course it will since ##4.0 = 4.0 × 10^{-10} = 4.0 × 10^{-100} = 4.0 × 10^{-1000} ~\dots~\rm {etc}## . But if you want the negative exponent of ##10^{~n}## to stop, nobody can ever do it. Exactly the value of 4.0 means ##4.0 × 10^{-\infty}##. If you count from 1 to ##\infty##, you will never stop counting.
So we assume deltaf/deltax value as 4. Because it’s not changing we have given it a name df/dx.
I want to clarify in terms of limits. Taking limit of 2x+deltax , we get values closer and closer to 2x. And we say limit is 2x. And this limit is defined as df/dx. 4 is the rate at x=2.
 
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  • #28
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23
Yes, its final value is 4.0 since ##f(x) = x^2~\Rightarrow~df/dx = 2x## and evaluating ##df/dx## at ##x_0 = 2## gives ##f'(x_0) = 2x_0 = 2(2) = 4 = f'(2)##.
 
  • #29
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Again, ##f(x) = x^2~\Rightarrow~df/dx = 2x## means that the rate of change of the function ##f(x)## with the independent variable ##x## is given by the function ##g(x) = 2x##.
When ##x = 2##, the rate of change of the function ##f(x)## is given by ##g(2) = 2(2) = 4##.
When ##x = 3##, the rate of change of the function ##f(x)## is given by ##g(3) = 2(3) = 6##.
We see that as ##x## increases, the rate of change of ##f(x)## with ##x## also increases.
 
  • #30
rudransh verma
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We see that as x increases, the rate of change of f(x) with x also increases.
Ok! Suppose this f(x) is f(t)=t^2 a distance function with respect to time t. Then the rate will be 2t. At t=2 sec then velocity will be 4 m/s ?
At t=3 sec velocity will be 6 m/s ?
 
  • #31
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Ok! Suppose this f(x) is f(t)=t^2 a distance function with respect to time t. Then the rate will be 2t. At t=2 sec then velocity will be 4 m/s ?
At t=3 sec velocity will be 6 m/s ?
##f(t) = t^2\Rightarrow~##a distance function of time ##\Rightarrow~x(t) = At^2~\Leftarrow~A = 1~\rm m/s^2 = constant##
##\Rightarrow~dx/dt = 2At = v(t)~\dots##
##\Rightarrow~v(2~{\rm s}) = 2A(2~\rm s) = 2(1~{\rm m⋅s^{-2}})(2~\rm s) = 4~\rm m/s~\dots ##
##\Rightarrow~v(3~{\rm s}) = 2A(3~\rm s) = 2(1~{\rm m⋅s^{-2}})(3~\rm s) = 6~\rm m/s~\dots ##
I included the constant ##A## with the proper unit so that I get the correct unit for ##x## in meter.
 
  • #32
jbriggs444
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But I don’t think it will ever stop at a fixed value.It will come as 4.000000000000000000000000000000000000….and then some digit….
This is not a correct intuition for how limits work.

However, you are correct that it never stops at a fixed value. That is also not how limits work.

So let us work on that. What does it mean when we talk about the limit of a function at a point? In particular, let us use the example at hand. What is the limit of ##f(x) = \frac{(4 + 4 \Delta x + (\Delta x)^2) - 4}{\Delta x}## as ##\Delta x## approaches zero?

You do agree, I hope, that the above limit matches the definition of the derivative of the function ##x^2## at the point 2. We can do a little simplification in the numerator of course so that our starting point is:
$$\lim_{\Delta x \to 0}\frac{4\Delta x + (\Delta x)^2}{\Delta x}$$If we were listening to the teacher, blindly accepting the rules of limits that we are taught, nodding our heads and saying "yes ma'am" we could glance at this limit and immediately say that it is equal to four. But that is not good enough. We want to know why it is exactly 4 and not something like 4.000<infinitely many zeroes>1.

The limit of a function at a point is not the value of the function somewhere very near that point.

The limit of a function at a point is not necessarily the value of the function at the point. [For a continuous function, the value of the function at a point will be equal to the limit of the function at that point].

The limit of a function at a point is the value (if there is one) that is approached ever more closely as one evaluates the function every more closely to the point.

The standard way of expressing this involves epsilons and deltas. One can quickly google up many copies of the definition. For example:
https://brilliant.org/wiki/epsilon-delta-definition-of-a-limit/ said:
Limit of a function (##\varepsilon-\delta## definition)

Let ##f(x)## be a function defined on an open interval around ##x_0## (##f(x_0)## need not be defined). We say that the limit of ##f(x)## as ##x## approaches ##x_0## is ##L##, i.e.
$$\lim _{ x \to x_{0} }{f(x) } = L$$
if for every ##\varepsilon > 0## there exists ##\delta >0## such that for all ##x##:$$0 < \left| x - x_{0} \right |<\delta \textrm{ } \implies \textrm{ } \left |f(x) - L \right| < \varepsilon$$
[Took a lot of work restoring the embedded ##\LaTeX## in that quote]

This epsilon-delta formulation can be difficult to grasp. Its advantage is that it takes the idea of time and "approaching" out of the picture. The idea of getting closer over time is OK for an intuition, but it is not OK for formal mathematics.

The idea is almost that of a game. You decide how close you want to the formula to get to the limit of 4. That is, you pick an epsilon (##\varepsilon##). I then tell you how close you need to stay to ##\Delta x## = 0. That is, I pick a delta (##\delta##).

If I can always successfully pick a delta (##\delta##) so that value of ##\frac{4\Delta x + \Delta x^2}{\Delta x}## stays within epsilon (##\varepsilon##) of 4 whenever ##\Delta x## is within delta (##\delta##) of zero then the I win and the limit is 4. [We never consider the case where ##\Delta x## is exactly zero]

If you can successfully pick an epsilon (##\varepsilon##) so that I cannot succeed then you win and the limit is not 4 (and may not even exist).

The derivative of x^2 at 2 is exactly 4. It is not an approximation. It is exact.
 
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  • #33
rudransh verma
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@kumusta we know (2x+deltax) which is deltafx/deltax will never stop at a fixed value as delta x approach 0. So we take limit of 2x+deltax and we assign a value to df/dx as 2x which is a rate we can never have. It’s close to 2x but never 2x.
 
  • #34
rudransh verma
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blindly accepting the rules of limits that we are taught, nodding our heads and saying "yes ma'am" we could glance at this limit and immediately say that it is equal to four. But that is not good enough.
Isn’t that applicable to everything we are taught in school and colleges.
And then there comes a time we don’t understand a single word of high level concepts taught in graduation and post graduation!
 
  • #35
jbriggs444
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Isn’t that applicable to everything we are taught in school and colleges.
And then there comes a time we don’t understand a single word of high level concepts taught in graduation and post graduation!
That has not been my experience. Certainly there are things that are difficult to grasp. But one can usually get at least a toehold on the edges.

In the case at hand, the teachers are correct. The limit is four exactly. The epsilons and deltas provide a way to define limits in the first place and then to prove this result rigorously, even though they might seem too esoteric to grasp at first.
 
  • #36
SammyS
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First I did d(rho)/dr which is equal to 35.4*10^-12/R. Then I integrated d(rho) by which I got rho=35.4*10^-12. And then the last eqn will be q=rho V. But the answer was wrong.
I have a doubt on the formula I am using for E because that formula is for a point charge or a charged shell.
Back to your Original Post. (OP)
You are given the charge density, ρ, as a function of r.

If you differentiate that, with respect to r, and then integrate over r, you get charge density back - in some form. If you find an indefinite integral, and the evaluate the constant of integration, you simply get the charge density back as a function of r. If you evaluate the definite integral from 0 to R, as you did, you simply get the value of the charge density at r = R, i.e., at the surface of the sphere.

To find the charge over some volume you need to do a volume integral, as has been mentioned. In your case, with spherically symmetric charge distribution, the volume element that's handy to use is ##dV=4\pi r^2dr## .

Then ##\displaystyle Q_{in}=35.4 (pC/m^3) \int_0^a \frac{r}{R} 4\pi r^2 dr ## is the amount of charge inside a sphere of radius ##a## for ##a \le R ## .
 

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