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blogfast25
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Quote: Originally posted by aga  |
The "lim delta t->0" part is a mystery.
Basically i do not know how it is applied to the delta x/delta v part, so cannot compute the whole thing.
Sorry if my Density appears to be increasing. |
You aren't supposed to be able to compute:
$$\lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t}$$
... because I haven't taught the limit theorem. But I have taught you the rules of derivation, so that you can compute:
$$v(t)=\frac{dx(t)}{dt}$$
... simply by applying the rules to the function x(t), which in the exercise was specified:
$$x(t)=3t^2+2t+4$$
[Edited on 26-3-2016 by blogfast25]
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aga
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Oh. That'd explain that then.
$$v(t)=\frac{dx(t)}{dt}$$
So does that mean, in words, that v (as a function of time) is equal to the difference in x (as a function of time) divided by the
difference in time ?
$$x(t)=3t^2+2t+4$$
then you're saying that the first derivative of x (as a function of time) =v(t) ?
$$v(t) = 6t+2$$
If that's right, still struggling to see how that relates to this part
$$v(t)=\frac{dx(t)}{dt}$$
I was off in the realms of
$$v(t) = \frac{x_2(t_2)-x_1(t_1)}{t_2-t_1}$$
$$v(t) = \frac{(3t_2 ^2+2t_2+4)-(3t_1 ^2+2t_1+4)}{t_2-t_1}$$
$$v(t) = \frac{3t_2 ^2+2t_2-3t_1 ^2-2t_1}{t_2-t_1}$$
$$v(t) = \frac{3(t_2^2-t_1^2) +2(t_2-t_1)}{t_2-t_1}$$
$$v(t) = \frac{3(t_2^2-t_1^2)}{t_2-t_1} + \frac{2(t_2-t_1)}{t_2-t_1}$$
$$v(t) = \frac{3(t_2^2-t_1^2)}{t_2-t_1} + 2$$
and getting very lost indeed.
Hang on, so if
$$v(t) = 6t+2$$
$$a(t) = 6$$
[Edited on 27-3-2016 by aga]
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blogfast25
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$$v(t)=6t+2$$
and:
$$a(t)=6$$
... are both correct. From a kinetic PoV, speed is the first derivative of position (in time) and acceleration is the first derivative of
speed (in time).
Regarding:
$$v = \frac{x(t_2)-x(t_1)}{t_2-t_1}$$
That would be the average value of v over the interval t<sub>1</sub> to t<sub>2</sub>.
To find the instantaneous speed v(t) we need to take the limit:
$$v(t)= \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}$$
$$v(t)= \lim_{\Delta t \to 0} \frac{x(t+\Delta t)-x(t)}{\Delta t}$$
Let's work out the numerator first:
$$x(t+\Delta t)-x(t)=3(t+\Delta t)^2+2(t+\Delta t)+4 - (3t^2+2t+4)$$
$$= 3(t^2+2t\Delta t+(\Delta t)^2)+2t+2\Delta t+4-3t^2-2t-4$$
$$=3t^2+6t\Delta t+3(\Delta t)^2+2t+2\Delta t+4-3t^2-2t-4$$
$$=6t\Delta t+3(\Delta t)^2+2\Delta t$$
$$=\Delta t(6t+3\Delta t+ 2)$$
Then divide by Δt, so:
$$v(t)=\lim_{\Delta t \to 0} (6t+3\Delta t+ 2)$$
For:
$$\Delta t \to 0$$
Then:
$$v(t)=6t+2$$
'On paper' we can work out:
$$v(t)=\frac{dx(t)}{dt}$$
... for any x(t) but you can imagine what kind of a kerfuffle that becomes for complicated functions x(t)!
And for that reason, aga, we simply apply The Rules! 
[Edited on 27-3-2016 by blogfast25]
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aga
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I see !
So with a limit,
f seriously> 0 where v(caught) > limit
where f is the Speeding Fine and v is the Speed 
'Limit' is certainly no theory ...
[Edited on 27-3-2016 by aga]
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blogfast25
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Yours isn't, that much is clear! 
"Dad, are we nearly there yet??"
"Integrals coming up soon, son!"
Take a deep breath...
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aga
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| Quote: Originally posted by blogfast25 |
$$q=\frac{\pi \Delta T}{\frac{1}{2k}\ln \frac{D}{d}+\frac{1}{hD}}$$
Determine for which value of D, in function of k and h, q is minimised.
Hint:
Set:
$$q=\frac{\pi \Delta T}{u}$$
Then see if:
$$\frac{1}{u}$$
... has an optimum. |
Do WHAT ?
Erm, so we want q'=0 then ?
OK. so
$$q=\frac{\pi \Delta T}{u}$$
$$q'=\frac{(\pi \Delta T)'(u) - (\pi \Delta T))(u)'}{(u)^2}$$
pi and deltaT are given as constants, so
$$q'=\frac{-\pi \Delta T}{(u)^2}$$
Eh ? (u)<sup>2</sup> can't be any number to make it end up as zero.
Briefly toyed with
$$q'=-\pi \Delta T(u)^{-2}$$
... fainted for a moment then gave up.
Is it a trick question, did i just get it wrong, or sometimes there is no optimum ?
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blogfast25
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$$q'=\frac{(\pi \Delta T)'(u) - (\pi \Delta T))(u)'}{(u)^2}$$
... is correct but should have become:
$$q'=-\pi \Delta T \frac{u'}{u^2}$$
And then:
$$u'=(\frac{1}{2k}\ln \frac{D}{d}+\frac{1}{hD})'$$
$$u'=\frac{1}{2kD}-\frac{1}{hD^2}=\frac{hD^2-2kD}{2khD^3}=\frac{hD-2k}{2khD^2}$$
$$\frac{dq}{dD}=-\frac{\pi \Delta T}{u^2} \frac{(hD-2k)}{2khD^2}$$
$$\frac{dq}{dD}=-\frac{\pi \Delta T}{2u^2khD^2} (hD-2k)$$
The factor:
$$-\frac{\pi \Delta T}{2u^2khD^2}$$
... never goes through zero, so that if there is optimum:
$$\frac{dq}{dD}=0$$
That can only mean:
$$hD-2k=0$$
So optimum is reached at:
$$D=\frac{2k}{h}$$
Loads of functions don't have optima but this isn't one of them.
[Edited on 27-3-2016 by blogfast25]
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yobbo II
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The v(t) = dx(t)/dt
is quite confusing when you throw it out. A new name is being put on dx/dt (called v) You then have to announce that v is a function of t and then
announce that x is also a function of t , hence the two (t)'s
Is that a sensible way of putting it?
Yob
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blogfast25
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$$v(t)=\frac{dx(t)}{dt}$$
and:
$$v=\frac{dx}{dt}$$
... are synonymous expressions that can be used interchangeably. The choice depends on context.
Some texts will even use:
$$x'(t)$$
... for speed (velocity).
[Edited on 28-3-2016 by blogfast25]
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blogfast25
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Integration (“anti-derivation”):
Woelen pointed out higher up and correctly that derivation is an operator that acts on a function f(x):
$$\frac{\mathrm d}{\mathrm d x} f(x)=f'(x)$$
There exist an anti-operator to derivation, called integration (rarely called anti-derivation), which achieves the opposite when
acting on a function f(x).
If F(x) (note capital F, not f!) is the integral of f(x), then:
$$\frac{\mathrm d}{\mathrm d x} F(x)=f(x)$$
Let's just give it a try, on:
$$f(x)=x^2$$
So we're looking for a function F(x), so that when we take the derivative of it, that derivative gives us x<sup>2</sup>. Let's try:
$$\frac13 x^3$$
Take the derivative:
$$\Big(\frac13 x^3\Big)'= \frac13 3 x^2=x^2$$
Bingo! So:
$$F(x)=\frac13 x^3$$
... is the integral of x<sup>2</sup>.
Notation:
If f(x) is a function and we call y its integral, then we can write as above:
$$\frac{\mathrm d}{\mathrm d x} y=f(x)$$
And:
$$dy=f(x)dx$$
The operator "integration" (or "to integrate" or "take the integral of") is symbolised by:
$$\int$$
Applying it to both sides we get:
$$\int dy=y=\int f(x)dx$$
Applied to our first example, if y is the integral of x<sup>2</sup>, then:
$$y=\int x^2dx= \frac13 x^3$$
Next: The Indefinite Integral.
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aga
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er, "whimper" seems appropriate.
I'm going to hide in the shed and boil ethanol over an open flame, pretending that i know what i'm doing so nothing bad happens ...
... prefereably not dis-integration.
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aga
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That done, i see now with my remaining good eye that Integration is basically the opposite of a Derivative.
Using the accelleration example, knowing the accelleration at time t, the integral of accelleration a(t) would tell you the speed v(t) ?
Are there anti-rules to go with integration, same as rules for derivatives ?
Edit:
Oh ! They're the same rules backwards ?
[Edited on 28-3-2016 by aga]
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blogfast25
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| Quote: Originally posted by aga | That done, i see now with my remaining good eye that Integration is basically the opposite of a Derivative.
Using the accelleration example, knowing the accelleration at time t, the integral of accelleration a(t) would tell you the speed v(t) ?
Are there anti-rules to go with integration, same as rules for derivatives ?
Edit:
Oh ! They're the same rules backwards ?
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Yes, integration will get us from a to v and from v to x. Simply put:
$$v=\int a dt$$
and:
$$x=\int v dt$$
The rules of integration are kind of the 'anti-rules' of derivation and that's very handy as we shall soon see!
<hr>
Indefinite integrals:
There is, when we wrote:
$$y=\int x^2dx= \frac13 x^3$$
... a little snake in the grass. Although it's true that:
$$\Big(\frac13 x^3\Big)'= \frac13 3 x^2=x^2$$
... there's another function that satisfies that condition also, because:
$$\Big(\frac13 x^3+C\Big)'= \frac13 3 x^2+0=x^2$$
... is also true, if C is a constant. We call it the integration constant.
So we really have to write:
$$y=\int x^2dx= \frac13 x^3+C$$
This is not limited to x<sup>2</sup>, in fact it's universal.
For the general case we can write:
$$y=\int f(x)dx=F(x)+C$$
... provided that:
$$\frac{\mathrm d}{\mathrm d x} F(x)=f(x)$$
Then:
$$F(x) + C$$
... is called the indefinite integral because at this point we do not know the value of C yet. Later we'll find out its value or how to
eliminate C.
Next up, simple integrals and the rules of integration. For now, just a few tasters (a is a constant):
$$\int dx=x+C$$
$$\int adx=a\int dx=ax+C$$
$$\int x^n dx= \frac{1}{n+1}x^{n+1}+C$$
$$\int ax^n dx= a\int x^n dx=\frac{a}{n+1}x^{n+1}+C$$
[Edited on 28-3-2016 by blogfast25]
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aga
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I, at least, am exremely grateful for the Effort you, blogfast25, have put into these Scientifically Educational threads, and the Time you have taken
to do so, despite seemingly little interest from the majority of the ScienceMadness community.
Why that is the case amongst supposed Scientists is a mystery, as maths is pretty much the Key to understanding the subject.
I noticed that with my own children that their attention span is quite a lot shorter than mine was at their age, so perhaps that's the reason for the
younger members' distinterest.
The older members may have studied this already and are scared of getting it wrong.
Either way, when the Fear of getting it Wrong stops you doing something, that's the point when you Stop Learning.
It is quivalent to the point where you think you know it all.
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aga
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That rings a dim and distant bell.
It sounds like 'bqoik' instead of 'Ding'.
Memory may be very defective on this, given the decades.
Edit:
Yes ! Constants, whatever their value when 'derived' they equal 0, so in reverse they're there, but unknown.
OK.
[Edited on 28-3-2016 by aga]
[Edited on 28-3-2016 by aga]
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blogfast25
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@aga:
Thank me when you've solved your first Real World differential equation.
As regards the interest in this thread, its page views/posts ratio is fairly normal, so I expect a lot of 'lurkers' and that's fine with me. Of course
it would have been nice to get a few more active participants but it is what it is. Among aspiring scientists the interest in theory is often
surprisingly low. 'They'll learn!' is my attitude to that.
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aga
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Does it Blend ? is my answer to that.
Edit:
no need for the @aga thing,.just us here.
I expect yobbo(x) was just visiting with some googled garbage and is highly unlikely to actually attempt the questions you have posed.
Maybe a future Probability Calculation lesson would be good, so we could calculate stuff like that.
[Edited on 28-3-2016 by aga]
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blogfast25
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| Quote: Originally posted by aga |
Maybe a future Probability Calculation lesson would be good, so we could calculate stuff like that.
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Ahhh... probability! One of the hardest, most counter-intuitive and most misunderstood concepts in mathematics.
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yobbo II
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$$ \int dx=x+C $$
is that the same thing as
$$ \int 1 dx=x+C $$
@ aga Name is x, yobbo x. Licenced to differentiate!
[Edited on 28-3-2016 by yobbo II]
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blogfast25
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Yes, it is.
So what is:
$$\int 0 dx=?$$
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yobbo II
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C
Do you see that aga, it's C
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blogfast25
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Correct. Of course that doesn't mean the integral of nothing is not nothing because C could be nothing . But it's possible that the
integral of the zero function is some number.
[Edited on 29-3-2016 by blogfast25]
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aga
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Thanks for joining in yobbo-eye-eye !
Not sure i do entirely C clearly.
Surely
$$\int n dx=x^n+C$$
The special case of n=0 appears to have a (possibly) important intermediate :-
$$\int 0 dx=x^0+C = 1+C$$
rolling that 1 into the Constant gets the lone C.
Whilst doing that is entirely valid, it warps the integration constant by 1.
[Edited on 29-3-2016 by aga]
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woelen
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Aga, here you make a mistake. What is the derivative of xn? It is not n, but nxn-1. The integeral of n equals nx + C and not
xn + C.
---------------------------------------------------------------------------------------------
The problem of integration of 0 leading to a non-zero value is a very subtle one. Only formally, it can be equal to a non-zero constant.
I'm quite sure blogfast25 will introduce the concept of definite integrals, which compute the surface of a function between its graph and the x-axis.
In such cases, the constant disappears and definite values are generated. In such cases, the integral of 0 always is zero, except for the special case
when integration occurs from a constant to infinity. Specially defined 0-functions can yield non-zero values in such cases (in fact, the inverse of a
dirac-function). A dirac function is a function which is zero everywhere, except for one single value, where it is infinite with the property that the
surface area of the single spike, which has zero thickness and infinite height, equals 1.
For now, just forget about the 0-integral being non-zero if you are a starter in this subject
[Edited on 29-3-16 by woelen]
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blogfast25
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@aga:
You need to distinguish sharply between n as an exponent and n as a coefficient.
<hr>
Basic Rules of integration (indefinite integrals):
Remember that the indefinite integrals of most simple functions can be found here (scroll down):
http://www.ambrsoft.com/Equations/Derivation/Derivation.htm
1. Constant rule (with a a constant):
$$\int af(x)dx=a\int f(x)dx$$
2. Power rules:
$$\int x^n dx=\frac{x^{n+1}}{n+1}+C$$
$$\int e^xdx=e^x+C$$
3. Sum/difference rule:
$$\int[f(x) \pm g(x)]dx=\int f(x)dx \pm \int g(x)dx$$
4. Chain Rule:
While the above three rules are somewhat self-evident, the chain rule of integration requires a bit of elaboration.
When taking derivatives we saw that the following chain rule applies:
If y is a function of u, with u a function of x, then:
$$y'(x)=\frac{dy}{dx}=y'(u) \times u'$$
Also:
$$dy=y'(u)u'dx$$
Since as:
$$u'=\frac{du}{dx}$$
$$u'dx=du$$
So:
$$dy=y'(u)du$$
Integrate both sides:
$$\int dy=y=\int y'(u)du$$
(Note that the chain rule for differentiation is sometimes also written in the following form: )
$$\frac{dy}{dx}=\frac{dy}{du} \times \frac{du}{dx}$$
If all this sounds like abstract gobbledygook, a few simple examples should soon clear things up. Applying the chain rule to integrals is also known
as ‘integration by substitution’.
1st example:
$$y=\int \sqrt{x-5}dx$$
Set:
$$u=x-5$$
Then:
$$du=(x-5)'dx=(1-0)dx=dx$$
Now substitute:
$$y=\int \sqrt u du=\frac{1}{1+1/2}u^{1+1/2}=\frac23u^\frac32$$
Substitute back:
$$y=\frac23 (x-5)^{\frac32}+C$$
2nd example:
$$y=\int \frac{x}{x^2+6}dx$$
Set:
$$u=x^2+6$$
$$du=(x^2+6)'dx=(2x+0)dx=2xdx$$
So:
$$xdx=\frac12 du$$
Now substitute:
$$y=\int \Big(\frac12\Big) \frac{du}{u}=\frac12 \int \frac{du}{u}$$
$$y=\frac12 \ln u$$
Substitute back:
$$y=\frac12 \ln(x^2+6)+C$$
3rd example:
$$y=\int \sin(7x)dx$$
Set:
$$u=7x$$
So:
$$du=(7x)'dx$$
$$du=7dx$$
So:
$$dx=\frac17 du$$
Now substitute:
$$y=\int \frac{\sin u du}{7}=\frac17 \int \sin u du=-\frac17 \cos u$$
Substitute back:
$$y=-\frac17 \cos (7x) + C$$
After ‘questions and answers’ on this section, a limited number of exercises will be put up.
[Edited on 29-3-2016 by blogfast25]
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