Calculus! For beginners, with a ‘no theorems’ approach!

Minimum distance problem:

A planner for Vulture Chemicals plots a grid on a map and determines that VC’s four most important customers are located at A(1, 5), B(0, 0), C(8, 0) and D(4,9) where units are in 100 miles. At what point W(x, y) should a warehouse be located in order to minimize the sum of the distances from W to A, B, C, and D?



Preamble:

In a Cartesian coordinate system the distance between two points:
$$(x_1,y_1),(x_2,y_2)$$
... can be calculated as follows:
$$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$
For this problem it's easier to use the squares of the distances:
$$d^2=(x_1-x_2)^2+(y_1-y_2)^2$$
If S is the sum of the squares of the distances for all four customers to the planned warehouse:
$$S=\Sigma d^2$$
Then:
$$S(x,y)=[(x-1)^2+(y-5)^2]+(x^2+y^2)+[(x-8)^2+y^2]+[(x-4)^2+(y-9)^2]$$
By means of the partial derivatives f_x and f_y, find the minimum for S.

(This is the last optimisation problem. Next: famous second order DVs)




[Edited on 3-5-2016 by blogfast25]

Quote: Originally posted by woelen

What you do is not equivalent. Minimizing the sum of distances in general need not give the same result as minimizing the sum of the squares of distances. It may be that in this particular case the answer is the same (I did not do any math on this), but in general this certainly is not the case.

You're correct.

If:
$$d_i=\sqrt{(x-x_i)^2+(y-y_1)^2}$$
Then:
$$S=\Sigma d_i=\Sigma_i \sqrt{(x-x_i)^2+(y-y_1)^2}$$
And:
$$f_x=\Sigma_i \frac{(x-x_i)}{\sqrt{(x-x_i)^2+(y-y_1)^2}}$$
$$f_y=\Sigma_i \frac{(y-y_i)}{\sqrt{(x-x_i)^2+(y-y_1)^2}}$$
Then that would not give the same results as we computed.

I found the exercise somewhere, made it a little harder but didn't stop to think. Grrr.

[Edited on 3-5-2016 by blogfast25]

Quote: Originally posted by blogfast25

I found the exercise somewhere, made it a little harder but didn't stop to think. Grrr.

Teaching is a 2 way thing.

Maybe a little has rubbed off

Quote: Originally posted by aga

Maybe a little has rubbed off

Don't get me worried...

Famous second order DEs:

Newtonian equations of motion:

Consider an object of mass m on which a number of forces act, shown below are 3 forces:



Then Newton’s Second Law tells us:
$$\Sigma_i \vec{F_i}=m\vec{a}$$
Worded: 'the vector sum of the acting forces equals the mass times the acceleration vector'

Analytically, this is usually solved by using the property that motions in the x, y and z directions are completely independent of each other. This allows to break down the vector problem into three independent non-vector equations:
$$\Sigma_i F_{x,i}=ma_x$$
$$\Sigma_i F_{y,i}=ma_y$$
$$\Sigma_i F_{z,i}=ma_z$$

From now on we’ll drop the indices x, y and z because it is usually very clear from the context of the problem which axis is meant. Also we’ll call:
$$\Sigma_i F_i=F$$
Where F is the sum of the force components along the intended axis, often also called the net force (along that axis).
So for a given axis we can now simply write:
$$ma=F$$
Also:
$$m\frac{dv}{dt}=F$$
And (for an x-axis):
$$m\frac{d^2x}{dt^2}=F$$
Specific examples:

1. F is constant:

Real world example: a free falling body (without drag).
$$m\frac{dv}{dt}=F$$
Integrate with boundary conditions t=0, v=v₀, x=x₀:
$$\int_{v_0}^v dv=\int_0^t \frac{F}{m}dt$$
$$\implies v(t)=v_0+\frac{F}{m}t$$
$$v(t)=\frac{dx(t)}{dt}=v_0+\frac{F}{m}t$$
Integrate again:
$$\int_{x_0}^xdx=\int_0^t(v_0+\frac{F}{m}t)dt$$
$$\implies x(t)=x_0+v_0t+\frac{F}{2m}t^2$$
2. F is a function of v:
$$ma=f(v)$$
$$m\frac{dv}{dt}=f(v)$$
Integrate:
$$m\int \frac {dv}{f(v)}=\int dt$$
Alternatively:
$$mx''(t)=f[x'(t)]$$
Real world example: projectile with drag. We derived such a problem higher up:
$$ma=mg-F_{drag}$$
Specifically:
$$v’=9.8-\frac{5v}{50}=9.8-\frac{v}{10}$$
We solved this by twice integrating, to obtain x(t).

But it can also be solved by writing it as:
$$\frac{d^2x}{dt^2}=9.8-\frac{1}{10}\frac{dx}{dt}$$
Or:
$$x''+0.1x'-9.8=0$$
The boundary conditions were:
$$t=0, x(0)=0, x'(0)=-10$$
I used Wolfram Alpha's DSolve function for this and obtained:
$$x(t)=98t+1080e^{-0.1t}-1080$$
Which is also what we found before.

[Edited on 3-5-2016 by blogfast25]

In
$$\Sigma_i \vec{F_i}=m\vec{a}$$
what does the i signify ?

The arrows mean Vectors, although quite how they work is a mystery.

A bit of utoob showed a tail-to-tip example which made sense in isolation (vector addition).

Putting them into an equation, Fwooosh ! Straight over head.

Edit:

It is nice to see that Wolfram got it right as well

Another thing: why is an accurate depiction of my brain being used in the example image ?


[Edited on 3-5-2016 by aga]

Quote: Originally posted by aga

In $$\Sigma_i \vec{F_i}=m\vec{a}$$ what does the i signify ? The arrows mean Vectors, although quite how they work is a mystery. A bit of utoob showed a tail-to-tip example which made sense in isolation (vector addition). Putting them into an equation, Fwooosh ! Straight over head. Edit: It is nice to see that Wolfram got it right as well Another thing: why is an accurate depiction of my brain being used in the example image ?

i is the index (number) of a force, as in:
$$F_1,F_2,...,F_i,...,F_n$$

Vector calculation:



Right: an example of two vectors being added up.

Left: F would be the resultant vector from adding up all vectors on the brain:

$$\Sigma_i \vec{F_i}=\vec{F}$$
The acceleration vector

$$\vec{a}$$

... is obtained by:

$$\vec{a}=\frac{\vec{F}}{m}$$

Both F and a vectors thus are parallel and point in the same direction, they're simply different in size.

But as explained, usually we kind of avoid all that vectoring by looking at the x,y,z force components.

I'll post a simple example.

[Edited on 3-5-2016 by blogfast25]

Quote: Originally posted by aga

So in $$\Sigma_i \vec{F_i}=\vec{F}$$ the meaning of $$\vec{F}$$ is : "The sum of all i involved vectors" Figuring out the Words is almost as hard as the maths.

Object sliding down an incline:

Consider an object of mass m sliding down a smooth incline (angle theta), without friction or drag:



Carefully note the choice of x and y-axis, please.

What are the forces and accelerations in play?

The weight mg is decomposed into an x and a y-component.

y-axis:

$$F_y=-mg\cos \theta$$

This force is countered by a reaction force F_N provided by the incline, so that:

$$F_N-mg\cos\theta=0$$
There's no NET force acting on the object in the y-direction.
$$\implies a_y=0$$

No acceleration in the y-direction.

x-axis:

$$F_x=mg\sin \theta$$

In the absence of friction or drag, F_x is also the ONLY force acting on the object along the x-axis, so:

$$ma_x=mg\sin\theta$$
$$\implies a_x=g\sin \theta$$

The object will accelerate along the x-axis with constant acceleration:
$$a_x=g\sin \theta$$
<hr>
Fun with a shopping trolley and two toddlers!

You're pushing a shopping trolley with force F₁. Toddler 1 want to pull the trolley towards the chocolates with force F₃, while toddler 2 want to pull it to the toys, with force F₂:



On the right we add up the force vectors, first F₃ + F₁, then add to that F₂, to get the resultant ('net') force F_R.

The trolley will experience an acceleration vector a:

$$\vec{a}=\frac{\vec{F_R}}{m}$$

[Edited on 4-5-2016 by blogfast25]

Quote: Originally posted by aga

Could the Distribution centre problem not be solved with vector maths ?

3. F is a function of x:

Real world example: Simple Harmonic Oscillator.



A massless Hookean spring with spring constant k is anchored at A and attached to an object of mass m. Ignore all friction or drag.

Initially the spring is neither extended nor compressed but at t=0 we move the object to x=x₀, then release it. The restoring force exerted by the spring is kx and the Newtonian equation is:

$$ma_x=-kx(t)$$
$$a_x=\frac{d^2x(t)}{dt^2}=x''(x)$$
$$mx''(t)+kx(t)=0$$
Or in 'shorthand':
$$mx''+kx=0$$
We assume the solution to be of the general form:
$$x=C_2\cos C_1 x$$
So:
$$x'=-C_2C_1 \sin C_1 x$$
$$x''=-C_2C_1^2\cos C_1 x$$
Plug it all into the DV:
$$-mC_2C_1^2\cos C_1 x+kC_2\cos C_1 x=0$$
C₂cosC₁x drops out and we get:
$$mC_1^2=k$$
$$C_1=\sqrt{\frac{k}{m}}$$
Now with the boundary condition:
$$t=0, x=x_0$$
$$\implies C_2=x_0\cos 0=x_0$$
So:
$$x(t)=x_0\cos \Big(\sqrt{\frac{k}{m}}t\Big)$$
C₁ is often called the angular velocity, omega:
$$\omega=\sqrt{\frac{k}{m}}$$
The frequency of oscillation is:
$$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$$
The period of oscillation is:
$$T=\frac1f$$
<hr>
Would it matter if we rotate the whole set up, axis included, clockwise by 90 degrees? The object would then be hanging off the spring.

Let's see. The Newtonian equation now is:

$$ma=mg-kx(t)$$
$$mx''(t)+kx(t)-mg=0$$
$$mx''(t)+k[x(t)-\frac{mg}{k}]=0$$
Substitute:
$$u=x(t)-\frac{mg}{k}$$
$$u'=x'(t)$$
$$u''=x''(t)$$
$$mu''+ku=0$$
The derivation is the same as above and yields:
$$x(t)=\Big(x_0-\frac{mg}{k}\Big)\cos \Big(\sqrt{\frac{k}{m}}t\Big)$$
So gravity doesn't affect angular velocity, frequency or period of oscillation. Only amplitude is affected.

[Edited on 4-5-2016 by blogfast25]

Quote: Originally posted by aga

Amazingly, over 10 years ago i posed this exact question on a freelance programmer's website to get a way to calculate the position of a mobile phone tower based on GPS co-ordinates with TA values in order to map where all the mobile towers are.

There's a numerical solution to the minimum distance problem.

Create a Cartesian x,y grid (shown in blue) and calculate the sum of distances for each x,y point. A visual inspection (or Excel triage) will show you where you are near to the optimum (x,y) point. Repeat with a finer mesh x,y grid in that area, for greater precision.

Quote: Originally posted by aga

Wow ! I assumed Gravity would have an effect on the frequency, although that may be due to naturally shoving friction in there. [Edited on 4-5-2016 by aga]

Quote: Originally posted by aga

I imagined the solution would be simpler in polar co-ordinates, seeing as the Location and Distance are the 'known' elements. Perhaps that's why it never got figured out properly.

Polar coordinates wouldn't help at all here. And distance is definitely an unknown element here...

More about the facility location problem, here: it's hard and has been from about 1860!

[Edited on 5-5-2016 by blogfast25]

Sorry, i was thinking about the mobile phone towers problem there.

The TA value is basically a discrete integer representing a 550m increment per step : basically the distance to the the tower with crappy granularity.

So the two known values would be GPS co-odinates of the A,B,C,D locations, each with a known TA (=crappy granuarity distance) from the central point = (The tower).

Problem was to calculate the GPS co-ordinates of that central point from a set of n GPS readings and measured TA values.

The geometry kind of suggested that at least 3 readings from varied positions would be needed to get an actual 'fix'.

Edit: these were the considered cases :



'X' marks the tower spot.

A,B, C are the known positions, the radius from each being the known distance to X.

F is the false positive, it being the same distance from A and B.

[Edited on 5-5-2016 by aga]

The Schrödinger equation (SE):

No mini-Pantheon of famous second order DEs can be complete without one of the most successful physics equations ever. Here it is again, time-independent and for one dimension only:
$$-\frac{\hbar^2}{2m}.\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x)=E\psi(x)$$
Now let’s see if with our new-found calculus ability we can analytically solve the simplest of all quantum problems: a particle in a one dimensional box (Pi1DB, see also the QP thread):



Mathematically:
$$-\frac{L}{2} \leq x\leq +\frac{L}{2}\:\text{then}\: V=V_0$$
$$|x| > \frac{L}{2} \:\text{then}\: V=+\infty$$
So the SE becomes:
$$-\frac{\hbar^2}{2m}.\frac{d^2\psi(x)}{dx^2}+V_0\psi(x)=E\psi(x)$$
Or:
$$-\frac{\hbar^2}{2m}.\frac{d^2\psi(x)}{dx^2}=(E-V_0)\psi(x)$$
So we’re looking for a function that twice derived kind of, sort of looks like the negative of itself. That’s quite reminiscent of:
$$(\sin x)'=\cos x$$
$$(\cos x)'=-\sin x$$
Or:
$$-(\sin x)''=\sin x$$
So we tentatively propose a solution of the form:
$$\psi=A\sin kx$$
$$\psi'=Ak\cos kx$$
$$\psi''=-Ak^2\sin kx$$
Plug it all into the SE:
$$-\frac{\hbar^2}{2m}(-Ak^2\sin kx)=(E-V_0)A\sin kx$$
$$\frac{\hbar^2}{2m}k^2=(E-V_0)$$
So it seems to be a fit but what are k and A?

We know that:
$$x=+\frac{L}{2}\implies \psi=0$$
$$\implies A\sin \Big(k\frac{L}{2}\Big)=0$$
And because:
$$A \neq 0$$
$$\implies \frac{kL}{2}=2\pi n$$
$$n=1,2,3,...$$
So:
$$k=\frac{4\pi n}{L}$$
$$\psi=A\sin \Big(\frac{4\pi n}{L}x\Big)$$
Also:
$$E_n=V_0+\frac{8\pi^2\hbar^2n^2}{mL^2}$$
To find A we need to apply the Normalisation condition:
$$\int_{-L/2}^{+L/2}\psi^2dx=1$$
$$\int_{-L/2}^{+L/2}\big[A\sin \Big(\frac{4\pi n}{L}x\Big)\Big]^2dx=1$$
It’s a banal integral which I’ll leave for readers to check, which yields:
$$A=\sqrt{\frac{2}{L}}$$
$$\psi_n(x)=\sqrt{\frac{2}{L}}\sin \Big(\frac{4\pi n}{L}x\Big)$$

First 8 wave functions (not to scale):



[Edited on 6-5-2016 by blogfast25]

[Edited on 6-5-2016 by blogfast25]

Quote: Originally posted by The Volatile Chemist

This thread suddenly passed my level of knowledge, haha. I just took my AP Calculus AB test for university credit yesterday. Good stuff.

Hope you pass!

Quote: Originally posted by aga

At this point, i can now follow large chunks of what is shown, which is Vast improvement.

At least that's a modest achievement, for both of us!

[Edited on 6-5-2016 by blogfast25]

Quote: Originally posted by blogfast25

At least that's a modest achievement, for both of us!

@aga:

Well, I'll put that feather in my cap then.

And I continue thoroughly, nay: streneously, to deny any involvement with any bodily thingymejibs of whatever kind.

[Edited on 6-5-2016 by blogfast25]

Heat Equation (one dimension only):

Consider a straight rod of length L, constant cross-section A, made of a homogeneous material of density ρ, heat capacity c_p and heat conductivity k.



At some point we start heating (or cooling) a section of the rod. This creates a dynamic situation: temperature T will now depend on position x, as well as time t:
$$T=f(x,t)$$
Before deriving the DV that describes the phenomenon, two more definitions are needed:

φ(x,t) is the heat flux (W/m² inside the rod due to heat conduction.

Q(dot)(x,t) is the heat input (or sink) per unit of volume per unit of time. We’ll keep Q(dot) a little vague for now, as its actual form will depend strongly from problem to problem.

To set up the DV, we set up the heat balance of an element of length Δx:
$$A\rho c_p \Delta x \frac{\Delta T}{\Delta t}=Aφ(x)-Aφ(x+\Delta x)+ A\Delta x \dot{Q}$$

Divide both sides by AΔx and rearrange slightly:
$$\rho c_p\frac{\Delta T}{\Delta t}=-\frac{φ(x+\Delta x)-φ(x)}{\Delta x}+\dot{Q}$$
Now take the limit of both sides to get derivatives:
$$\rho c_p \lim_{\Delta t \to 0} \frac{\Delta T}{\Delta t}=-\lim_{\Delta x \to 0}\frac{φ(x+\Delta x)-φ(x)}{\Delta x}+\dot{Q}$$
$$\rho c_p\frac{\partial T}{\partial t}=-\frac{\partial φ}{\partial x}+\dot{Q}$$
Looks ‘pretty’ perhaps but it isn’t very useful: we need to eliminate φ(x,t). To do so we use Fourier’s law on heat conduction:
$$φ(x)=-k\frac{\partial T}{\partial x}$$
Insert into the previous equation:
$$c_p\rho \frac{\partial T}{\partial t}=-\frac{\partial}{\partial x}\Big(-k\frac{\partial T}{\partial x}\Big)+\dot{Q}(x,t)$$
$$c_p\rho \frac{\partial T}{\partial t}=k\frac{\partial^2T}{\partial x^2}+\dot{Q}(x,t)$$
Call κ (kappa) the heat diffusivity:
$$\kappa=\frac{k}{c_p\rho}$$
$$\frac{\partial T}{\partial t}=\kappa \frac{\partial^2T}{\partial x^2}+\frac{\dot{Q}(x,t)}{c_p\rho }$$

After some time of constant heating (or cooling) a steady state will have been achieved, where T(x,t) will no longer depend on t, so:
$$\frac{\partial T}{\partial t}=0$$
The heat equation then reduces to the steady state version:
$$\kappa \frac{d^2T}{dx^2}+\frac{\dot{Q}(x,t)}{c_p\rho }=0$$
<hr>
Next up: two applications of the heat equation.

[Edited on 7-5-2016 by blogfast25]

Quote: Originally posted by aga

We still have not done limit theory yet, have we ? How does one 'take the limit' ?

No, because this thread is a 'theorem free zone' (kind of), remember?

But we have seen the basics.

In essence, is we have a function f(x), then limit theory tells us:

$$\lim_{\Delta x \to x}\frac{f(x+\Delta x)-f(x)}{\Delta x}=\frac{df(x)}{dx}=f'(x)$$

For simple functions limit taking is easy. Take e.g. f(x) = 3x² - 4.

$$\frac{[3(x+\Delta x)^2-4]-[3x^2-4]}{\Delta x}$$
$$=\frac{3(x^2+2x\Delta x+\Delta x^2)-4-3x^2+4}{\Delta x}$$
$$=\frac{3x^2+6x\Delta x+3\Delta x^2-3x^2}{\Delta x}$$
$$=\frac{\Delta x(6x+3\Delta x)}{\Delta x}$$
$$=6x+3\Delta x$$
$$\Delta x \to 0$$
$$\implies 6x=\frac{df(x)}{dx}$$
Which is correct, as we know.

For more complicated functions, limit taking can be a lot harder, though...


[Edited on 7-5-2016 by blogfast25]

Quick reminder of what is a derivative: definition of linear velocity



An object travels along an x-axis. At time t its position is x(t) and some time Δt later it is x(t+Δt).

Between these two points the average velocity would be:

$$v_{av}=\frac{\Delta x}{\Delta t}=\frac{x(t+\Delta t)-x(t)}{\Delta t}$$
But the True velocity is found by:
$$v(t)=\lim_{\Delta t \to 0}\frac{x(t+\Delta t)-x(t)}{\Delta t}=\frac{dx}{dt}$$

Halleluyah!

[Edited on 7-5-2016 by blogfast25]

This limit taking lark, as in the heat equation set up, lies at the heart of setting up many DVs.

Take this simple example of a tank emptying itself through a bottom hole:



The tank with constant cross-section A and a sharply cut bottom hole of cross-section a obeys Torricelli's law (inertias assumed zero), so that volumetric rate through the bottom hole is:

$$Q_v=a\sqrt{2gh}$$

In a time interval Δt the height of fluid in the tank changes by:
$$\Delta y=y(t+\Delta t)-y(t)$$
That's a volume ΔV (negative sign is needed because ΔV is negative):

$$\Delta V=-A\Delta y$$
Divide by Δt:

$$\frac{\Delta V}{\Delta t}=-A\frac{y(t+\Delta t)-y(t)}{\Delta t}$$
Take the limit for Δt to 0, so we get:

$$\frac{dV}{dt}=-A\frac{dy}{dt}$$
This is of course another expression for Q_v, so:
$$-A\frac{dy}{dt}=a\sqrt{2gy}$$
From this DV the total emptying time can then be calculated by integration.

[Edited on 8-5-2016 by blogfast25]

Quote: Originally posted by aga

Very clever having a hole in the top of the tank at the top right hand side. I suspect the equations would be a bit more complex otherwise !

Yes. Pressure above the water would be dropping, reducing throughput.

If p₀ is atmospheric pressure and V₀ the volume of air in the tank at the start, then with Bernoulli and the Ideal Gas Law:

$$\frac{v^2}2+\frac{p_0}{\rho}=gy+\frac{p_0V_0}{\rho(V_0+Ay)}$$
$$\frac{v^2}2=gy+\frac{p_0V_0}{\rho(V_0+Ay)}-\frac{p_0}{\rho}$$
$$\frac{v^2}2=gy+\frac{p_0V_0}{\rho(V_0+Ay)}-\frac{p_0(V_0+Ay)}{\rho(V_0+Ay) }$$
$$\frac{v^2}2=gy-\frac{p_0Ay}{\rho(V_0+Ay)}$$
$$\frac{v^2}2=\frac{\rho gy(V_0+Ay)-p_0Ay}{\rho(V_0+Ay)}$$
$$\frac{v^2}2=\frac{[\rho g(V_0+Ay)-p_0A]y}{\rho(V_0+Ay)}$$
$$Q_v=a\sqrt{2\frac{[\rho g(V_0+Ay)-p_0A]y}{\rho(V_0+Ay)}}$$
$$-A\frac{dy}{dt}=a\sqrt{2\frac{[\rho g(V_0+Ay)-p_0A]y}{\rho(V_0+Ay)}}$$

Whether that's even analytically integratable remains to be seen!
<hr>
An additional consideration is needed here. In order for liquid to be able to flow out of the opening, the pressure on the tank side must always be higher than on the other side:
$$\frac{p_0V_0}{V_0+Ay}+\rho g y>p_0$$
The point where steady flow would stop is:
$$y=\frac{p_0A-\rho g V_0}{\rho gA}$$
Beyond that point, liquid would still be trying to leave and air trying to enter at the same time. This is what happens when you try and empty a full bottle of liquid by turning it up-side down!


[Edited on 8-5-2016 by blogfast25]

Cooling fin: Steady State

A cooling fin is attached to a wall at T₀, surrounded by a fluid at T_∞:



We assume steady state and heat loss by convection only. For a segment of length Δx, convection heat loss is acc. Newton:
$$- p\Delta xh(T-T_{\infty})$$
(p is the perimeter of the fin, for a slab e.g. p = 2 x width + 2 x thickness)

The heat balance of that segment now is:
$$A\rho c_p \Delta x \frac{\Delta T}{\Delta t}=Aφ(x)-Aφ(x+\Delta x)- p\Delta xh(T-T_{\infty})$$
Deriving as for the general heat equation and for steady state, we obtain:
$$\frac{\partial^2T}{\partial x^2}-\frac{ph}{kA}(T-T_{\infty})=0$$
Solution:

Firstly a simple substitution:
$$\tau=T-T_{\infty}$$
$$\implies d\tau=dT,d^2\tau=d^2T$$
And call:
$$m^2=\frac{ph}{Ak}$$
Plugging into the DV:
$$\tau''-m^2\tau=0$$
This a linear, second order homogeneous DV and its solution is of the general form:
$$\boxed{\tau=c_1e^{mx}+c_2e^{-mx}}$$
Where c₁ and c₂ are two integration constants. We find these by applying the boundary conditions.

1st boundary condition:

For x=0, T=T₀, which gives us:

$$T_0-T_{\infty}=c_1+c_2$$

2nd boundary condition:

Here we assume that at the end of the fin (x=L), no more heat conduction is possible and with Fourier's law that means:
$$\Big(\frac{d\tau}{dx}\Big)_{x=L}=0$$
$$\implies \frac{d\tau}{dx}=mc_1e^{mx}-mc_2e^{-mx}$$
$$c_1e^{mL}-c_2e^{-mL}=0$$
Together with:
$$T_0-T_{\infty}=c_1+c_2$$
... it's a simple system of simultaneous equations, from which c₁ and c₂ can be computed (I won't do that here).

Inserting these constants into the proposed solution and very considerable rework then yields:
$$\frac{T-T_{\infty}}{T_0-T_{\infty}}=\frac{\cosh\Big(\big(1-\frac{x}{L}\big)\sqrt{\frac{hp}{kA}}L\Big)}{\cosh\Big(\sqrt{\frac{hp}{kA}}L\Big)}$$
Note that the hyperbolic cosine means:
$$\cosh f(x)=\frac{e^{f(x)}+e^{-f(x)}}{2}$$

Heat dissipated by the cooling fin:

More important than the temperature profile of the fin is how much heat it withdraws from the wall (that is after all its purpose).

Because we're in steady state, we can calculate this as the heat conducted into the fin at the interface with the wall. That heat is then also what is dissipated by the fin because Q_in=Q_out. With Fourrier's law:

$$\dot{Q}_{x=0}=\left.-kA\frac{d}{dt}(T-T_{\infty})\right|_{x=0}$$
I'll only present the result here:
$$\dot{Q}_{x=0}=\sqrt{khpA} (T-T_{\infty})\tanh mL$$
With:
$$\tanh mL=\frac{e^{mL}-e^{-mL}}{e^{mL}+e^{-mL}}$$
An inspection of tanh ml shows that it increases with L but that there's hardly any more gain in Q for mL > 3. Excessively long fins hardly dissipate more heat than reasonably shorter ones.

[Edited on 9-5-2016 by blogfast25]

Quote: Originally posted by Eosin Y

@Blogfast25 are you a maths teacher? I'm in top maths set (out of 14) in the first year of British boarding school, and this goes right over my head.

Experimental determination of heat conductivity k:

Finally another example of heat applied but this one without second order DEs!

The following set up allows the determination of the heat conductivity k on uniform, non-metallic materials:



A slab of the sample S is sandwiched between a heat source at constant T₀ and a thin slab of copper. The copper slab is held against a highly insulating wall (very low _k), so no heat can escape from the copper slab. The sample and copper slab thus heat up in time. A thermocouple measures the temperature of the copper slab during the experiment.

The heat flux through the sample slab can be calculated using Fourier:

$$\dot{Q}=-kA\frac{dT}{dx}$$
$$-kA\int_{T_0}^{T_{Cu}}dT=\dot{Q}\int_0^Ldx$$
$$\dot{Q}L=-kA(T_{Cu}-T_0)$$
$$\dot{Q}=-\frac{kA}{L}(T_{Cu}-T_0)$$

For the copper slab, assuming it's fairly thin, we assume a quasi-steady state, where:

$$\frac{dT}{dx}=0$$

Because Cu is such an excellent heat conductor and because the Cu is thinish that's a reasonable assumption.

With no heat lost, the copper slab will now heat up acc.:
$$\dot{Q}=mc_p\frac{dT_{Cu}}{dt}$$
With m the mass of Cu and c_p the specific heat capacity of Cu. So we have an identity:
$$mc_p\frac{dT_{Cu}}{dt}=-\frac{kA}{L}(T_{Cu}-T_0)$$
This simple first order DV with separation of variables solves as:
$$\frac{dT_{Cu}}{T_{Cu}-T_0}=-\frac{kA}{mc_pL}dt$$
$$\int_{T_{Cu,i}}^{T_{Cu,t}}\frac{dT_{Cu}}{T_{Cu}-T_0}=-\frac{kA}{mc_pL}\int_0^tdt$$
$$\ln \frac{T_{Cu,t}-T_0}{T_{Cu,i}-T_0}=-\frac{kA}{mc_pL}t$$
Where T_Cu,i is the initial (t = 0) temperature of the copper slab.

Measuring T_Cu,t over time and plotting the left hand expression versus time gives a straight line with slope:

$$-\frac{kA}{mc_pL}$$

... from which k can then be calculated.

One thing that's neat about the method is that it requires no knowledge of the specific heat capacity of the sample material.

[Edited on 10-5-2016 by blogfast25]

Quote: Originally posted by Eosin Y

How tf do you get those equations in your posts?

Quote: Originally posted by Eosin Y

Is my equation correct?

Well, you sure learned how to use LaTex quickly, well done!

I can't check your result because I don't know whether the following is true:

Quote: Originally posted by Eosin Y

Finally, enthalpy of formation. In RDX, bond enthalpy = 7,149kj/mol Given that this is true: \[\Delta H_{reaction}^{o}=\sum \Delta H_{f}^{o}(products)-\sum \Delta H_{f}^{o}(reagents)\] And the bond enthalpy of the products=8,817: \[\Delta H_{RDX decomposition}^{o}=7,149-8,817\] Finally, 7,149-8,817=-1668.

Quote: Originally posted by Eosin Y

All my data about RDX, its products and its enthalpy came from here:https://www.youtube.com/watch?v=HpkGhuV7RPU My LaTeX equations have been made here:https://www.codecogs.com/latex/eqneditor.php You just plug LaTeX into the drop-down at the bottom, and lo and behold!

Quote: Originally posted by Eosin Y

I doubt that the error with my bond enthalpy formula would really amount to 300%+? There's something else I could try...

Quote: Originally posted by Eosin Y

That's what I tried to do for RDX, though I can't find any proper values for X/delta. That might be my problem. Can you find these values anywhere?

Quote: Originally posted by Eosin Y

Fair enough Blogfast. Once I've drawn a graph, I'll post my results. Anyone know of any decent graph drawing software?

Flow of a fluid between parallel plates:

This is an interesting problem because of how the differential equation is set up.

We have two very large parallel plates, both stationary. Between them a Newtonian fluid flows in one direction. From prior art we can expect the velocity profile u(y) to look like this:



Now we consider a slice of fluid with thickness Δy:



To derive u(y) we first note that there is no acceleration, only uniform motion:

$$\frac{\partial u}{\partial t}=0$$

This means acc. Newton's second law that the balance of forces acting on a slice of moving fluid of thickness Δy must be zero::

$$\Sigma F_x=0$$

The forces acting on the slice are:

1. P₁ΔyW and P₂ΔyW. This pressure difference is of course the driving force for the flow.

2. Two shear forces (top and bottom): LWσ(y) and LWσ(y+Δy).

The balance of forces is:

$$P_1\Delta y W-P_2\Delta y W+LW[\sigma(y+\Delta y)-\sigma(y)]=0$$

Rearranged slightly:
$$\frac{\Delta P}{L}=-\frac{\sigma(y+\Delta y)-\sigma(y)}{\Delta y}$$
Take the limit to get a derivative:
$$\frac{\Delta P}{L}=-\lim_{\Delta y \to 0}\frac{\sigma(y+\Delta y)-\sigma(y)}{\Delta y}$$

$$\frac{\Delta P}{L}=-\frac{d\sigma}{dy}$$
The shear stress for a Newtonian fluid is:
$$\sigma=\mu\frac{du}{dy}$$
Plugged in we get:
$$\frac{\Delta P}{L}=-\mu\frac{d^2u}{dy^2}$$
This is a simple second order differential equation which on integration gives:
$$u=-\frac{\Delta P}{2\mu L}y^2+c_1y+c_2$$
c₁ and c₂ are integration constants, the values of which are obtained by applying the following boundary conditions:
$$y=-\frac{h}{2} \implies u=0, y=+\frac{h}{2} \implies u=0$$
Plugging in the values for c₁ and c₂ we get:
$$u=\frac{\Delta Ph^2}{2L\mu}\Big[\frac14-\Big(\frac{y}{h}\Big)^2\Big]$$
For y = 0, we get:
$$u_{max}=\frac{\Delta Ph^2}{8L\mu}$$
For a infinitely long plates we can optionally write:
$$\frac{\Delta P}{L}=\frac{\partial P}{\partial x}$$
Now we need to find the average velocity
$$\bar{u}$$
from:
$$\bar{u}=\frac{1}{h}\int_{-h/2}^{+h/2}u(y)dy$$
This yields:
$$\bar{u}=\frac{\Delta Ph^2}{12L\mu}$$
And so with a substitution:
$$\large{u(y)=6\bar{u}\Big[\frac14-\Big(\frac{y}{h}\Big)^2\Big]}$$

[Edited on 15-5-2016 by blogfast25]

Thread table of content:

WiFi woes?

The field strength of a WiFi signal in a house is governed by the Helmholtz Equation (here in two dimensions):

$$\nabla^2E+\frac{k^2}{n^2}E=f(x,y)$$

Where E is the signal strength, k the wave number of the used radiation and n the refractive index of the (x,y) space to that radiation. f(x,y) is the source function of the radiation.

$$\nabla^2E=\frac{\partial^2E}{\partial x^2}+\frac{\partial^2E}{\partial y^2}$$

If the (x,y) space is sub-divided into:

$$N\times M$$

... grid points, then the signal strengths can be calculated from a giant system of equations of the form:

$$\frac{E(i+1,j)+E(i-1,j)-2E(i,j)}{\Delta x^2}+\frac{E(i,j+1)+E(i,j-1)-2E(i,j)}{\Delta y^2}+\frac{k^2}{n(i,j)^2}E(i,j)=f(i,j)$$

i and j are the index numbers of the grid points.

Here's a fellow who did just that (and more!)



Look at the complicated room plan, the position of the (red point) WiFi source and the tendrils of 'internet goodness' as they snake through the available space!

The link also points to a related Android app by the same author.

Second order DEs (continued):

Quote: Originally posted by blogfast25

That was all kind of easy

Quote: Originally posted by aga

Quote: Originally posted by blogfast25   That was all kind of easy Speak for yourself !

Quote: Originally posted by aga

$$y_p=e^{\lambda x}$$ $$y_p'=\lambda e^{\lambda x}=\lambda y_p$$ The e^x bit basically. Natural logs in general.

Quote: Originally posted by aga

Bollocks! That sodding Chain Rule ! So, if i can ever get even close to 'getting it'. the function of x is $$\lambda x$$ and the function of the function of x is $$e^{\lambda x}$$ hence, following derivative chain rule : $$(e^{\lambda x})'=e^{\lambda x}\times (\lambda x)'$$ Why is it not a Single un-chained function of x, where $$f(x) = (e^{\lambda x})$$ ? Edit: Is it as simple as the fact that x gets times by lambda = one function, then raising 'e' by the result of that function constitues another function ? [Edited on 19-5-2016 by aga]

Quote: Originally posted by aga

So, if i can ever get even close to 'getting it'. the function of x is $$\lambda x$$ and the function of the function of x is $$e^{\lambda x}$$

aga, are you interested in solving your chain rule problem or did you not see the previous post?

Well, these were all 100 % correct, showing you have no problem seeing which is the 'master' and which is the 'subordinate' function.

To derive we now apply the Chain Rule. For my examples so far we had:

1.
$$f=\sqrt{x^2+5}$$
$$g=x^2+5$$
$$f=\sqrt{g}$$
2.
$$f=(2\ln x+8)^{1/3}$$
$$g=2\ln x+8$$
$$f=g^{1/3}$$
3.
$$f=e^{3x^2}$$
$$g=3x^2$$
$$f=e^g$$
<hr>

The Chain Rule says:
$$\frac{df}{dx}=\frac{df}{dg}\times \frac{dg}{dx}$$
So let's apply it to these examples:

1.
$$\frac{df}{dg}=(\sqrt{g})'=\frac12g^{-1/2}=\frac{1}{2\sqrt{x^2+5}}$$
(Edit: changed 'x' to 'g')
$$\frac{dg}{dx}=(x^2+5)'=2x+0=2x$$
$$\implies \frac{df}{dx}=\frac{df}{dg}\times \frac{dg}{dx}=f'(x)=\frac{2x}{2\sqrt{x^2+5}}=\frac{x}{\sqrt{x^2+5}}$$
2.
$$\frac{df}{dg}=(g^{1/3})'=\frac13 g^{-2/3}=\frac{1}{3g^{2/3}}=\frac{1}{3(2\ln x+8)^{2/3}}$$
$$\frac{dg}{dx}=(2\ln x+8)'=2\frac{1}{x}+0=\frac{2}{x}$$
$$\implies f'(x)=\frac{df}{dg}\times \frac{dg}{dx}=\frac{2}{3x(2\ln x+8)^{\frac23}}$$
3.
$$\frac{df}{dg}=\big(e^g\big)'=e^g=e^{3x^2}$$
$$\frac{dg}{dx}=(3x^2)'=3(x^2)'=3(2x)=6x$$
$$\implies f'(x)=6xe^{3x^2}$$
<hr>
Note that for very simple composite functions of the form (where a is a constant):
$$f(ax)$$
... the chain rule also becomes very simply:
$$f=(ax)^n \implies f'=an(an)^{n-1}$$
$$f=\cos(ax) \implies f'=-a\sin(ax)$$
$$f=\sin(ax) \implies f'=a\cos(ax)$$
$$f=e^{ax} \implies f'=ae^{ax}$$
<hr>
Now smash the chain gremlins and apply the Chain Rule to the previous exercises!

'Cheatsheet':

http://www.ambrsoft.com/Equations/Derivation/Derivation.htm

[Edited on 23-5-2016 by blogfast25]

Quote: Originally posted by aga

Assuming a typo, i'll now Baldly go where no Baldy has gone before ...

AAAAaaaarrghh!!! TERRIBLE typo alert. Harakiri being contemplated now.

[Edited on 23-5-2016 by blogfast25]

Quote: Originally posted by blogfast25

AAAAaaaarrghh!!! TERRIBLE typo alert. Harakiri being contemplated now.

1. is entirely correct. Well done.

5 to go.

Quote: Originally posted by aga

. Out of interest, which form did you end up with as the 'correct' answer ?