Uživatel:Dean/Pískoviště

Dokažte, že napětí na pseudounipolárním Golbergově svodu je 1,5-násobek napětí klasického unipolárního svodu.

Řešení

 Podrobnější informace naleznete na stránce Elektrokardiografie/Fyzika .

Napětí mezi dvěma body je definováno jako rozdíl jejich potenciálů. Označme R, L, F potenciál na pravé horní, levé horní a levé dolní končetině. Bez újmy na obecnosti si zvolíme elektrodu na levé dolní končetine. V klasickém unipolárním svodu je její měřena vůči Wilsonově svorce. Potenciál Wilsonovy svorky (značme W) je aritmetickým průměrem potenciálů všech končetinových svodů, aneb:

$W={\frac {R+L+F}{3}}$

Napětí klasického unipolární svodu dolní končetiny (značme VF) je tedy rovno:

$VF=F-W=F-{\frac {R+L+F}{3}}$

Oddělením členu F ze zlomku dostáváme:

$VF=F-{\frac {R+L}{3}}-{\frac {F}{3}}$

Což je rovno:

$VF={\frac {2}{3}}F-{\frac {R+L}{3}}$

Vynásobme rovnicí zlomkem 3/2 aby jsme oslobodili člen F.

${\frac {3}{2}}VF=F-{\frac {R+L}{2}}$

Pravá strana tohoto výrazu je ale z definice napětí na Goldberově psedounipolárním svodu.

$aVF=F-{\frac {R+L}{2}}$

Tedy skutečně platí:

$aVF={\frac {3}{2}}VF$

$\square$

Derive the formula for pH of a weak acid with initial concentration c and acid disocciation constant K. Neglect the self-ionization of water.

Solution

General equation for acid dissociation is:

${\ce {HA-> H+ + A-}}$

Equilibrium constant, or acid dissociation constant, is defined as:

$K={\frac {[\mathrm {H} ^{+}][\mathrm {A} ^{-}]}{[\mathrm {HA} ]}}$

Note that these concentrations are in equilibrium. Fraction of initial acid concentration c disassociates. Amount which disassociates is equal to $[\mathrm {A} ^{-}]$. Remaining (equilibirum) amount is the aforementioned ${\ce {[HA]}}$. Therefore:

$c=[\mathrm {HA} ]+[\mathrm {A} ^{-}]$

Since we neglect self-ionization of water, the only source of protons is the acid, therefore it holds:

$[\mathrm {A} ^{-}]=[\mathrm {H} ^{+}]$

For simplicity, let us denote either of these $x$. What we are trying to find is the pH, that is by definition:

$\mathrm {pH} =-\log {[\mathrm {H} ^{+}]}=-\log {x}$

Plugging in to the formula for the equilibrium constant we obtain:

$K={\frac {x^{2}}{c-x}}$

Rewriting into the standard form of quadratic equation:

$x^{2}+Kx-Kc=0$

Solutions to this equation are:

$x={\frac {-K\pm {\sqrt {K^{2}+4Kc}}}{2}}$

We choose the solution with positive sign, for we assume only non-negative concentration. Overall result is therefore:

$\mathrm {pH} =-\log {\left({\frac {-K+{\sqrt {K^{2}+4Kc}}}{2}}\right)}$

Remark: Let's assume K is very small compared to c. That is a reasonable assumption, sice we are dealing with weak acid (small K) of a reasonable concentration ("big" c). Then disregarding all the terms in the numerator except the one with c, we would obtain:

$\mathrm {pH} \approx -\log {\left({\frac {\sqrt {4Kc}}{2}}\right)}=-\log {({\sqrt {Kc}})}=-{\frac {1}{2}}\log {(Kc)}=-{\frac {1}{2}}\log {(K)}-{\frac {1}{2}}\log {(c)}={\frac {1}{2}}(\mathrm {pK_{a}} -\log {c})$

Which is the familiar simplified formula. Alternative (and simpler) way te derive it, would be to assume $c\approx [\mathrm {HA} ]$, that is to assume essentially all of the acid would remain undissociated. Taking log of acid dissociation constant would then yield the result. Because of the added assumption, this formula is less accurate than the one we derived.

Prove the identity log(xy)=log(x)+log(y).

Solution

Let's begin with the definition of log. Assume that x is a positive number expressed as a power of some base b (also positive, distinct from 1), that is

$x=b^{n}$

then logarithm (of base b) is a function that outputs this exponent:

$\log _{b}{x}=n$

We will omit the base of log in the following text. Plugging this for n into the first equation we conclude that every such number x can be expressed as:

$x=b^{\log {x}}$

Let's express in this way both terms in the xy product:

$xy=b^{\log {x}}b^{\log {y}}$

From the property of exponentials we can equivalently write:

$xy=b^{\log {x}+\log {y}}$

Taking log of both sides yields:

$\log {(xy)}=\log {(b^{\log {x}+\log {y}})}$

But from the definition of log, we can cancel exponential and log on the right side:

$\log {(xy)}=\log {x}+\log {y}$

Which proves the original identity.

$\square$

Derive the formula for expressing log(x) in the terms of ln(x).

Solution

From previous problem we know that every positive number x can be expressed as:

$x=b^{\log {x}}$

But the base itself can also be written in this form. Let's choose Euler's number for the base on the right side. Log with this base is called natural logarithm and denoted ln(x).

$b=e^{\ln {b}}$

Let's combine these two expression, that is plug in the second equation for b into the first equation.

$x={(e^{\ln {b}})}^{\log {x}}$

But from the property of exponents, this expression is equal to:

$x=e^{\ln {(b)}\log {(x)}}$

Taking natural log of both sides:

$\ln {x}=\ln {(b)}\log {(x)}$

Rearranging we get the final result:

$\log _{b}{x}={\frac {\ln {x}}{\ln {b}}}$

Derive the formula for hydrostatic pressure of incompressible fluid.

Solution

Pressure in general is the force acting on a surface, divided by that surface area:

$p={\frac {F}{A}}$

With hydrostatic pressure, the acting force is the weight of the fluid. Weight of a body is calculated as:

$F_{g}=mg$

Where m is the mass of the body and g is the gravitational acceleration. Combining these two equations we have:

$p_{h}={\frac {mg}{A}}$

We put lower index to denote the pressure is hydrostatic. Let us introduce density into the equation. Density is by definition mass per unit volume:

$\rho ={\frac {m}{V}}$

Solving for m and plugging into the previous equation we obtain:

$p_{h}={\frac {\rho \cdot Vg}{A}}$

We will assume that the volume of interest is prism or cylinder. Its height h is then perpendicular to its base, whose surface area is again A. Volume can therefore be calculated as:

$V=A\cdot h$

Combining with the previous we get:

$p_{h}={\frac {\rho \cdot A\cdot h\cdot g}{A}}$

We can see that the surface areas cancel out, thus the hydrostatic pressure truly does not depend on the shape of the contatiner, only its height.

$p_{h}=\rho \cdot h\cdot g$

Derive the formula for atmospheric pressure as a function of altitude. Assume the air is ideal gas with constant temperature.

Solution

In the previous problem we built on the fact, that the liquid is incompressible, thus its density is independent of pressure. Hydrostatic pressure was then a linear function of height. However, this does not hold in gases. Yet it is still true, that the small change in pressure is proportional to the small change in height, or:

$dp=\rho \cdot g\cdot dh$

In contrast to the previous problem, we choose such coordinate system, that height (altitude) is increasing upwards, in centrifugal direction. Gravitational acceleration is then negative, yet we will employ absolute value and put negative sign before.

$dp=-\rho \cdot |g|\cdot dh$

We will need to integrate this function to get p(h), keeping in mind that the density is dependent on the pressure. We will need to find this function ρ(p) first. State equation of ideal gas tells us that:

$pV=nRT$

where the right side of the equation is the number of moles, universal gas constant and absolute temperature respectively. With slight modification, we get:

${\frac {p}{RT}}={\frac {n}{V}}$

Dimensions are mole per unit volume, but the density is mass per unit volume. If we therefore multiply both sides by molar mass M, the dimensions would be as desired.

${\frac {pM}{RT}}={\frac {nM}{V}}=\rho$

Plugging the density into our original equation:

$dp=-{\frac {pM}{RT}}\cdot |g|\cdot dh$

Rearranging:

${\frac {dp}{p}}=-{\frac {|g|M}{RT}}\cdot dh$

Now we can integrate the equation with appropriate integral bounds. Lower bound is the reference height (let's choose 0) with pressure denoted $p_{0}$. Upper bound is the height of interest h with presure of interest p.

$\int _{p_{0}}^{p}{\frac {1}{p}}dp=\int _{0}^{h}-{\frac {|g|M}{RT}}dh$

Performing the integration and assuming the right side contains only the constants, we get:

$\ln {\frac {p}{p_{0}}}=-{\frac {|g|M}{RT}}\cdot h$

After slight rearranging, we get out final result:

$p=p_{0}e^{-{\frac {|g|M}{RT}}h}$

We can see that the pressure decreases in exponential manner with altitude.

Poznámky

$^{*}$ Korekční faktor plyne z velikosti výšky strany v rovnostranném trojúhelníku. Výška na libovolnou stranu je 2/√3 krát menší než samotná strana (Pythagorova věta), tudíž je napětí na svodu aVF (platí taky pro aVR i aVF) 2/√3 krát podhodnoceno. Tato korekce se někdy opomíjí. 2/√3 ≐ 1.155

$\gamma =\lim _{N\to \infty }\left(\sum _{k=1}^{N}{\frac {1}{k}}-\log {N}\approx 0.577\right)$

Kalkulačka srdeční osy - Python

import math

print("QRS amplitude in lead I [mV]:")
I=float(input())
print("QRS amplitude in lead aVF [mV]:")
aVF=float(input())

if I!=0:
    if I>0:
        axis=math.atan((aVF*2)/(math.sqrt(3)*I))

    if I<0 and aVF>=0:
        axis=math.atan((aVF*2)/(math.sqrt(3)*I))+math.pi

    if I<0 and aVF<0:
        axis=math.atan((aVF*2)/(math.sqrt(3)*I))-math.pi

    print("Cardiac axis is",int(180*axis/math.pi), "degrees.")

if I==0:
    if aVF>0:
        print("Cardiac axis is",90, "degrees.")
    
    if aVF<0:
        print("Cardiac axis is",-90, "degrees.")

    if aVF==0:
        print("Cardiac axis is undefined. Verify input values or electrode placement.")

Formátovanie matematiky

• Lead sentence should include article title in bold, along with alternative names also in bold.
• Use $\textstyle {\frac {1}{2}}$ to display fraction inline without compromising line width. Similarly ${\textstyle \sum _{n=1}^{\infty }1/n^{2}=\pi ^{2}/6}$ and ${\textstyle {\sqrt {x+y}}}$.
• Differential operator $d$ can be italic or upright, but consistently so.
• Both formulas and formulae are acceptable.

Viac info: https://en.wikipedia.org/wiki/Wikipedia:Manual_of_Style/Mathematics

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