Anecdotes please - quadratic equations in real life

[EdS2]

Nearby:

(05-05-2024 12:48 PM)Maximilian Hohmann Wrote:  ... I keep wondering why the quadratic equation and it's memorized solution is attributed such a high significance by schools. I have never since finishing school come across a quadratic equation in real life.

Can anyone share any experiences when they came across a quadratic equation in real life? (Let's not count the very simple case where only a square root is needed.)

Edit: Just to note - this enquiry is for interest only, not intended to be controversial or confrontational.

[Dave Britten]

Apart from occasionally dealing with acceleration, I can't think of a whole lot of/any situations where I run across them. And even when doing something involving acceleration, I don't think I've ever needed to employ the quadratic formula to find the zeroes.

[dm319]

Not for me. I've spent my career in medical sciences, where statistics, probability, survival statistics, high dimensional data has been of more use.

[carey]

Here are a few examples from my experience.

- projectile motion problems. Parabolic trajectories result from quadratic kinematic equations so solving for the zeros is needed to solve for time. (However, the reason it often appears that solving for roots of a quadratic equation isn't needed in projectile motion problems is that if one of the two roots is negative (negative time) and therefore can be discarded on physical grounds, the quadratic equation is then able to be reduced to a linear equation with one root.)

- finding current in even a simple series circuit consisting of a power supply, lamp and resistor (given the voltage, power, and resistance respectively), leads, after applying Kirchhoff's loop rule, to a quadratic equation (not just taking a square root) because of quadratic dependence of power on current.

- analyzing biological processes (most of which are nonlinear) often involves Taylor expansions, but if only linear terms are retained, nonlinear properties are lost, so, at a minimum, 2nd order terms (quadratics) need to be kept.

[Johnh]

In structural engineering design, there is a daily need to calculate the area of steel reinforcement required to achieve a given bending strength in a reinforced concrete beam of known size. This involves the area and yield strength of the steel, multiplied by a lever arm from the centroid of the steel to that of the compression block in the concrete. The more steel required the deeper is the compression block required to balance it and so the shorter is the lever arm. Hence there are depth-squared terms in the relationship. To solve it so that steel area is the solution, it becomes a quadratic.

TBH, a design engineer will not be punching numbers into the quadratic solution formula every day, but such a solution will be implied within whatever software is used.It will also be required when coding such software or setting up spreadsheets (or calc programs for the oldies). And it's also a necessary insight when learning the principles of concrete design in engineering education.

[Steve Simpkin]

[JoJo1973]

https://www.smbc-comics.com/comic/learn

[Thomas Klemm]

In summer I often go swimming in the river. First \(s_1 = 300\text{m}\) upstream along the river bank and then back \(s_2 = 450\text{m}\) in the middle of the river.
I don't have a watch with me, but there is a big clock by the changing rooms where I start.
This shows me that I need \(t = 25\text{min}\) in total.

What is my own speed \(v\) if the flow speed on the river bank is \(u_1 = 35\text{m}/\text{min}\) and in the middle \(u_2 = 40\text{m}/\text{min}\)?

\(
\begin{align}
t &= \frac{s_1}{v - u_1} + \frac{s_2}{v + u_2} \\
25 &= \frac{300}{v - 35} + \frac{450}{v + 40} \\
25 &= \frac{750(v-5)}{(v - 35)(v + 40)} \\
\end{align}
\)

\(
\begin{align}
v^2 + 5v - 1400 &= 30v - 150 \\
v^2 - 25v - 1250 &= 0 \\
\end{align}
\)

\(
\begin{align}
v
&= 12.5 + \sqrt{12.5^2 + 1250} \\
&= 12.5 + \sqrt{1406.25} \\
&= 12.5 + 37.5 \\
&= 50 \\
\end{align}
\)

Therefore my speed is \(v = 50\text{m}/\text{min}\).
The other solution is \(-25\text{m}/\text{min}\) and doesn't make sense.

I can measure the distances \(s_1\) and \(s_2\) pretty accurately using Google Maps.
However, it is not easy to estimate the flow speed of the river.
The power plant located a little further down makes the measurements available online.
But they make no distinction between the river bank and the middle of the river.
You can still follow a leaf along the river and measure how far it travels in a minute or 10 seconds.

Thus I made up some numbers for \(u_1\) and \(u_2\) to get an integer result for \(v\).
However, the times \(t_1 = 20\text{min}\) upstream and \(t_2 = 5\text{min}\) downstream feel reasonable to me.

[Gjermund Skailand]

Perhaps only to teach that there is more to life than average and linear equations?
- Gjermund

[EdS2]

Thanks everyone for your replies so far.

Just to note - and I'll add this to the head post - my enquiry is for interest only, not intended to be controversial or confrontational.

[floppy]

(05-05-2024 01:24 PM)EdS2 Wrote:  Can anyone share any experiences when they came across a quadratic equation in real life? (Let's not count the very simple case where only a square root is needed.)

2 times:
1. creating a net for simulating the head of a forming rotating press into a steel mass, for producing pipes for use in atomic plant (the head was a circle with tangent lines)
2. calculating the interference point between a ray tracing and a glas ocular

[Allen]

useful for factoring certain special form numbers.

guess near 854 (any number from 854-859 will factor n using the formula below),
by converting n to base 854 then solve quadratic.

\(
\begin{align}
n= 859*997 \\

guess = 854 \\
n = (1,148,715) _{854} \\
(a,b,c) = (1,148,715) \\

g - \frac{-b + \sqrt{b^2 -4ac}} {2} = 859\\
g - \frac{-b - \sqrt{b^2 -4ac}} {2} = 997\\

\end{align}
\)

Even if your guess is elsewhere, you can make fairly large steps (knowing the number has no factor in that range) by solving the quadratic equation there on the number in base g.

For example, guessing 883, the quadratic eq returns 894, then guessing 894 it returns 914... it's not practical for many circumstances, but this method works well when \( n \) has a factor that's around \( \sqrt n - a \sqrt {\sqrt{n}} \) for some "small" a.

[Albert Chan]

(05-05-2024 11:42 PM)Thomas Klemm Wrote:  \(
\begin{align}
t &= \frac{s_1}{v - u_1} + \frac{s_2}{v + u_2} \\
25 &= \frac{300}{v - 35} + \frac{450}{v + 40} \\
\end{align}
\)
...

Therefore my speed is \(v = 50\text{m}/\text{min}\).
The other solution is \(-25\text{m}/\text{min}\) and doesn't make sense.

It may not be obvious why negative velocity is wrong.
If we play the film backward, even time can be negative.

However, time cannot be positive and negative at the same time.
RHS time terms must have the same sign: (v > 35) or (v < -40)
This is why v = -25 is wrong.

Another way, |x| < 0.5, to ensure both time terms have same sign.

1 = 12/(v-35) + 18/(v+40) = (.5+x) + (.5-x)
v = 12/(.5+x)+35 = 18/(.5-x)-40

18/(.5-x) - 12/(.5+x) = 75
.24/(.5-x) - .16/(.5+x) = 1
(.24-.16)*.5 + (.24+.16)*x = .5^2 - x^2

x^2 + .4*x - .21 = 0
x = -.2 ± √(.2^2 + .21) = -.2 ± .5 = .3, -.7
v = 12/(.5+x)+35 = 50

Or, we can just iterate for solution, without quadratic formula
This may be faster than getting quadratic coefficients.

v = 35 + 12 / (1 - 18/(v+40))
v: 35+12=47 → 50.130 → 49.994 → 50.000

[Albert Chan]

(12-30-2022 11:27 PM)Albert Chan Wrote:  e^W(a) = y = (y+a) / (ln(y)+1)

Let r = 1/e
If a = -r + h (tiny h), we have y = r + z (tiny z)

e^W(a) = (r+z) = (h+z) / ln(1+e*z)

Let H = e*h (known), Z = e*z (unknown), we have:

(1+Z) = (H+Z) / ln(1+Z)

H = (1+Z) * ln(1+Z) - Z = Z^2/2 - Z^3/6 + Z^4/12 - Z^5/20 + ...

2H ≈ Z^2 / (1+Z/3)
Z^2 - 2*(H/3)*Z - 2H ≈ 0
Z ≈ H/3 ± √((H/3)^2 + 2H) ≈ H/3 ± √(2H)

Divide both side by e, we have: y = r + z ≈ r + h/3 ± √(2*r*h)

e^W₀(z ≈ -1/e)   ≈   1/e + (z+1/e)/3 + sqrt ((2/e)*(z+1/e))

http://www.finetune.co.jp/~lyuka/technot...w-42s.html