Hi, there!
When I started feeling I got into an old rut at a trading company,
I thought it was high time I quit my job, not because of human
relations, but because of my strong desire to spend more time
doing a creative and giving an impact to society.
It was not until my fifth mission to be an expatriate in South Af-
rica that I submitted a resignation letter to the senior executive
director. He seemingly got shocked and asked me why. I didn’t
want to procrastinate my decision and couldn’t let this opportu-
nity slip by.
It was time for me to burn the bridges behind me.
As a businessman, I worked directly under the executive direc-
tor; he taught me a lot about business negotiations, presenta-
tions, and the dos and don’ts. I have made it this far, thanks to
his draconian mentoring.
While working with him, I had seldom been praised by him, but
on the last day I was about to leave the company, he spoke to
me with appreciation, “Without doubt, you are a great ‘Shousha-
man’ (businessman).”
Today, I’ll solve a word problem related to a bridge, recalling my
bygone turning point.
Between the two towers of a suspension bridge, each of the two
main cables has the shape of the parabola \(y = \frac{1}{5}x^{2}\) (units are
kilometers). The two towers are 2 km. apart; the vertical cables
from main cable to the horizontal roadway are closely and equal-
ly spaced. (A) Set up a definite integral which gives the length of
each main cable between the two towers. (B) What is the average
length to the nearest meter of the vertical cables? (Source: Mas-
sachusetts Institute of Technology, Final exams [Note] Slightly
changed its equation)
Here are my solutions to the problem.
(A) \(y = \frac{x^{2}}{5}, y’ = \frac{2}{5}x\)
Therefore, Arch length = \(\int_{-1}^{1}\sqrt{1 + \frac{4}{25}x^{2}}\)
(B) Average = \(\int_{0}^{1}\frac{x^{2}}{5}dx = [\frac{x^{3}}{15}]_{0}^{1} = \frac{1}{15}\)km \(≓ 67\) meters, or
\(\frac{1}{2}\int_{-1}^{1}\frac{x^{2}}{5}dx\)
\(= \frac{1}{2}[\frac{x^{3}}{15}]_{-1}^{1}\)
\(= \frac{1}{2}\displaystyle\left\{\frac{1}{15} – (-\frac{1}{15})\right\}\)
\(= \frac{1}{2}(\frac{1}{15} + \frac{1}{15})\)
\(= \frac{1}{2}(\frac{2}{15})\)
\(= \frac{1}{15}\)
\(≓ 67\) meters
Are you the person who crosses that bridge when you come to it?
Stay tuned, and expect to see my next post.
Keep well.
Frank Yoshida
￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣
【グローバルビジネスで役立つ数学】でもっと学習する b＾＾)
【参考図書】『もう一度高校数学』（著者：高橋一雄氏）株式会社日本実業出版社
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