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6.5-bo'lim : Ko'proq ovoz balandligi muammolari

1. Asosi $L$ uzunlikdagi teng yonli uchburchak bo‘lgan $h$ balandlikdagi piramidaning hajmini toping.

Yechim

Mayli, keling, piramidaning eskizidan boshlaylik. Ularning eskizini chizish qiyin bo'lishi mumkin, ammo eskizga ega bo'lish muammoning o'rnatilgan qismini hal qilishda katta yordam beradi.

$Three sketches of the pyramid with a typical cross-section shown in red on each and labeled $A\left( y \right)$. The pyramid has been set with its point at the origin and drawn upwards so that the base is at $y = h$, and each of the three equal sides of the base is labeled $L$. Two of the sketches show the pyramid in three dimensions from different angles and the third, at the bottom, shows it from the front as a triangle of base $L$ and height $h$ with the cross-section drawn in as a horizontal red line of length $s$ at a height $y$.$

Bu erda bizda bir nechta eskizlar bor. Har bir eskizda biz aniq tasavvurlar maydonini ko'rsatdik (qizil rangda ko'rsatilgan). Kesma $y$ o'qining istalgan nuqtasiga joylashtirilishi mumkinligi sababli, kesmaning maydoni rasmda ko'rsatilganidek, $y$ funktsiyasi bo'ladi.

Yuqori o'ngdagi eskiz biz "old" dan piramidani va yuqori chapdagi eskizni "yuqoridan" piramidani ko'ramiz. E'tibor bering, biz piramidaning nuqtasini boshiga o'rnatdik va piramidani yuqoriga tortdik. Bu muammoni sozlashni biroz osonlashtirish uchun qilingan. Shuningdek, biz piramidani shunday chizdikki, piramidaning bir tomoni $x$ o'qiga parallel bo'lsin. Bu biz pastki eskizni chizishimiz (bir soniyadan keyin erishamiz) va tasvirlar mos kelishi uchun qilingan.

Pastki eskiz - piramidaning $x$ o'qiga parallel bo'lgan tomonining eskizi. Shuningdek, unda biz ko'rsatishimiz kerak bo'lgan barcha turli miqdorlar mavjud. Bu erda vakillik kesimi eskizdagi qizil chiziq bilan ko'rsatilgan.

2-qadam

Keling, odatiy kesma qanday ko'rinishga ega bo'lgan eskizdan boshlaylik.

$A sketch of a typical cross-section, which is an equilateral triangle with each side labeled $s$. Each of the three interior angles is labeled $\frac{\pi }{3}$ and a dashed line drawn from the top vertex straight down to the base is labeled $h$. It splits the base into two pieces each of length $\frac{s}{2}$.$

Bu holda biz kesmalar teng qirrali uchburchaklar ekanligini bilamiz va shuning uchun barcha ichki burchaklar $\frac{\pi }{3}$ va barcha tomonlar bir xil uzunlikda ekanligini bilamiz, aytaylik $s$. Yuqoridagi eskizda bizda teng qirrali uchburchak borligi sababli, biz bilamizki, kesilgan chiziq (uchburchakning balandligini ifodalovchi) uchburchak asosini teng uzunlikdagi qismlarga bo'ladi, i.e. $\frac{s}{2}$. Bundan tashqari, asosiy to'g'ri burchakli uchburchak trigisidan (kesmaning har bir "yarmi" to'g'ri burchakli uchburchakmi?) biz balandlikni $s$ ko'rinishida quyidagicha yozishimiz mumkinligini ko'rishimiz mumkin,

\[\tan \left( {\frac{\pi }{3}} \right) = \frac{h}{{{}^{s}/{}_{2}}}\hspace{0.5in} \Rightarrow \hspace{0.5in} h = \frac{s}{2}\tan \left( {\frac{\pi }{3}} \right) = \frac{{\sqrt 3 }}{2}s\]

Shuning uchun, $s$ nuqtai nazaridan har bir kesmaning maydoni,

\[{\mbox{Area}} = \frac{1}{2}\left( s \right)\left( {\frac{{\sqrt 3 }}{2}s} \right) = \frac{{\sqrt 3 }}{4}{s^2}\]

Endi biz 1-bosqichdagi eskizlardan bilamizki, kesma maydoni $y$ funksiyasi bo‘lishi kerak. Shunday qilib, agar biz $s$ va $y$ o'rtasidagi munosabatni aniqlay olsak, bizga kerak bo'lgan narsaga ega bo'lar edik. Keling, 1-bosqichdagi eskizlardan birini qayta ko'rib chiqaylik.

$A sketch of the pyramid seen from the front, which is a triangle with base $L$ at the top and its point at the bottom, with the overall height labeled $h$. The cross-section is drawn in as a horizontal red line of length $s$ sitting a distance $y$ above the point. The whole triangle and the smaller triangle below the red line are similar triangles.$

Bundan bizda ikkita o'xshash uchburchak borligini ko'rishimiz mumkin. Umumiy tomoni ($L$ bazasi va balandligi $h$), shuningdek, kesma maydonini ifodalovchi qizil chiziqdan tashkil topgan “pastki” qism ($s$ bazasi va balandligi $y$).

Bu ikki uchburchak o'xshash uchburchaklar bo'lgani uchun biz quyidagi nisbatlar teng bo'lishi kerakligini bilamiz.

\[\frac{s}{y} = \frac{L}{h}\hspace{0.5in} \Rightarrow \hspace{0.5in} s = \frac{L}{h}y\]

Endi bizda $s$ va $y$ o‘rtasida bog‘liqlik bor, shuning uchun $y$ nuqtai nazaridan kesmaning maydonini olish uchun buni yuqoridan maydon formulasiga ulang.

\[A\left( y \right) = \frac{{\sqrt 3 }}{4}{\left( {\frac{L}{h}y} \right)^2} = \frac{{\sqrt 3 {L^2}}}{{4{h^2}}}{y^2}\]

3-qadam

Nihoyat, bizga hajmning o'zi kerak. Quyidagi integralni baholash orqali hajm topilishini bilamiz.

\[V = \int_{c}^{d}{{A\left( y \right)\,dy}}\]

Bizda allaqachon 2-bosqichdan $A\left( y \right)$ formulasi bor va 1-bosqichdagi eskizlardan biz “birinchi” kesma $y = 0$ da, “oxirgi” kesma esa $y = h$ da bo‘lishini ko‘rishimiz mumkin va shuning uchun bu integral uchun chegaralardir.

Keyin hajm,

\[V = \int_{0}^{h}{{\frac{{\sqrt 3 {L^2}}}{{4{h^2}}}{y^2}\,dy}} = \frac{{\sqrt 3 {L^2}}}{{4{h^2}}}\int_{0}^{h}{{{y^2}\,dy}} = \left. {\frac{{\sqrt 3 {L^2}}}{{4{h^2}}}\left( {\frac{1}{3}{y^3}} \right)} \right|_0^h = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{\sqrt 3 \,{L^2}h}}{{12}}}}\]

Integral va maydon formulasida $h$ va $L$ haqida hayajonlanmang. Bu shunchaki doimiylar. Haqiqatan ham o'zgaruvchan yagona harf - $y$. $h$ va $L$ konstantalar bo'lgani uchun biz haqiqiy raqamlarda bo'lgani kabi ularni integraldan chiqarib tashlashimiz mumkin.